Sol Cap 06 - Edicion 8

COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 1.
\
Joint FBDs:
Joint B:
FAB 800 lb FBC
=
=
15
8
17
so
FAB = 1500 lb T W
FBC = 1700 lb C W
Joint C:
FAC Cx 1700 lb
=
=
8
15
17
FAC = 800 lb T W
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 2.
Joint FBDs:
Joint B:
ΣFx = 0:
1
4
FAB − FBC = 0
5
2
ΣFy = 0:
1
3
FAB + FBC − 4.2 kN = 0
5
2
7
FBC = 4.2 kN
5
so
Joint C:
FAB =
ΣFx = 0:
12 2
kN
5
FBC = 3.00 kN C !
FAB = 3.39 kN C !
4
12
(3.00 kN) −
FAC = 0
5
13
FAC =
13
kN
5
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
FAC = 2.60 kN T !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 3.
Joint FBDs:
Joint B:
FAB FBC 450 lb
=
=
12
13
5
FAB = 1080 lb T W
so
FBC = 1170 lb C W
Joint C:
ΣFx = 0:
3
12
FAC − (1170 lb ) = 0
5
13
FAC = 1800 lb C W
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 4.
Joint FBDs:
Joint D:
FCD FAD 500 lb
=
=
8.4 11.6
8
FAD = 725 lb T W
FCD = 525 lb C W
Joint C:
ΣFx = 0:
FBC − 525 lb = 0
FBC = 525 lb C W
This is apparent by inspection, as is FAC = C y
ΣFx = 0:
8.4
3
(725 lb) − FAB − 375 lb = 0
11.6
5
Joint A:
FAB = 250lb T W
ΣFy = 0:
FAC −
4
8
(250 lb) −
(725 lb) = 0
5
11.6
FAC = 700 lb C W
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 5.
FBD Truss:
ΣFx = 0 :
Cx = 0
By symmetry: C y = D y = 6 kN
Joint FBDs:
Joint B:
ΣFy = 0:
− 3 kN +
1
FAB = 0
5
FAB = 3 5 = 6.71 kN T W
Joint C:
ΣFx = 0:
ΣFy = 0:
Joint A:
ΣFx = 0:
ΣFy = 0:
2
FAB − FBC = 0
5
FBC = 6.00 kN C W
3
FAC = 0
5
FAC = 10.00 kN C W
6 kN −
6 kN −
4
FAC + FCD = 0
5
FCD = 2.00 kN T W
 1

3

− 2
3 5 kN  + 2  10 kN  − 6 kN = 0 check
5

 5

By symmetry:
FAE = FAB = 6.71 kN T W
FAD = FAC = 10.00 kN C W
FDE = FBC = 6.00 kN C W
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 6.
FBD Truss:
ΣM A = 0:
(10.2 m ) C y + ( 2.4 m )(15 kN ) − ( 3.2 m )( 49.5 kN ) = 0
C y = 12.0 kN
Joint FBDs:
Joint FBDs:
Joint C:
FBC
F
12 kN
= CD =
7.4
7.4
8
FBC = 18.50 kN C W
FCD = 18.50 kN T W
Joint B:
ΣFX = 0:
4
7
(18.5 kN) = 0
FAB −
5
7.4
FAB = 21.875 kN;
ΣFy = 0:
FAB = 21.9 kN C W
3
2.4
(21.875 kN) − 49.5 kN +
(18.5 kN) + FBD = 0
5
7.4
FBD = 30.375 kN;
FBD = 30.4 kN C W
Joint D:
ΣFx = 0: −
4
7
FAD +
(18.5 kN ) + 15 kN = 0
5
7.4
FAD = 40.625 kN;
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
FAD = 40.6 kN T W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 7.
Joint FBDs:
Joint E:
FBE FDE 3 kN
=
=
5
4
3
FBE = 5.00 kN T W
FDE = 4.00 kN C W
Joint B:
ΣFx = 0:
− FAB +
4
(5 kN) = 0
5
FAB = 4.00 kN T W
ΣFy = 0:
FBD − 6 kN −
3
(5 kN) = 0
5
FBD = 9.00 kN C W
Joint D:
ΣFy = 0:
3
FAD − 9 kN = 0
5
FAD = 15.00 kN T W
ΣFx = 0: FCD −
4
(15 kN) − 4 kN = 0
5
FCD = 16.00 kN C W
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 8.
Joint FBDs:
Joint B:
FAB = 12.00 kips C W
By inspection:
FBD = 0 W
Joint A:
FAC FAD 12 kips
=
=
5
13
12
FAC = 5.00 kips C W
FAD = 13.00 kips T W
Joint D:
ΣFx = 0: FCD −
12
(13 kips) − 18 kips = 0
13
FCD = 30.0 kips C W
ΣFy = 0:
5
(13 kips) − FDF = 0
13
FDF = 5.00 kips T W
Joint C:
ΣFx = 0: 30 kips −
12
FCF = 0
13
FCF = 32.5 kips T W
ΣFy = 0: FCE − 5 kips −
5
(32.5 kips)
13
FCE = 17.50 kips C W
Joint E:
by inspection:
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
FCF = 0 W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 9.
FCH = 0 !
First note that, by inspection of joint H:
and
FCG = 0 !
then, by inspection of joint C:
and
Joint D:
FBC = FCD
FBG = 0 !
then, by inspection of joint G:
and
Joint FBDs:
FDH = FGH
FFG = FGH
FBF = 0 !
then, by inspection of joint B:
and
FAB = FBC
FCD FDH 10 kips
=
=
12
13
5
Joint A:
FCD = 24.0 kips T !
so
FDH = 26.0 kips C !
FAB = FBC = 24.0 kips T !
and, from above:
FGH = FFG = 26.0 kips C !
FAF
F
24 kips
= AE =
5
4
41
FAF = 30.0 kips C !
Joint F:
FAE = 25.6 kips T !
ΣFx = 0: FEF −
12
(26 kips) = 0
13
FEF = 24.0 kips C !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 10.
FBD Truss:
ΣFx = 0: H x = 0
By symmetry: A y = H y = 4 kips
FAC = FCE and FBC = 0 W
by inspection of joints C and G :
FEG = FGH and FFG = 0 W
also, by symmetry FAB = FFH , FBD = FDF , FCE = FEG and FBE = FEF
Joint FBDs:
Joint A:
FAB FAC 3 kips
=
=
5
4
3
FAB = 5.00 kips C W
so
FAC = 4.00 kips T W
FFH = 5.00 kips C W
and, from above,
and
Joint B:
FCE = FEG = FGH = 4.00 kips T W
4
4
10
(5 kips) − FBE −
FBD = 0
5
5
109
ΣFx = 0:
ΣFy = 0:
3
3
3
FBD + FBE = 0
( 5 kips ) − 2 −
5
5
109
so
FBD = 3.9772 kips, FBE = 0.23810 kips
FBD = 3.98 kips C W
or
Joint E:
FBE = 0.238 kips C W
FDF = 3.98 kips C W
and, from above,
FEF = 0.238 kips C W
ΣFy = 0 :
FDE − 2
3
(0.23810 kips) = 0
5
FDE = 0.286 kips T W
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 11.
FBD Truss:
ΣFx = 0:
Ax = 0
ΣM G = 0:
3a Ay − 2a (3 kN) − a (6 kN) = 0
A y = 4 kN
by inspection of joint C,
FAC = FCE and FBC = 0 W
by inspection of joint D,
FBD = FDF and FDE = 6.00 kN C W
Joint FBDs:
Joint A:
FAC FAB 4 kN
=
=
21
29
20
FAB = 5.80 kN C W
FAC = 4.20 kN C W
from above,
Joint B:
ΣFy = 0:
20
20
( 5.80 kN ) − 3 kN − FBE = 0
29
29
FBE =
ΣFx = 0:
FCE = 4.20 kN C W
29
20
FBE = 1.450 kN T W
21 
29

kN  − FBD = 0
 5.80 kN +
29 
20

FBD = 5.25 kN C W
FDF = 5.25 kN C W
from above,
Joint F:
ΣFx = 0:
5.25 kN −
21
FEF = 0
29
FEF = 7.25 kN T W
ΣFy = 0:
FFG −
20
(7.25 kN) − 1 kN = 0
29
FFG = 6.00 kN C W
by inspection of joint G,
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
FEG = 0 W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 12.
FBD Truss:
ΣFx = 0:
Ax = 0
By symmetry: A y = B y = 4.90 kN
FAB = FEG , FAC = FFG , FBC = FEF
and
FBD = FDE , FCD = FDF
5
4
FAC − FAB = 0
5
29
2
3
FAC − FAB + 4.9 kN = 0
5
29
ΣFx = 0:
Joint FBDs:
Joint A:
ΣFy = 0:
FAC = 2.8 29 kN
FAC = 15.08 kN T FAD = 17.50 kN C Joint B:
ΣFx = 0:
ΣFy = 0:
4
1
FBC = 0
(17.5 kN − FBD ) −
5
2
3
1
FBC − 2.8 kN = 0
(17.5 kN − FBD ) +
5
2
FBD = 15.50 kN C FBC = 1.6 2 kN;
Joint C:
ΣFy = 0:
FBC = 2.26 kN C 4
1
1.6 2 kN −
FCD −
5
2
ΣFx = 0: FCF
2
(2.8 29 kN) = 0
29
FCD = 9.00 kN T 1
3
5
1.6 2 kN + (9 kN) −
(2.8 29 kN) = 0
+
5
2
29
FCF = 7.00 kN T (
(
)
)
from symmetry,
FEG = 17.50 kN C FFG = 15.08 kN T FEF = 2.26 kN C FDE = 15.50 kN C FDF = 9.00 kN T Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 13.
FBD Truss:
ΣFx = 0: A x = 0
ΣM A = 0: (8 m) Gy − (4 m)(4.2 kN) − (2m)(2.8 kN) = 0
G y = 2.80 kN
ΣFy = 0:
Ay − 2.8 kN − 4.2 kN + 2.8 kN = 0
A y = 4.2 kN
Joint FBDs:
5
4
FAC − FAB = 0
5
29
2
3
FAC − FAB + 4.2 kN = 0
5
29
ΣFx = 0:
Joint A:
ΣFy = 0:
FAB = 15.00 kN C !
FAC = 12.92 kN T !
FAC = 2.4 29
Joint B:
ΣFx = 0:
ΣFy = 0:
4
1
FBC = 0
(15.00 kN − FBD ) −
5
2
3
1
FBC − 2.8 kN = 0
(15.00 kN − FBD ) +
5
2
FBD = 13.00 kN C !
FBC = 1.6 2 kN,
Joint C:
ΣFy = 0:
(
4
FCD −
5
FBC = 2.26 kN C !
)
2
1
2.4 29 kN −
(1.6 2 kN) = 0
29
2
FCD = 8.00 kN T !
ΣFx = 0:
FCF +
3
(8.00 kN ) −
5
+
5
(2.4 29 kN)
29
1
(1.6 2 kN) = 0
2
FCF = 5.60 kN T !
continued
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
By inspection of joint E,
FDE = FEG
and
FEF = 0 !
Joint F:
ΣFy = 0:
ΣFx = 0:
4
FDF −
5
2
FFG = 0
29
3
5
− 5.6 kN − FDF +
FFG = 0
5
29
FDF = 4.00 kN T !
FFG = 1.6 29 kN
Joint G:
ΣFx = 0:
4
FEG −
5
FFG = 8.62 kN T !
5
(1.6 29 kN) = 0
29
FEG = 10.00 kN C !
from above (joint E)
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
FDE = 10.00 kN C !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 14.
FBD Truss:
ΣFx = 0: A x = 0
ΣM A = 0:
4a H y − 3a (1.5 kN) − 2a (2 kN)
− a (2 kN) = 0
ΣFy = 0:
Ay − 1 kN − 2 kN − 2 kN − 1.5 kN − 1 kN
+ 3.625 kN = 0
Joint FBDs:
Joint A:
A y = 3.875 kN
FAB FAC
2.625 kN
=
=
1
29
26
FAB = 15.4823 kN,
FAB = 15.48 kN C !
FAC = 14.6597 kN,
FAC = 14.66 kN T !
By inspection of joint C:
Joint B:
H y = 3.625 kN
ΣFy = 0:
FCE = FAC = 14.66 kN T,
2
(15.4823 kN − FBD ) − 2 kN = 0
29
FBD = 10.0971 kN,
ΣFx = 0:
FBC = 0 !
FBD = 10.10 kN C !
5
(15.4823 kN − 10.0971 kN ) − FBE = 0
29
FBE = 5.0000 kN,
FBE = 5.00 kN C !
continued
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Joint D:
FDF = 10.0971 kN,
By symmetry:
ΣFy = 0:
FDF = 10.10 kN C !
 2

10.0971 kN  − 2 kN = 0
− FDE + 2 
 29

FDE = 5.50 kN T !
FGH FFH
2.625 kN
=
=
1
26
29
Joint H:
By inspection of joint G:
ΣFx = 0 :
Joint F:
FFH = 14.1361 kN
FFH = 14.14 kN C !
FGH = 13.3849 kN
FGH = 13.38 kN T !
FEG = FGH = 13.38 kN T
FEF +
and
FFG = 0 !
2
(10.0971 kN − 14.1361 kN ) = 0
29
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
FEF = 3.75 kN C !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 15.
FBD Truss:
ΣFx = 0: A x = 0
ΣM A = 0:
8a ( J y − 1 kN) − 7a (1 kN) − 6a (2.8 kN)
− 4a (4.5 kN) − 2a (4 kN) − a(1 kN) = 0
J y = 6.7 kN
ΣFy = 0:
− 1 kN − 1 kN + 6.7 kN = 0
Joint FBDs:
Joint A:
Ay − 1 kN − 1 kN − 1.4 kN − 4.5 kN − 2.8 kN
ΣFy = 0:
5 kN −
A y = 6.0 kN
3
5 13
FAB = 0, FAB =
kN
3
13
FAB = 6.01 kN C !
ΣFx = 0:
Joint B:
ΣFx = 0:
FAC −
2
13
 5 13

kN  = 0,

 3

FAC = 3.33 kN T !

2  5 13
kN − FBC − FBD  = 0,

13  3

5 13
kN
3

3  5 13
kN + FBC − FBD  − 1 kN = 0,

13  3

4 13
FBD − FBC =
kN
3
3
13 kN
FBD =
FBD = 5.41 kN C !
2
1
13 kN
FBC =
FBC = 0.601 kN C !
6
FBC + FBD =
ΣFy = 0:
continued
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
FEF = 0 !
By inspection of joint F:
Joint E:
FDE = FEG
By symmetry:
ΣFy = 0:
2
1
FDE − 4.5 kN = 0,
17
FDE =
9
17 kN
4
FDE = 9.28 kN C !
ΣFx = 0:
2 3
4 9


13 kN  −
17 kN  = 0


13  2
17  4


FDF +
Joint D:
FDF = 6.00 kN T !
ΣFy = 0:
3 3
1 9


13 kN  − 1.4 kN − FCD −
17 kN  = 0


13  2
17  4


FCD = 0.850 kN T !
Joint C:
ΣFy = 0:
0.850 kN −
FCG =
ΣFx = 0:
FCI −
3
13
 13
 3
kN  − FCG = 0

 5
 6

1.75
kN
3
FCG = 0.583 kN C !
 10
4  1.75
2  13

kN  −
kN  −
kN = 0


5 3
13  6

 3
FCI = 3.47 kN !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 16.
\
FBD Truss:
ΣFx = 0:
ΣM A = 0:
Ax = 0
8a( J y − 1 kN) − 7a(1 kN) − 6a(2.8 kN)
− 4a(4.5 kN) − 2a(4 kN) − a(1 kN) = 0
J y = 6.7 kN
ΣFy = 0:
Ay − 1 kN − 1 kN − 1.4 kN − 4.5 kN − 2.8 kN
−1 kN − 1 kN + 6.7 kN = 0
Joint FBDs:
Joint J:
ΣFy = 0:
(6.7 − 1) kN −
3
FHJ = 0,
13
A = 6.0 kN
FHJ = 1.9 13 kN
FHJ = 6.85 kN C !
ΣFx = 0:
Joint H:
ΣFx = 0:
ΣFy = 0:
2
(1.9 13 kN) − FIJ = 0,
13
2
( FGH + FHI − 1.9 13 kN) = 0
13
3
( FHI − FGH + 1.9 13 kN) − 1 kN = 0
13
26
13 kN,
FGH =
FGH = 6.25 kN C !
15
FHI =
Joint I:
ΣFx = 0:
3.80 kN −
FGI −
13
kN,
6
FHI = 0.601 kN C !

2  13
kN  − FCI = 0

13  6

FCI =
ΣFy = 0:
FIJ = 3.80 kN T !
10.4
kN,
3

3  13
kN  = 0,


13  6

FCI = 3.47 kN T !
FGI = 0.500 kN T !
continued
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
FEF = 0 !
By inspection of joint F,
By symmetry FDE = FEG
Joint E:
ΣFy = 0:
 1

FEG  − 4.5 kN = 0,
2
17


FEG =
9
17 kN
4
FEG = 9.28 kN C !
Joint G:
ΣFy = 0:
3  26
3
1 9
 1

13 kN  − kN − FCG −
17 kN 


5
13  15
17  4
 2

− 2.8 kN = 0
FCG = −
ΣFx = 0:
1.75
kN
3
FCG = 0.583 kN C !
4 9
4  1.75 

17 kN  − FFG −  −


5 3 
17  4

−
2  26

13 kN  = 0

13  15

FFG = 6.00 kN T !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 17.
FBD Truss:
By load symmetry,
ΣFx = 0:
θ = tan −1
Joint FBDs:
A y = H y = 1600 lb
Ax = 0
6.72 ft
= 16.2602°
23.04 ft
ΣFy′ = 0:
(1600 lb − 400 lb) cosθ − FAC sin θ = 0
FAC = 4114.3 lb
Joint A:
ΣFx = 0:
FAC cos 2θ − FAB cosθ = 0
FAB = 3613.5 lb
ΣFx′ = 0:
Joint B:
Joint C:
FBC = 0.768 kips C !
FCD sin 2θ − (768 lb) cosθ = 0
FCD = 1371.4 lb
ΣFx′′ = 0:
FBD = 3.84 kips C !
FBC − (800 lb) cosθ = 0
FBC = 768.00 lb
ΣFy′′ = 0:
FAB = 3.61 kips C !
3613.5 lb − FBD + (800 lb)sin θ = 0
FBD = 3837.5 lb
ΣFy′ = 0:
FAC = 4.11 kips T !
FCD = 1.371 kips T !
FCE + (1371.4 lb) cos2θ − (768 lb)sin θ − 4114.3 lb = 0
FCE = 2742.9 lb
FCE = 2.74 kips T !
continued
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Joint E:
ΣFy = 0:
(2742.9 lb)sin 2θ − FDE cosθ = 0
FDE = 1536.01 lb
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
FDE = 1.536 kips C !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 18.
FBD Truss:
A y = H y = 1600 lb
By load symmetry,
ΣFx = 0:
Ax = 0
θ = tan −1
6.72 ft
= 16.2602°
23.04 ft
Joint FBDs:
Joint H:
ΣFy = 0:
1600 lb − 400 lb − FFH sin θ = 0
FFH = 4285.7 lb,
( 4285.7 lb ) cosθ
ΣFx = 0:
Joint F:
− FGH = 0
FGH = 4114.3 lb
ΣFy′ = 0:
FGH = 4.11 kips T
FFG − ( 800 lb ) cosθ = 0
FFG = 768.0 lb
ΣFx′ = 0:
FFH = 4.29 kips C
FFG = 0.768 kips C
FDF + ( 800 lb ) sin θ − 4285.7 lb = 0
FDF = 4061.7 lb
FDF = 4.06 kips C
Joint G:
ΣFy = 0:
FDG sin 2θ − (768 lb) cosθ = 0
FDG = 1371.4 lb
ΣFx = 0:
FDG = 1.371 kips T
4114.3 lb − (1371.4 lb ) cosθ − FEG − (768 lb) sin θ = 0
FEG = 2742.9 lb
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
FEG = 2.74 kips T
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 19.
FBD Truss:
ΣFx = 0:
ΣM L = 0:
Ax = 0
(1.5 m ) (6.6 kN) + (3.0 m)(2.2 kN)
+ (5.5 m)(6 kN) − (11 m) Ay = 0
A y = 4.5 kN
Joint FBDs:
Joint A:
ΣFy = 0:
4.5 kN − 6 kN − 2.2 kN − 6.6 kN + L = 0
L = 10.3 kN
4.5 kN FAC
F
=
= AB ,
1
2
5
FAC = 9.00 kN T !
FAB = 4.5 5,
Joint C:
FAB = 10.06 kN C !
9 kN FBC
F
=
= CE ,
16
5
281
FBC =
45
kN
16
FBC = 2.81 kN C !
FCE =
Joint B:
ΣFx = 0:
ΣFy = 0:
Solving:
9
281,
16
FCE = 9.43 kN T !
2 
16
FBD + (4.5 5) kN  +
FBE = 0


5
265
1 
FBD + (4.5 5) kN  −

5
3
45
kN = 0
FBE +
16
265
72
5 kN,
11
45
265 kN,
FBE =
176
FBD = −
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
FBD = 14.64 kN C !
FBE = 4.16 kN T !
COSMOS: Complete Online Solutions Manual Organization System
Joint E:
ΣFx = 0:
16 
 9
 FEG −  16
281 

FEG =
ΣFy = 0:

16  45


281  kN  −
265  kN = 0

265  176



9
281 kN,
11
FEG = 13.72 kN T !
5 9
9

281 kN −
281 kN  +

16
281  11

3  45

265 kN 

265  176

+ FDE = 0,
FDE = −
Joint D:
ΣFx = 0:
ΣFy = 0:
Solving:
45
kN,
22
FDE = 2.05 kN C !
2 
72
10

FDG = 0
5 kN  +
 FDF +
11
5
101

1 
72
1
45

FDG +
5 kN  −
kN = 0
 FDF +
11
22
5
101

7.5
101 kN,
22
FDG = 3.43 kN T !
FDF = − 8.25 5 kN,
FDF = 18.45 kN C !
FDG =
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 20.
FBD Truss:
ΣFx = 0:
ΣM L = 0:
Ax = 0
(1.5 m ) (6.6 kN) + (3.0 m)(2.2 kN) + (5.5 m)(6 kN)
− (11 m) Ay = 0
A y = 4.5 kN
Joint FBDs:
ΣFy = 0:
4.5 kN − 6 kN − 2.2 kN − 6.6 kN + L = 0
Joint L:
L = 10.3 kN
10.3 kN FKL FJL
=
=
,
1
2
5
FKL = 20.6 kN T !
FJL = 10.3 5 kN,
FJL = 23.0 kN C !
20.6 kN FJK
F
=
= IK ,
16
5
281
Joint K:
FJK =
51.5
kN
8
FJK = 6.44 kN C !
FIK =
Joint J:
ΣFx = 0:
ΣFy = 0:
Solving:
−
10.3
281 kN,
8
(
FIK = 21.6 kN T!
)
2
16
FHJ + 10.3 5 kN −
FIJ = 0
5
265
(
)
1
FHJ + 10.3 5 kN −
5
FHJ = −
FIJ = −
3
51.5
kN = 0
FIJ − 6.6 kN +
8
265
112
5 kN,
11
1.3
265 kN,
88
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
FHJ = 22.8 kN C !
FIJ = 0.240 kN C !
COSMOS: Complete Online Solutions Manual Organization System
Joint I:
16 
10.3
16  1.3


281 kN  −
265 kN  = 0
 − FGI +

8
88
281 
265 


ΣFx = 0:
FGI =
ΣFy = 0:
ΣFx = 0:
ΣFy = 0:
−
10.4
kN,
88
1 
112
1
10.4

FGH − 2.2 kN −
5 kN  −
kN = 0
 FFH +
11
88
5
101

FFH = − 8.25 5 kN,
FGH = −
34
101 kN,
88
FFH = 18.45 kN C !
FGH = 3.88 kN C !
FDF = 8.25 5 kN
By symmetry:
ΣFy = 0:
FHI = 0.1182 kN T !
2 
112
10

FGH = 0
5 kN  −
 FFH +
11
5
101

Solving:
Joint F:
FGI = 21.3 kN T !
5  112
10.3

281 kN −
281 kN  + FHI

8
281  88

3  1.3

265 kN  = 0
−

265  88

FHI =
Joint H:
112
281 kN,
88
(
)
2
8.25 5 kN − 6 kN − FFG = 0
5
FFG = 10.50 kN T !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 21.
Joint FBDs:
FDF
F
500 lb
= EF =
916 916
480
Joint F:
FDF = FEF = 954.17 lb
FDF = 954 lb T !
or
FEF = 954 lb C !
Joint D:
FBD FDE
954.17 lb
=
=
884 240
916
FBD = 920.84 lb
FDE = 250.00 lb
FBD = 921 lb T !
or
FDE = 250 lb C !
By symmetry of joint A vs. joint F
FAB = 954 lb T !
FAC = 954 lb C !
Joint B:
ΣFx = 0:
920.84 lb −
884
4
( 954.17 lb ) + FBE = 0
916
5
FBE = 0 !
ΣFy = 0:
FBC −
240
( 954.17 lb ) = 0
916
FBC = 250.00 lb
or
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
FBC = 250 lb C !
COSMOS: Complete Online Solutions Manual Organization System
Joint C:
ΣFx = 0:
884
( 954.17 lb ) − FCE = 0
916
FCE = 920.84 lb
ΣFy = 0:
FCH −
240
(954.17 lb) − 250.00 lb = 0
916
FCH = 500.00 lb
ΣFx = 0:
FCE = 921 lb C !
or
920.84 lb −
or
FCH = 500 lb C !
884
4
( 954.17 lb ) − FHE = 0
916
5
FHE = 0 !
Joint E:
ΣFy = 0:
FEJ −
240
(954.17 lb) − 250.00 lb = 0
916
FEJ = 500.00 lb
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
or
FEJ = 500 lb C !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 22.
Joint FBDs:
FJL
F
500 lb
= KL =
1212 1212
480
Joint L:
FJL = FKL = 1262.50 lb
FJL = 1263 lb T !
FKL = 1263 lb C !
Comparing joint G to joint L:
FGH = 1263 lb T !
FGI = 1263 lb C !
Joint J:
ΣFx = 0:
1188
(1262.50 lb ) − FHJ = 0
1212
FHJ = 1237.50 lb,
ΣFy = 0:
FJK −
240
(1262.50 lb ) − 500 lb = 0
1212
FJK = 750.00 lb,
Joint H:
ΣFx = 0:
FHJ = 1238 lb T !
1237.50 lb −
FJK = 750 lb C !
1188
4
(1262.50 lb ) + FHK = 0
1212
5
FHK = 0 !
ΣFy = 0:
FHI −
240
3
(1262.50 lb) − 500 lb − (0) = 0
1212
5
FHI = 750 lb C !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Joint I:
ΣFx = 0:
1188
(1262.50 lb ) − FIK = 0
1212
FIK = 1237.50 lb,
ΣFy = 0:
FIN −
240
(1262.50 lb ) − 750.00 lb = 0
1212
FIN = 1000.00 lb,
Joint K:
ΣFx = 0:
FIK = 1238 lb C !
1237.50 lb −
FIN = 1000 lb C !
1188
4
(1262.50 lb ) − FKN = 0
1212
5
FKN = 0 !
ΣFy = 0:
FKO −
240
(1262.50 lb) − 750.00 lb = 0
1212
FKO = 1000 lb C !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 23.
FBD Truss:
ΣFx = 0:
ΣM A = 0:
Ax = 0
(12 m ) (M y − 1 kN)
− (2.4 m + 4.8 m + 6 m + 8.4 m + 10.8 m) (1.5 kN) = 0
M y = 5.05 kN
ΣFy = 0:
Ay − 2(1 kN) − 5(1.5 kN) + M y = 0
A y = 4.45 kN
Joint FBDs:
ΣFy = 0:
4.45 kN − 1 kN −
Joint A:
ΣFx = 0:
ΣFx = 0:
Joint C:
ΣFy = 0:
ΣFy = 0:
Joint B:
FAC −
5
FAB = 0,
13
12
(8.97 kN ) = 0,
13
FAC = 8.28 kN T !
12
FCE − 8.28 kN = 0,
13
FCE = 8.97 kN T !
5
(8.97 kN) − FBC = 0,
13
FBC = 3.45 kN C !
5
5
(8.97 kN ) − 1.5 kN + 3.45 kN − FBD = 0
13
13
FBD = 14.04 kN
ΣFx = 0:
FAB = 8.97 kN C !
FBD = 14.04 kN C !
12
12
(8.97 kN) − (14.04 kN) + FBE = 0
13
13
FBE = 4.68 kN
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
FBE = 4.68 kN T !
COSMOS: Complete Online Solutions Manual Organization System
Joint E:
ΣFx = 0:
6
12
FEH − 4.68 kN − (8.97 kN) = 0
13
37
FEH = 13.1388 kN
ΣFy = 0:
FDE −
5
(8.97 kN ) −
13
or
1
(13.1388) = 0
37
FDE = 5.6100 kN
Joint D:
ΣFx = 0:
ΣFy = 0:
FEH = 13.14 kN T !
or
FDE = 5.61 kN T !
12
12
1
(14.04 kN ) − FDG − FDH = 0
13
13
2
5
5
(14.04 kN ) − FDG − 1.5 kN − 5.61 kN
13
13
1
+
FDH = 0
2
Solving:
FDG = 8.60 kN C !
FDH = 7.10 kN C !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 24.
FBD Truss:
ΣFx = 0:
Ax = 0
ΣM A = 0:
(12 m ) (M y − 1 kN)
− (2.4 m + 4.8 m + 6 m + 8.4 m + 10.8 m) (1.5 kN) = 0
M y = 5.05 kN
Joint FBDs:
ΣFy = 0:
Ay − 2(1 kN) − 5(1.5 kN) + M y = 0
A y = 4.45 kN
Joint M:
ΣFy = 0:
5.05 kN − 1 kN −
5
FKM = 0,
13
ΣFx = 0:
12
(10.53 kN) − FLM = 0,
13
ΣFx = 0:
9.72 kN −
FKM = 10.53 kN C !
FLM = 9.72 kN T !
6
FJL = 0,
37
FJL = 1.62 37
Joint L:
FJL = 9.85 kN T !
ΣFy = 0:
ΣFx = 0:
Joint K:
ΣFy = 0:
1
(1.62 37 kN) − FKL = 0,
37
12
FIK −
13
FKE = 1.620 kN C !
24
12
FJK − (10.53 kN) = 0
13
577
5
5
(10.53 kN) −
FIK −
13
13
1
FJK
577
− 1.5 kN + 1.62 kN = 0
Solving:
FIK = 10.8136 kN,
FIK = 10.81 kN C !
FJK = 0.26205 kN,
FJK = 0.262 kN T !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Joint J:
ΣFx = 0:
24
(0.26205 kN) −
577
6
( FHJ − 9.85 kN) = 0
37
FHJ = 10.1154 kN,
ΣFy = 0:
1
(10.1154 kN ⋅ 9.85 kN) −
37
FHJ = 10.12 kN T !
1
(0.26205 kN) − FIJ = 0
577
FIJ = 0.054541 kN,
Joint I:
ΣFx = 0:
ΣFy = 0:
Solving:
Joint G:
12
24
( FGI − 10.8136 kN ) + FHI = 0
13
25
5
7
(10.8136 kN − FGI ) + FHI − 1.5 kN + 0.05454 kN = 0
13
25
FGI = 8.6029 kN,
FGI = 8.60 kN C !
FHI = 2.1257 kN,
FHI = 2.13 kN C !
By symmetry:
ΣFy = 0:
FIJ = 54.5 N C !
FOG = 8.60 kN C !
 5
2   (8.6029 kN) − 1.5 kN − FGH = 0
 13 
FGH = 5.12 kN T !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 25.
FBD Truss:
ΣFx = 0: 180 lb − Ax = 0
A x = 180 lb
ΣM A = (12 ft ) G y − ( 6 ft )( 480 lb ) − ( 2 ft )(180 lb )
− (8 ft)(120 lb) = 0,
ΣFy = 0:
G y = 350 lb
Ay − 480 lb − 120 lb + 350 lb = 0
A y = 250 lb
Joint FBDs:
Joint A:
ΣFx = 0:
1
FAC − 180 lb = 0, FAC = 180 2 lb
2
FAC = 255 lb T !
ΣFy = 0:
Joint B:
ΣFx = 0:
ΣFy = 0:
Solving:
1
(180 2 lb) + 250 lb − FAB = 0,
2
FAB = 430 lb C !
4
2
FBD +
FBC = 0
5
5
3
1
430 lb − FBD +
FBC = 0
5
5
180 lb −
FBC = 590 5 lb,
FBC = 1319 lb T !
FBD = 1700 lb C !
Joint C:
ΣFy = 0:
5
1
1
(590 5 lb) −
(180 2 lb) = 0
FCD −
41
5
2
FCD = 154 41 lb,
ΣFx = 0:
FCD = 986 lb T !
4
2
(154 41 lb) −
(590 5 lb)
41
5
1
−
(180 2 lb) = 0
12
FCE = 744 lb T !
FCE +
By inspection of joint G:
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
FFG = 350 lb C and FEG = 0 !
COSMOS: Complete Online Solutions Manual Organization System
Joint F:
FDF
F
350
= EF =
;
5
1
20
FDF = 1750 lb C !
FEF = 700 5 lb,
FEF = 1565 lb T !
Joint E:
ΣFy = 0:
5
1
(700 5 lb) = 0
FDE − 120 lb −
41
5
FDE = 164 41 lb,
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
FDE = 1050 lb T !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 26.
Joint FBDs:
Joint A:
ΣFy = 0:
9
FAC − 1.8 kips = 0,
41
40
FAB − 8.20 kips = 0,
41
ΣFx = 0:
DE =
ΣFy = 0:
FBD = 8.00 kips C !
FBC = 0.6 kips,
and then
Note:
FAB = 8.00 kips C !
FBD = FAB
By inspection of joint B,
Joint C:
FAC = 8.20 kips T !
FBC = 0.600 kips C !
9.2
(4.5 ft) = 2.07 ft,
20
CE = 4.14 ft = 2 DE
1
FCD − 0.6 kips = 0,
5
FCD = 0.6 5 kips
FCD = 1.342 kips T !
ΣFx = 0:
8.2 kips +
(
)
2
0.6 5 kips − FCE = 0
5
FCE = 9.20 kips C !
Joint D:
ΣFy = 0:
(
)
40
2
0.6 5 kips = 0
( FDG − 8.2 kips ) −
41
5
FDG = 9.43 kips T !
ΣFx = 0:
FDE −
(
)
9
1
0.6 5 kips = 0
( 9.43 kips − 8.2 kips ) −
41
5
FDE = 0.330 kips C !
Joint E:
ΣFy = 0:
ΣFx = 0:
5
FEG − 0.33 kips = 0,
13
FEG = 0.858 kips T !
12
( 0.858 kips ) + 9.2 kips − FEF = 0
13
FEF = 9.992 kips,
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
FEF = 9.99 kips C !
COSMOS: Complete Online Solutions Manual Organization System
Joint F:
By vertical symmetry FFG = FFH
ΣFx = 0:
4

9.992 kips − 2  FFG  = 0,
5


FFG = 5.995 kips
FFG = FFH = 6.00 kips C !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 27.
P6.14
Starting with
ABC, add, in order, joints E, D, F, G, H
∴simple truss
P6.15
Starting with
DEF, add, in order, G, C, B, A, I, H, J
∴simple truss
P6.23
Starting with
ABC, add, in order, E, F, D, H, G, I, J, K, L, M
∴simple truss
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 28.
P6.21
Starting with ABC, add, in order, joints E, D, F, H, J, K, L, I, G, N,
P, Q, R, O, M, S, T
∴simple truss
P6.25
Starting with
ABC, add, in order, joints D, E, F, G
∴simple truss
P6.29
Starting with
ABD, add, in order, joints H, G, F, E, I, G, J
∴simple truss
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 29.
ΣFx = 0:
Fx = 0
Then, by inspection of joint F,
FFG = 0
Then, by inspection of joint G,
FGH = 0
By inspection of joint J,
FIJ = 0
Then, by inspection of joint I,
FHI = 0
FEI = 0
Then, by inspection of joint E,
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
FBE = 0
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 30.
By inspection of joint D,
FDI = 0
By inspection of joint E,
FEI = 0
Then, by inspection of joint I,
FAI = 0
By inspection of joint F,
FFK = 0
By inspection of joint G,
FGK = 0
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 31.
\
By inspection of joint C:
FBC = 0 W
Then, by inspection of joint B:
FBE = 0 W
Then, by inspection of joint E:
FDE = 0 W
By inspection of joint H :
FFH = 0 W
and FHI = 0 W
By inspection of joint Q :
FOQ = 0 W
and FQR = 0 W
By inspection of joint J :
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
FLJ = 0 W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 32.
By inspection of joint C :
FBC = 0 W
By inspection of joint G :
FFG = 0 W
Then, by inspection of joint F :
FFE = 0 W
By inspection of joint I :
FIJ = 0 W
By inspection of joint M :
FMN = 0 W
Then, by inspection of joint N :
FKN = 0 W
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 33.
By inspection of joint C:
FBC = 0 !
By inspection of joint M :
FLM = 0 !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 34.
By inspection of joint A:
FAF = 0 W
By inspection of joint C :
FCH = 0 W
By inspection of joint E :
FDE = FEI = 0 W
By inspection of joint L :
FGL = 0 W
By inspection of joint N :
FIN = 0 W
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 35.
(a)
By inspection of joint H :
FCH = 0 W
Then, by inspection of joint C :
FCG = 0 W
Then, by inspection of joint G :
FBG = 0 W
Then, by inspection of joint B :
FBF = 0 W
(b)
By inspection of joint J :
Then, by inspection of joint I :
FIJ = 0 W
FEI = 0 W
FHI = 0 W
Then, by inspection of joint E :
also,
ΣFx = 0;
FBE = 0 W
Fx = 0
So, by inspection of joint F :
FFG = 0 W
And by inspection of joint G :
FGH = 0 W
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 36.
FBD Truss:
ΣFz = 0:
Bz = 0
ΣM x = 0:
− ( 0.47 m ) C y + ( 0.08 m )( 940 N ) = 0
C y = 160 Nj
Joint FBDs:
Joint D:
where FAD = FAD
FBD = FBD
= FBD
FCD = FCD
ΣFy = 0:
ΣFx = 0:
ΣFz = 0:
4
FBD − 940 N = 0,
21
10
80
−
FAD −
FCD −
89
101
1
39
−
FAD +
FCD −
89
101
Solving:
− 0.80 m i − 0.08 m k
( 0.8)
2
= FAD
2
+ ( 0.08 ) m
−10i − k
101
− 0.8 m i − 0.16 m j − 0.2 m k
( 0.8 m )2 + ( 0.16 m )2 + ( 0.2 m )2
− 20i + 4 j − 5k
21
− 0.8 m i + 0.39 m k
( 0.8 m )
2
+ ( 0.39 m )
FBD = 4935 N,
2
= FCD
− 80i + 39k
89
FBD = 4.94 kN T !
20
( 4935 N ) = 0
21
5
( 4935 N ) = 0
21
FAD = − 590 101 N,
FAD = 5.93 kN C !
FCD = 1335 N,
FCD = 1.335 kN T !
continued
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Joint C:
FBC = FBC
0.16 m j − 0.59 m k
( 0.16 m )2 + ( 0.59 m )2
= FBC
16 j − 59k
3737
16
FBC + 160 N = 0
3737
ΣFy = 0:
FBC = −10 3737 N
ΣFz = 0:
− FAC −
(
FBC = 611 N C !
)
59
39
−10 3737 N −
(1335 N ) = 0
89
3737
FAC = 5.00 N T !
Joint B:
FAB = FAB
ΣFy = 0:
− 0.16 m j + 0.12 m k
( 0.16 m )
−
4
FAB +
5
2
+ ( 0.12 m )
2
(
− 4 j + 3k
5
= FAB
)
16
4
10 3737 N −
( 4935 N ) = 0
21
3737
FAB = 975 N C !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 37.
FBD Truss:
ΣFz = 0:
Bz + 987 N = 0
B z = − 987 N k
ΣM x = 0:
− ( 0.47 m ) C y − ( 0.16 m )( 987 N )
+ ( 0.08 m )( 940 N ) = 0
C y = −176.0 N j
Joint FBDs:
Joint D:
FAD = FAD
FBD = FBD
FCD = FCD
ΣFy = 0:
ΣFx = 0:
ΣFz = 0:
− 0.8 m i − 0.08 m k
( 0.8 m )2 + ( 0.08 m )2
FAD
( −10i − k )
101
=
− 0.8 m i − 0.16 m j − 0.2 m k
( 0.8 m )
2
2
+ ( 0.16 m ) + ( 0.2 m )
− 0.8 m i + 0.39 m k
( 0.8 m )2 + ( 0.39 m )2
=
2
=
FCD
( − 80i + 39k )
89
4
FBD − 940 N = 0, FBD = 4935 N,
21
10
80
20
−
FAD −
FCD −
( 4935 N ) = 0
89
21
101
1
39
5
−
FAD +
FCD −
( 4935 N ) + 987 N = 0
89
21
101
Solving:
FBD
( − 20i + 4 j − 5k )
21
FCD = − 534 N,
FAD = − 422 101 N,
FBD = 4.94 kN T !
FCD = 534 N C !
FAD = 4.24 kN C !
continued
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Joint C:
FBC = FBC
0.16 m j − 0.59 m k
( 0.16 m )
2
+ ( 0.59 m )
2
=
FBC
(16 j − 59k )
3737
16
FBC − 176 N = 0, FBC = 11 3737 N
3737
ΣFy = 0:
FBC = 672 N T !
ΣFz = 0:
− FAC −
(
)
59
39
11 3737 N +
( 534 N ) = 0
89
3737
FAC = − 415 N,
FAC = 415 N C !
Joint B:
FAB = FAB
ΣFy = 0:
− 0.16 m j + 0.12 m k
( 0.16 m )2 + ( 0.12 m )2
−
4
FAB −
5
(
=
FAB
( − 4 j + 3k )
5
)
16
4
11 3737 N −
( 4935 N ) = 0
21
3737
FAB = −1395 N, FAB = 1.395 kN C !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 38.
FBD Truss:
ΣM BD = 0:
ΣM x = 0:
ΣFz = 0:
( 4 ft )( 50 lb ) + ( 4 ft )( Ez ) = 0
E z = − ( 50 lb ) k
( 4 ft )( 300 lb ) + ( 4 ft )( − 50 lb ) + ( 4 ft )( Dz ) = 0
D z = − ( 250 lb ) k
Bz − 50 lb − 250 lb = 0
B z = ( 300 lb ) k
ΣM Bz = 0:
( 2 ft )( 300 lb ) − ( 4 ft ) ( C y ) = 0
C y = (150 lb ) j
AB = AC = 20 ft
AD = AE = 6 ft
CD = 4 2 ft
ΣFx = 0:
ΣFy = 0:
B x = − ( 50 lb ) i
Bx + 50 lb = 0
By + 150 lb − 300 lb = 0
B y = (150 lb ) j
FAE = FAE
=
2 ft i − 4 ft j + 4 ft k
( 2 ft )2 + ( 4 ft )2 + ( 4 ft )2
FAE
( 2i − 4 j + 4k )
3
Joint FBDs:
Joint E:
ΣFz = 0:
ΣFx = 0:
2
FAE − 50 lb = 0,
3
1
FDE + ( 75 lb ) = 0,
3
FAE = 75.0 lb T !
FDE = − 25 lb
FDE = 25.0 lb C !
ΣFy = 0:
− FCE −
2
( 75 lb ) = 0,
3
FCE = − 50 lb
FCE = 50.0 lb C !
continued
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
− 2 ft i − 4 ft j + 4 ft k
FAD = FAD
( 2 ft )
Joint D:
ΣFx = 0:
2
+ ( 4 ft ) + ( 4 ft )
− 4 ft i − 4 ft j
FCD = FCD
ΣFz = 0:
2
( 4 ft )
2
+ ( 4 ft )
2
=
2 ft i + 4 ft k
FAC = FAC
( 2 ft )
)
Joint B:
FAB = FAB
2
+ ( 4 ft )
2
=
FAC = 0 !
FBC − 100 lb = 0,
− 2 ft i + 4 ft k
( 2 ft )
2
ΣFz = 0:
+ ( 4 ft )
2
=
FBD = 150.0 lb C !
FAC
( i + 2k )
5
2
FAC = 0,
5
ΣFz = 0:
ΣFx = 0:
FAD
( − i − 2 j + 2k )
3
2
FAD − 250 lb = 0, FAD = 375 lb,
FAD = 375 lb T !
3
1
1
FCD − ( 375 lb ) = 0, FCD = −100 2 lb
25 lb −
3
2
FCD = 141.4 lb C !
1
2
− FBD −
−100 2 lb − ( 375 lb ) = 0
3
2
FBD = − 150 lb
Joint C:
=
FCD
( − i − j)
2
(
ΣFy = 0:
2
FBC = 100.0 lb T !
FAB
( − i + 2k )
5
2
FAB + 300 lb = 0
5
FAB = −150 5 lb,
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
FAB = 335 lb C !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 39.
(a)
FBD Truss:
Bx = 0
ΣM y = 0:
Bx = 0
ΣM z = 0:
(1.7 m ) Az + ( 0.6 m )(1700 N ) = 0,
− (1.7 m ) Ax − (1.125 m )(1700 N ) = 0
ΣFx = 0:
Cx − 1125 N = 0
Cx = 1125 N
C y − 1700 N = 0
C y = 1700 N
Bz − 600 N = 0
Bz = 600 N
ΣM x = 0:
ΣFy = 0:
ΣFz = 0:
Az = − 600 N
Ax = −1125 N
A = − (1125 Ν ) i − ( 600 N ) k
so
B = ( 600 Ν ) k
C = (1125 Ν ) i + (1700 N ) j
(b)
Find zero force members:
by inspection of joint B,
FAB = 0 !
by inspection of joint D,
FAD = 0 !
by inspection of joint C ,
FBC = 0 !
Joint FBDs:
Joint E:
FAE = FAE
ΣFy = 0:
−1.125 m i + 1.7 m j − 0.6 m k
(1.125 m )
2
2
+ (1.7 m ) + ( 0.6 m )
1.7
FAE − 1700 N = 0,
2.125
2
=
FAE
( −1.125i + 1.7 j − 0.6k )
2.125
FAE = 2125 N
FAE = 2.13 kN T !
ΣFx = 0:
− FBE −
1.125
( 2125 N ) , FBE = −1125 N
2.125
FBE = 1.125 kN C !
ΣFz = 0:
− FDE −
0.6
( 2125 N ) = 0,
2.125
FDE = − 600 N
FDE = 600 N C !
continued
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Joint D:
FBD = FBD
ΣFz = 0:
−1.125 m i + 0.6 m k
(1.125 m )2 + ( 0.6 m )2
600 N −
0.6
FBD = 0,
1.275
=
FBD
( −1.125i + 0.6k )
1.275
FBD = 1275 N
FBD = 1.275 kN T !
ΣFx = 0:
− FCD −
1.125
(1275 N ) = 0,
1.275
FCD = −1125 N
FCD = 1.125 kN C !
Joint C:
By inspection:
FAC = −1700 N
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
FAC = 1.700 kN C !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 40.
(a)
(b)
To check for simple truss, start with ABDE and add three members at a time which meet at a single new
joint, successively adding joints G, F, H and C.
∴ This is a simple truss !
There are six reaction force components, Ax, Ay, Az, By, Bz, and Gy, no two of which are colinear, so all
can be determined with the six equilibrium equations. Motion is prevented by the constraints.
∴ Truss is completely constrained, and statically determinate. !
FBD Truss:
ΣM AB = 0:
( 9.60 ft ) G y + (10.08 ft )( 240 lb ) = 0
G y = − 252 lb
ΣM z = 0:
G y = − 252 lb j
(11.00 ft ) ( By − 252 lb ) − (10.08 ft )( 275 lb ) = 0
B y = 504 lb j
ΣFx = 0:
ΣM y = 0:
− Ax + 275 lb = 0,
A x = − 275 lb i
− ( 9.6 ft )( 275 lb ) − (11.0 ft )( Bz ) = 0
Bz = − 240 lb,
B z = − 240 lb k
ΣFy = 0:
− Ay + 504 lb − 252 lb = 0,
A y = − 252 lb j
ΣFz = 0:
Az − 240 lb + 240 lb = 0
Az = 0
Determine zero force members:
By inspection of joint C :
FBC = FCD = FGC = 0
By inspection of joint F :
FBF = FEF = FFG = 0
By inspection of joint A:
with Az = 0,
By inspection of joint H :
FDH = 0
FAD = 0
And finally, considering joint D knowing that FAD = FCD = FDH = 0, and recognizing that the three
remaining members are not co-planar, they must also carry zero load, FBD = FDE = FGH = 0.
The only load bearing members are thus AB, AE, BE, BG, EG, EH, GH.
continued
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Joint FBDs
FAB = 275 lb T,
By inspection:
from above:
Joint A:
FAE = 252 lb T !
FDE = 0 !
FEF = 0 !
FEH = 240 lb C !
By inspection of joint H
ΣFy = 0:
− 252 lb −
10.08
FBE = 0
1492
FBE = − 373 lb,
Joint E:
ΣFz = 0:
240 lb −
FBE = 373 lb C !
9.6
FEG = 0
14.6
FEG = 365 lb T !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 41.
(a)
(b)
To check for simple truss, start with ABDE and add three members at a time which meet at a single new
joint, successively adding joints G, F, H and C.
∴ This is a simple truss !
There are six reaction force components, Ax, Ay, Az, By, Bz, and Gy, no two of which are colinear, so all
can be determined with the six equilibrium equations. Motion is prevented by the constraints.
∴ Truss is completely constrained, and statically determinate. !
FBD Truss:
ΣM AB = 0:
( 9.60 ft ) G y + (10.08 ft )( 240 lb ) = 0
G y = − 252 lb,
ΣM z = 0:
G y = − 252 lb j
(11.00 ft ) ( By − 252 lb ) − (10.08 ft )( 275 lb ) = 0
B y = 504 lb j
ΣFx = 0:
ΣM y = 0:
− Ax + 275 lb = 0,
Ax = − 275 lb i
− ( 9.6 ft )( 275 lb ) − (11.0 ft )( Bz ) = 0
Bz = − 240 lb,
B z = − 240 lb k
ΣFy = 0:
− Ay + 504 lb − 252 lb = 0,
A y = − 252 lb j
ΣFz = 0:
Az − 240 lb + 240 lb = 0
Az = 0
Determine zero force members:
By inspection of joint C :
FBC = FCD = FGC = 0
By inspection of joint F :
FBF = FEF = FFG = 0
By inspection of joint A:
with Az = 0,
By inspection of joint H :
FDH = 0
FAD = 0
And finally, considering joint D knowing that FAD = FCD = FDH = 0, and recognizing that the three
remaining members are not co-planar, they must also carry zero load, FBD = FDE = FGH = 0.
The only load bearing members are thus AB, AE, BE, BG, EG, EH, GH.
continued
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Now, by inspection of joint H,
FGH = 275 lb C !
FCG = 0 !
from above,
FDG = 0 !
FFG = 0 !
FBD Joint G:
FBG = FBG
−10.08j + 9.6k
13.92
FEG = FEG
−11i + 9.6k
14.6
ΣFx = 0:
ΣFy = 0:
275 lb −
−
11
FEG = 0,
14.6
10.08
FBG − 252 lb = 0,
13.92
FEG = 365 lb T !
FBG = − 348 lb
FBG = 348 lb C !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 42.
FBD Truss:
ΣFx = 0:
Kx = 0
(
)
6a K y − 125 lb − 5a ( 250 lb )
ΣM A = 0:
− 4a ( 250 lb ) − 3a ( 375 lb )
− 2a ( 500 lb ) − a ( 500 lb ) = 0
K y = 937.5 lb
Ay − 3 ( 250 lb ) − 2 ( 500 lb )
ΣFy = 0:
− 375 lb − 125 lb + 937.5 lb = 0
A y = 1312.5 lb
FBD Section ABEC:
( 2 ft ) FCF + ( 4 ft )( 500 lb )
ΣM E = 0:
+ ( 8 ft )( 250 lb − 1312.5 lb ) = 0
FCF = 3250 lb,
ΣFy = 0:
FCF = 3.25 kips T W
1312.5 lb − 250 lb − 2 ( 500 lb ) −
FEF = 62.5 5 lb,
ΣFx = 0:
3250 lb +
(
1
FEF = 0
5
FEF = 139.8 lb T W
)
2
62.5 5 lb − FEG = 0
5
FEG = 3375 lb,
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
FEG = 3.38 kips C W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 43.
FBD Truss:
ΣFx = 0:
ΣM A = 0:
Kx = 0
(
)
6a K y − 125 lb − 5a ( 250 lb )
− 4a ( 250 lb ) − 3a ( 375 lb )
− 2a ( 500 lb ) − a ( 500 lb ) = 0
K y = 937.5 lb
ΣFy = 0:
Ay − 3 ( 250 lb ) − 2 ( 500 lb )
− 375 lb − 125 lb + 937.5 lb = 0
A y = 1312.5 lb
FBD Section IJLK:
ΣM z = 0:
( 4 ft )( 937.5 lb −125 lb ) − ( 2 ft ) FHJ = 0
FHJ = 1625 lb,
ΣFy = 0:
937.5 lb − 125 lb − 250 lb −
FHI = 562.5 5 lbs,
ΣFx = 0:
1.625 kips +
(
FHJ = 1.625 kips C W
1
FHI = 0
5
FHI = 1.258 kips C W
)
2
0.5625 5 kips − FFI = 0
5
FFI = 2.75 kips T W
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 44.
FBD Truss:
ΣFx = 0:
Ax = 0
By load symmetry:
A y = L y = 840 N
ΣM E = 0:
2
24
(1.75 m ) FBD + ( 2 m )( 240 N )
3
25
− ( 4 m )( 840 N − 120 N ) = 0
FBD =
FBD Section ABC:
ΣM B = 0:
FBD = 2.14 kN C W
 1.75 
m  FCE − ( 2 m )( 840 N − 120 N ) = 0

 3

FCE =
ΣM A = 0:
15000
N,
7
17280
N,
7
FCE = 2.47 kN T W
( 4 m ) 
7

FBE  − ( 2 m )( 240 N ) = 0
 25

FBE =
3000
N,
7
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
FBE = 429 N C W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 45.
FBD Truss:
ΣFx = 0:
Ax = 0
By load symmetry
A y = L y = 840 N
0.4
2
=
3
15
3.5
7
IK = 3 =
4
24
2.3 23
IJ =
=
3
30
slope: GI =
FBD Section JKL:
Resolve FIK and FGJ at L
ΣM J = 0:
( 3 m ) 
7

 24

FIK  − ( 0.4 m ) 
FIK 
25
25




+ ( 3 m )( 840 N − 120 N ) − (1 m )( 240 N ) = 0
FIK = 4210.53 N,
ΣM I = 0:
FIK = 4.21 kN C W

  3.5   15

2
FGJ  − 
FGJ 
m
  229
 229
  3

(4 m) 
+ ( 4 m )( 840 N − 120 N ) − ( 2 m )( 240 N ) = 0
FGJ = 3823.0 N,
ΣFx = 0:
FGJ = 3.82 kN T W
7
15
30
FIJ = 0
( 4210.5 N ) −
( 3823.0 N ) −
25
229
1429
FIJ = 318 N T W
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 46.
\
FBD Truss:
ΣFx = 0:
By symmetry,
Ax = 0
A y = M y = 8.5 kN
FBD Section ABDC:
ΣM D = 0:
(1 m )(1.5 kN − 8.5 kN ) + ( 0.5 m )
40
FCE = 0
41
FCE = 14.35 kN T W
ΣM E = 0:
(1 m )(1.5 kN ) + ( 2 m )(1.5 kN − 8.5 kN )
+ ( 0.5 m )
40
FDF = 0,
41
FDF = 25.625 kN
FDF = 25.6 kN C W
ΣFx = 0:
40
40
FDE = 0
(14.35 kN − 25.625 kN ) +
41
1721
FDE = 11.4084 kN,
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
FDE = 11.41 kN T W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 47.
FBD Truss:
ΣFx = 0:
By symmetry,
Ax = 0
A y = M y = 8.5 kN
FBD Section IMNJ:
ΣM J = 0:
( 2 m )(8.5 kN − 1.5 kN ) − (1 m )(1.5 kN )
− ( 0.5 m )
40
FGI = 0
41
FGI = 25.625 kN,
ΣM G = 0:
FGI = 25.6 kN T W
( 3 m )(8.5 kN − 1.5 kN ) − ( 2 m )(1.5 kN )
40
FHJ = 0
41
FHJ = 30.8 kN C W
− (1 m )( 3 kN ) − ( 0.5 m )
FHJ = 30.75 kN,
ΣFx = 0:
40
( 30.75 kN − 25.625 kN ) −
41
40
FGJ = 0
2441
FGJ = 6.18 kN T W
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 48.
FBD Truss:
ΣFx = 0: A x = 0
By symmetry: A y = L y = 6 kN
FBD Section:
Notes:
15
m
6
2 5 5
yD = ⋅ = m
3 2 3
2
2
yE = ⋅ 1 = m
3
3
5
yF − yD = m
6
yG = 1 m
yF =
yD − yG =
2
m
3
continued
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
ΣM D = 0: (1 m )
6
FEG + ( 2 m )( 2 kN ) + ( 4 m )(1 kN − 6 kN ) = 0
37
FEG =
8
37 kN
3
FEG = 16.22 kN T !
 1

 3

ΣM A = 0: ( 2 m )( 2 kN ) + ( 4 m )( 2 kN ) − ( 6 m ) 
FDG  − (1 m ) 
FDG  = 0
 10

 10

FDG =
ΣFx = 0:
4
10 kN
3
6
3
12
FEG −
FDG −
FDF = 0
13
37
10
FDG = 4.22 kN C !
16 − 4 −
FDF = 13 kN
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
12
FDF = 0
13
FDF = 13.00 kN C !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 49.
FBD Truss:
ΣFx = 0: A x = 0
By symmetry: A y = L y = 6 kN
FBD Section:
Notes:
so
2
3
2
=
3
yI =
m
yH
⋅
5 5
= m
2 3
yH − yI = 1 m
continued
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
ΣM I = 0:
( 4 m )( 6 kN − 1 kN ) − ( 2 m )( 2 kN ) − (1 m ) 
12

FHJ  = 0
 13

FHJ =
ΣM H = 0:
FHJ = 17.33 kN C !
 6

FGI  = 0
 37

( 4 m )( 6 kN − 1 kN ) − ( 2 m )( 2 kN ) − (1 m ) 
FGI =
ΣM L = 0:
52
kN
3
8
37 kN
3
( 2 m )( 2 kN ) − ( 4 m ) FHI
FGI = 16.22 kN T !
=0
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
FHI = 1.000 kN T !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 50.
FBD Truss:
Distance between loads = 1.5 m
ΣFx = 0:
By symmetry,
Ax = 0
A y = K y = 18 kN
FBD Section ABC:
FBD Section ABC:
ΣM D = 0:
(1.5 m ) FCE + (1.5 m )( 6 kN ) − ( 3 m )(18 kN − 3 kN ) = 0
FCE = 22.5 kN T !
ΣM A = 0:
(1.8 m )
4
FCD − (1.5 m )( 6 kN ) = 0
5
FCD = 6.25 kN T !
ΣFy = 0:
18 kN − 3 kN − 6 kN −
8
4
FBD + ( 6.25 kN ) = 0
17
5
FBD = 29.8 kN C !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 51.
FBD Truss:
Distance between loads = 1.5 m
ΣFx = 0:
By symmetry,
Ax = 0
A y = K y = 18 kN
FBD Section ABC:
FBD Section GHK:
FBD section GHK:
ΣM F = 0:
( 4.5 m )(18 kN − 3 kN ) − ( 3 m )( 6 kN )
− (1.5 m )( 6 kN ) − ( 2.4 m ) FEG = 0
FEG = 16.875 kN,
ΣM K = 0:
 8

FFG  = 0
 73

(1.5 m )( 6 kN ) + ( 3 m )( 6 kN ) − ( 3.6 m ) 
FFG = 8.0100 kN,
ΣFx = 0:
FEG = 16.88 kN T !
15
FFH −
17
FFG = 8.01 kN T !
3
(8.0100 kN ) − 16.875 kN = 0
73
FFH = 22.3 kN C !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 52.
FBD Truss:
ΣFX = 0:
A x + 4 kips − 4 kips = 0
Ax = 0
By symmetry,
Ay = Ny = 0
FBD Section ABDC:
ΣM D = 0:
(18 ft )( 4 kips ) − ( 9 ft )
4
FCE = 0
5
FCE = 10.00 kips C !
ΣM E = 0:
(18 ft )( 4 kips ) − (12 ft )
3
FDF = 0
5
FDF = 10.00 kips C
ΣFx = 0:
4 kips +
4
(10 kips − 10 kips ) − FDE = 0
5
FDE = 4.00 kips C !
FBD Joint E:
ΣFy = 0:
4
3
FEF − ( 4 kips ) = 0
5
5
FEF = 3.00 kips T !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 53.
FBD Truss:
ΣFx = 0:
A x + 4 kips − 4 kips = 0
Ax = 0
Ay = Ny = 0
By symmetry,
FBD Section IJNM:
ΣM G = 0:
4
− ( 27 ft )( 4 kips ) + ( 9 ft ) FHI = 0
5
FHI = 15.00 kips T !
ΣM N = 0:
4
5
( 9 ft ) (15 kips ) − ( 27 ft ) FGI
=0
FGI = 4.00 kips C !
FBD Joint I:
ΣFx = 0:
3
4
( 4 kips ) − FIJ = 0
5
5
FIJ = 3.00 kips T !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 54.
FBD Truss:
ΣFx = 0:
Ax = 0
A y = M y = 7.5 kips
By symmetry,
FBD Section IJNM:
( 9 ft )( 7.5 kips − 0.5 kips ) − ( 4.5 ft )(1 kip )
ΣM J = 0:
 4

− ( 5.25 ft ) 
FGI  = 0,
 17

FGI = 11.4858 kips.
FGI = 11.49 kips T !
(12 ft )( 7.5 kip − 0.5 kips ) − ( 7.5 ft )(1 kip )
ΣM G = 0:
− ( 3 ft )( 3 kips ) + ( 4.5 ft ) FHJ = 0
FHJ = 15 kips C
By inspection of joint H,
ΣFy = 0:
FFH = 15.00 kips C !
1
(11.4858 kips ) + 7.5 kips − 4.5 kips
17
−
3
FGJ = 0
13
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
FGJ = 6.95 kips T !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 55.
FBD Truss:
ΣFx = 0: A x = 0
By symmetry:
A y = v y = 4.5 kips
FBD Joint K:
ΣFx = 0:
ΣFy = 0:
1
( FIK − FKN ) = 0
2
 1

FIK  − 1 kip = 0
2
 2

so
FIK = FKN
FIK =
2
kip
2
FIK = 0.707 kip C !
FBD Section ABEIJ:
ΣFy = 0:
ΣM I = 0:
4.5 kips − 1 kip − 2 (1.5 kips ) −
1 2
3
FJL = 0,
kips +
2
2 2
( 3 ft )(1.5 kips ) + ( 6 ft )(1 kip − 4.5 kips )
+
3
( 3 ft ) ( FJM ) = 0
2
FJM =
FJL = 0 !
11
kips
3
FJM = 6.35 kips T !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 56.
FBD Truss:
ΣFx = 0:
ΣM N = 0:
Ax = 0
( 4.8 ft )( 300 lb ) + ( 7.2 ft )( 750 lb )
+ (11.52 ft )( 750 lb ) + (16.8 ft )(1950 lb )
+ (19.2 ft )( 900 lb ) + ( 26.4 ft )(1050 lb )
(
)
+ ( 33.6 ft ) 150 lb − Ay = 0
Ay = 2925 lb
ΣFy = 0:
2925 lb − 2 (150 lb ) − 2 ( 750 lb ) − 1050 lb
− 900 lb − 1950 lb − 300 lb + N y = 0
FBD Section ABC:
N y = 3075 lb
ΣM B = 0:
( 3.6 ft ) FCE − ( 7.2 ft )( 2925 lb − 150 lb ) = 0
FCE = 5550 lb,
ΣM A = 0:
( 7.2 ft )
FCE = 5.55 kips T !
1
2
FBE + ( 3.6 ft )
FBE − ( 7.2 ft )(1050 lb ) = 0
5
5
FBE = 525 5 lb,
ΣFx = 0:
(
FBE = 1.174 kips C !
)
2
FBD − 525 5 lb + 5550 lb = 0
5
FBD = −5031.2 lb
By inspection of joint D, FDF = FBD , so
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
FDF = 5.03 kips C !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 57.
FBD Truss:
ΣFx = 0:
Ax = 0
( 4.8 ft )( 300 lb ) + ( 7.2 ft )( 750 lb )
+ (11.52 ft )( 750 lb ) + (16.8 ft )(1950 lb )
+ (19.2 ft )( 900 lb ) + ( 26.4 ft )(1050 lb )
+ ( 33.6 ft ) (150 lb − Ay ) = 0
ΣM N = 0:
Ay = 2925 lb
FBD Section KJMN:
2925 lb − 2 (150 lb ) − 2 ( 750 lb ) − 1050 lb
ΣFy = 0:
− 900 lb − 1950 lb − 300 lb + N y = 0
N y = 3075 lb
( 7.2 ft )( 3075 lb − 150 lb ) − ( 2.4 ft )( 300 lb )
ΣM J = 0:
− (1.8 ft ) FIK = 0
FIK = 11300 lb,
ΣM I = 0:
FIK = 11.30 kips T
(11.52 ft )( 3075 lb − 150 lb ) − ( 6.72 ft )( 300 lb )
 2

− ( 4.32 ft )( 750 lb ) + ( 3.96 ft ) 
FHJ  = 0
 5

ΣFx = 0:
FBD Joint I:
FHJ = −3590.5 5 lb,
FHJ = 8.03 kips C !
2
12
−
−3590.9 5 lb −
FIJ − 11300 lb = 0
13
5
(
)
FIJ = − 4461.4 lb,
ΣFx = 0:
11300 lb −
FIJ = 4.46 kip C !
12
( 4461.4 lb ) − FGI = 0
13
FGI = 7181.8 lb,
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
FGI = 7.18 kips T !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 58.
FBD Truss:
Notes: α = 20°, β = 40°, γ = 60°, δ = 80°, φ = 22.5°, θ = 45°, ψ = 60°
outer members AC, CE, etc. are each 1.0 m
radial members AB, CD, etc. are each 0.4 m
By symmetry , Ay = By = 2.6 kN
ΣFx = 0: A x = 0
FBD Section ABDC:
ΣM F = 0:
( resolve FCE at E )
 1

 1

 2 ( 0.4 m )  FCE sin 40° +  2 ( 0.4 m )  ( FCE cos 40° )




1


+ (1 m ) sin 40° +
( 0.4 m ) ( 0.4 kN )
2


1


− (1 m ) sin 20° + (1 m ) sin 40° +
( 0.4 m ) ( 2.6 kN ) = 0,
2


FCE = 7.3420 kN,
FCE = 7.34 kN C !
(1 m ) cos 40° − ( 0.4 m ) sin 45° + ( 0.4 m ) sin 22.5°
(1 m ) sin 40° + ( 0.4 m ) cos 45° − ( 0.4 m ) cos 22.5°
(1 m ) cos 40° − ( 0.4 m ) sin 45° = 27.566°
= tan −1
(1 m ) sin 40° + ( 0.4 m ) cos 45°
note σ = tan −1
and ε
Then,
ΣM C = 0:
= 48.848°
( 0.4 m )  FDF cos ( 90° − 22.5° − 48.848°)
− (1 m ) sin 20° ( 2.6 kN ) = 0,
FDF = 2.3464 kN,
and,
ΣFx = 0:
FDF = 2.35 kN T !
FCF ( cos 27.566° ) + ( 2.3464 kN )( cos 48.848° )
− ( 7.3420 kN ) sin 40° = 0
FCF = 3.5819 kN,
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
FCF = 3.58 kN T !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 59.
FBD Truss:
Notes: α = 20°, β = 40°, γ = 60°, δ = 80°, φ = 22.5°, θ = 45°, ψ = 60°
outer members AC, CE, etc. are each 1.0 m
radial members AB, CD, etc. are each 0.4 m
By symmetry, Ay = By = 2.6 kN
ΣFx = 0: A x = 0
FBD Section ABDFHGEC:
( resolve FGI at I )
( 0.4 m ) FGI sin 80° + (1 m )( sin 80°)(1 kN )
ΣM J = 0:
+ (1 m ) sin 80° + (1 m ) sin 60°  ( 0.6 kN )
+ (1 m ) sin 80° + (1 m ) sin 60° + (1 m ) sin 40° ( 0.4 kN )
− (1 m ) sin 80° + (1 m ) sin 60° + (1 m ) sin 40° + (1 m ) sin 20° ( 2.6 kN ) = 0
FGI = 10.8648 kN,
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
FGI = 10.86 kN C !
COSMOS: Complete Online Solutions Manual Organization System
Note:
so,
σ = tan −1
ΣM H = 0:
0.4 m − (1 m ) cos80°
= 12.9443°
(1 m ) sin 80°
( 0.4 m ) cos 60° ( FGJ sin12.9443° )
− ( 0.4 m ) sin 60° ( FGJ cos12.9443° )
− ( 0.4 m ) sin 60° (10.8648 kN ) sin 80°
+ ( 0.4 m ) cos 60° (10.8648 kN ) cos80°
+ ( 0.4 m ) cos 60° (1 kN ) + (1 m ) sin 60° + ( 0.4 m ) cos 60° ( 0.6 kN )
+ (1 m ) sin 60° + (1 m ) sin 40° + ( 0.4 m ) cos 60° ( 0.4 kN )
− (1 m )( sin 60° + sin 40° + sin 20° ) + ( 0.4 m ) cos 60° ( 2.6 kN ) = 0
FGJ = 0.93851 kN,
FGJ = 939 N T !
FBD Joint I:
By symmetry:
ΣFy = 0:
FIK = 10.8648 kN
− FIJ − 1.2 kN + 2 (10.8648 kN ) cos80° = 0
FIJ = 2.5733 kN,
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
FIJ = 2.57 kN T !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 60.
FBD Truss:
ΣFx = 0: A x = 0
By symmetry, Ay = Ny = 1.6 kips
FBD Section using cut a–a:
ΣM D = 0:
( 9 ft ) FFH + ( 6 ft )( 0.5 kip ) − (12 ft )(1.6 kips ) = 0
FFH = 1.800 kips T !
ΣFx = 0:
1.800 kips − FDG = 0
FDG = 1.800 kips C !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 61.
\
FBD Truss:
ΣFx = 0: A x = 0
By symmetry Ay = Ny = 1.6 kips
FBD Joint J:
ΣFx = 0:
− FGJ − FHJ = 0, FHJ = − FGJ
By inspection of joint I, FIJ = 0.6 kip C
FBD Section using cut b-b:
3
 FGJ − ( − FGJ )  − 0.6 kip − 0.5 kip + 1.6 kips = 0
5
ΣFy = 0:
FGJ = −
2.5
kip,
6
ΣM J = 0:
FGJ = 0.417 kip C !
(12 ft )(1.6 kips ) − ( 6 ft )( 0.5 kip ) − ( 4.5 ft )( FHK )
− ( 4.5 ft )( FIL ) = 0
ΣFx = 0:
FIL − FHK = 0 so FIL = FHK
and
FIL = 1.800 kips C !
FHK = 1.800 kips T !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 62.
FBD Section above a-a:
ΣM G = 0:
 27

FIK  − ( 2.7 m )( 40 kN ) − ( 5.4 m )( 40 kN ) = 0
 793

( 5.9 m ) 
FIK = 57.275 kN,
ΣFy = 0:
FIK = 57.3 kN C W
27
( 57.275 kN − FGJ ) = 0
793
FGJ = 57.275 kN T
FBD Section ACIG:
27
( 57.275 kN − 57.275 kN )
793
ΣFy = 0:
+
ΣFx = 0:
18
( FHK − FHJ ) = 0,
949
3 ( 40 kN ) − 2
FHJ = FHK
25
FHK = 0
949
FHK = 53.884 kN
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
FHK = 53.9 kN C W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 63.
FBD Section ACFBD:
ΣM D = 0:
( 4.3 m )
27
FFI − ( 2.7 m )( 40 kN ) = 0
793
FFI = 26.196 kN,
ΣFy = 0:
FFI = 26.2 kN C !
27
( 26.196 kN − FDG ) = 0
793
FDG = 26.196 kN T
FBD Section ACFD:
ΣFy = 0:
27
( 26.196 kN − 26.196 kN )
793
+
ΣFx = 0:
54
( FEI − FEG ) = 0,
6397
−2
−
FEI = FEG
59
( FEG ) + 2 ( 40 kN )
6397
8
( 26.196 kN + 26.196 kN ) = 0
793
FEG = 44.136 kN,
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
FEG = 44.1 kN T !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 64.
FBD Truss:
ΣFx = 0:
ΣM F = 0:
Ax = 0
( 2.4 m )(8 kN ) + ( 5.1 m )(12 kN )
− ( 7.5 m ) Ay = 0,
A y = 10.72 kN
Since only CD can provide an upward force necessary for
equilibrium, it must be in tension, and FBE = 0
FBD Section ABC:
ΣFy = 0:
10.72 kN − 12 kN +
FCD = 2.4229 kN,
ΣM C = 0:
1.68
FCD = 0
3.18
FCD = 2.42 kN T (1.68 m ) FBD − ( 2.4 m )(10.72 kN ) = 0
FBD = 15.3143 kN,
ΣFx = 0: FCE +
FBD = 15.31 kN C 2.7
( 2.4229 kN ) − 15.3143 kN = 0
3.18
FCE = 13.26 kN T Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
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COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 65.
FBD Truss:
ΣFx = 0:
ΣM F = 0:
Ax = 0
( 2.4 m )(8 kN ) + ( 5.1 m )( 6 kN ) − ( 7.5 m ) Ay
=0
A y = 6.64 kN
FBD Section ABC:
Since only BE can provide the downward force necessary for
equilibrium,
it must be in tension, so FCD = 0
ΣFy = 0:
6.64 kN − 6 kN −
FBE = 1.21143 kN,
ΣM = 0:
FBE = 1.211 kN T !
(1.68 m ) FCE − ( 2.4 m )( 6.64 kN ) = 0
FCE = 9.4857 kN,
ΣFx = 0:
1.68
FBE = 0
3.18
9.4857 kN +
FCE = 9.49 kN T !
2.7
(1.21143 kN ) − FBD = 0
3.18
FBD = 10.51 kN C !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 66.
FBD Truss:
ΣFx = 0:
ΣM C = 0:
Cx = 0
(8 ft )( 6 kips ) − (8 ft )( 9 kips ) + (16 ft ) G y
− ( 24 ft )(12 kips ) = 0
G y = 19.5 kips
FBD Section ABC:
ΣFy = 0: 19.5 kips + C y − 6 kips − 9 kips − 12 kips = 0
C y = 7.5 kips
Since only BE can provide the downward force necessary for
equilibrium, it must be in tension, so CD is slack, FCD = 0
ΣFy = 0: − 6 kips + 7.5 kips −
3
FBE = 0
5
FBE = 2.50 kips T !
FBD Section FGH:
Since only EF can provide the downward force necessary for equilibrium,
it must be in tension, so DG is slack, FDG = 0
ΣFy = 0: 19.5 kips − 12 kips −
3
FEF = 0
5
FEF = 12.00 kips T !
Knowing that FCD = FDG = 0, inspection of joint D gives
FDE = 0 !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 67
FBD Truss:
( 32 ft ) H y − ( 24 ft )(12 kips ) − (16 ft )( 9 kips )
ΣM A = 0:
− ( 8 ft )( 6 kips ) = 0,
ΣFy = 0:
H y = 15 kips
Ay − 6 kips − 9 kips − 12 kips + 15 kips = 0
A y = 12 kips
ΣFx = 0:
FBD Section ABC:
Ax = 0
Since only BE can provide the downward force necessary for
equilibrium, it must be in tension, so CD is slack, FCD = 0
ΣFy = 0: 12 kips − 6 kips −
3
FBE = 0
5
FBE = 10.00 kips T !
Since only EF can provide the downward force necessary for equilibrium,
it must be in tension, so DG is slack, FDG = 0
FBD Section FGH:
ΣFy = 0: 15 kips − 12 kips −
3
FEF = 0
5
FEF = 5.00 kips T !
Knowing that FCD = FDF = 0, inspection of joint D gives
FDE = 0 !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 68
FBD Section ABDC:
Since only DE can provide the leftward force necessary for equilibrium, it must be in tension, and CF must be
slack, FCF = 0
ΣFx = 0:
ΣM D = 0:
20 kN −
4
FDE = 0,
5
FDE = 25.0 kN C !
( 3.2 m ) FCE − ( 2.4 m )( 20 kN ) = 0
FCE = 15.00 kN T !
ΣFy = 0: FDF − 15.00 kN −
3
( 25 kN ) = 0
5
FDE = 30.0 kN C !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 69.
FBD Section ABFE:
It is not apparent which counter is active, so we guess that FEH = 0;
ΣM F = 0:
 24

FEG + 40 kN  − ( 4.8 m )( 20 kN ) = 0
 745

( 3.2 m ) 
FEG = −11.3728 kN
FEG = 11.37 kN C !
ΣM G = 0:
 24

FFH  − ( 7.2 m )( 20 kN ) − (1.3 m )( 40 kN ) = 0
 745

( 5.8 m ) 
FFH = 38.432 kN,
ΣFx = 0:
20 kN −
FFH = 38.4 kN C !
13
15
( 38.432 kN − 11.3728 kN ) − FFG = 0
17
745
FFG = 8.0604 kN,
FFG = 8.06 kN T !
Since FEG is in tension, our guess was correct. A negative answer would be impossible, indicating
an incorrect guess.
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 70.
Structure (a):
Non-simple truss with r = 3, m = 16, N = 10,
so 2N > m + r
∴ partially constrained !
Structure (b):
Non-simple truss with r = 3, m = 15, N = 10,
so 2N > m + r
∴ partially constrained !
Structure (c):
Non-simple truss with r = 4, m = 16, N = 10,
so N = 2m + r
Examine Forces:
FHJ = 0, FIJ = P C
By inspection of joint J,
Then, by inspection of joint H,
By inspection of joint I,
FFH = 0,
FGI = 0,
FGH = P C
Iy = 0
Although several members carry no load with the given truss loading, they do
constrain the motion of GH and IJ. Truss ABFG is a simple truss with r = 3, m = 11,
N = 7 (2N = r + m)
so the structure is completely constrained, and determinate. !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 71.
Structure (a):
Structure (b):
Structure (c):
Simple truss with r = 4, m = 16, n = 10
So m + r = 20 = 2n so
completely constrained and determinate !
Compound truss with r = 3, m = 16, n = 10
so
So m + r = 19 < 2n = 20
partially constrained !
Non-simple truss with r = 4, m = 12, n = 8
So m + r = 16 = 2n but must examine further, note that reaction forces
A x and H x are aligned, so no equilibrium equation will resolve them.
Also
consider
∴ Statically indeterminate !
For ΣFy = 0: H y = 0, but then
ΣM A ≠ 0 in FBD Truss,
∴ Improperly constrained !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 72.
Structure (a):
Simple truss (start with ABC and add joints alphabetical to complete
truss), with
r = 4, m = 13, n = 8
r + m = 17 > 2n = 16
so
Structure is completely constrained but indeterminate. !
Structure (b):
From FBD II:
FBD I:
FBD II:
ΣM G = 0 ⇒
Jy
ΣFx = 0 ⇒
Fx
ΣM A = 0 ⇒
Fy
ΣFy = 0 ⇒
Ay
ΣFx = 0 ⇒
Ax
ΣFy = 0 ⇒
Gy
Thus have two simple trusses with all reactions known,
so structure is completely constrained and determinate. !
Structure (c):
Structure has r = 4, m = 13, n = 9
so
r + m = 17 < 2n = 18,
structure is partially constrained !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 73.
Non-Simple truss with r = 4, m = 12, N = 8, so 2N = r + m,
but we must examine further:
Structure (a):
ΣFx = 0: E x = 0
By symmetry, FBD = FDF and FCD = FDG
FBD joint D:
Vertical equilibrium cannot be satisfied with FCD = FDG,
Improperly constrained !
Structure (b):
Non-simple truss with r = 3, m = 10, N = 7,
so 2N = m + r
∴ Partially constrained !
Structure (c):
Non-simple truss with r = 3, m = 13, N = 8,
but we must examine further
so 2N = r + m,
ΣM A = 0: 3a G y − ( a + 3a + 4a ) P = 0
Gy =
Joint H:
ΣFy = 0:
Joint F:
1
FFH − P = 0
2
FFH = P 2 T
ΣFy = 0: FFG − P −
8
P
3
Joint G:
(
)
1
P 2 =0
2
FFG = 2P C
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
ΣFy ≠ 0
Improperly constrained !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 74.
Structure (a):
No. of members
m = 12
No. of joints
n=8
m + r = 16 = 2n
No. of react. comps.
r =4
unks = eqns
FBD of EH:
ΣM H = 0 → FDE ; ΣFx = 0 → FGH ; ΣFy = 0 → H y
Then ABCDGF is a simple truss and all forces can be determined.
This example is completely constrained and determinate. !
Structure (b):
No. of members
m = 12
No. of joints
n=8
m + r = 15 < 2n = 16
No. of react. comps.
r =3
unks < eqns
partially constrained !
Note: Quadrilateral DEHG can collapse with joint D moving downward:
in (a) the roller at F prevents this action.
Structure (c):
No. of members
m = 13
No. of joints
n=8
m + r = 17 > 2n = 16
No. of react. comps.
r =4
unks > eqns
completely constrained but indeterminate !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 75.
Structure (a):
Rigid truss with r = 3, m = 14, n = 8
so
r + m = 17 > 2n = 16
so completely constrained but indeterminate !
Structure (b):
Simple truss (start with ABC and add joints alphabetically), with
r = 3, m = 13, n = 8
so
r + m = 16 = 2n
so completely constrained and determinate !
Structure (c):
Simple truss with r = 3, m = 13, n = 8 so r + m = 16 = 2n, but
horizontal reactions ( Ax and Dx ) are collinear so cannot be resolved by
any equilibrium equation.
∴ structure is improperly constrained !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 76.
\
FBD member AE:
Ax = 120 lb
ΣFx = 0: − Ax + 120 lb = 0,
on ABCD
ΣM A = 0:
A x = 120.0 lb
!
(8 in.)( C ) − ( 2 in.)(120 lb ) = 0
C = 30 lb
ΣFy = 0: 30 lb − Ay = 0,
Ay = 30 lb
A y = 30.0 lb !
on ABCD
FBD member ABCD:
ΣFx = 0: 120 lb − Bx = 0,
ΣM B = 0:
B x = 120.0 lb
!
( 6 in.) D − ( 4 in.)( 30 lb ) − ( 4 in.)( 30 lb )
− ( 2 in.)(120 lb ) = 0
D = 80.0 lb !
ΣFy = 0: 30 lb − By − 30 lb + 80 lb = 0
B y = 80.0 lb !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 77.
FBD member ABC:
Note that BD is a two-force member
(a)
ΣΜ C = 0:
(16 in.) 
4

FBD  − ( 24 in.)( 80 lb ) = 0
5

FBD = 150 lb,
(b)
ΣFy = 0:
ΣFx = 0:
Cy −
Cx −
FBD = 150.0 lb
3
(150 lb ) = 0,
5
C y = 90 lb
4
(150 lb ) + 80 lb = 0,
5
Cx = 40 lb
C = 98.5 lb
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
36.9° W
66.0° W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 78.
FBD Frame:
ΣM A = 0:
( 0.25 m ) Dx − ( 0.95 m )( 480 N ) = 0
D x = 1824 N
W
E x = 1824 N
W
FBD member DF:
Note that BE is a two-force member, Ex = Ey
ΣFx = 0:
−1824 N + Ex = 0,
so
ΣM D = 0:
ΣFy = 0:
E y = 1824 N
W
( 0.50 m )(1824 N ) − ( 0.75 m ) C + ( 0.95 m )( 480 N ) = 0
C = 1824 N
W
D y = 480 N
W
− D y + 1824 N − 1824 N + 480 N = 0
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 79.
FBD Frame:
ΣM A = 0:
( 0.25 m ) Dx − 400 N ⋅ m = 0
D x = 1600 N
E x = 1600 N
∴ E y = 1600 N
FBD member DF:
Note BE is a two-force member, so Ex = Ey
ΣFx = 0: 1600 N − E x = 0,
ΣM D = 0: − ( 0.50 m )(1600 N ) + ( 0.75 m ) C − 400 N ⋅ m = 0
C = 1600 N
ΣFy = 0: Dy − 1600 N + 1600 N = 0
Dy = 0 Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 80.
FBD Ring:
(a)
ΣM A = 0:
(8 in.) ( FBC cos 35° ) − (8 in.)( 6 lb ) = 0
FBC = 7.3246 lb,
(b)
ΣFx = 0:
FBC = 7.32 lb C W
Ax − ( 7.3246 lb ) cos 35° = 0
Ax = 6 lb
ΣFy = 0:
Ay + ( 7.3246 lb ) sin 35° − 6 lb = 0
Ay = 1.79876 lb
A = 6.26 lb
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
16.69° W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 81.
FBD Ring:
(a)
FBC =
(b)
(8 in.) ( FBC cos 20° − 6 lb ) = 0
ΣM A = 0:
6 lb
,
cos 20°
ΣFx = 0:
ΣFy = 0:
FBC = 6.39 lb C W
Ax −
Ay −
6 lb
cos 20° = 0,
cos 20°
Ax = 6 lb
6
sin 20° − 6 lb = 0,
cos 20°
Ay = 8.1838 lb
A = 10.15 lb
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
53.8° W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 82.
FBD Frame:
ΣM A = 0:
 2

180 N  = 0
 2

( 0.2 m ) Bx − ( 0.36 m ) 
Bx = 162 2 N,
Ax = 252 2 N,
ΣM C = 0:
A x = 356 N
W
2
(180 N ) = 0
2
B y = 127.3 N
B y = 90 2 N,
FBD member ABD:
W
2
(180 N ) = 0
2
ΣFx = 0: −162 2 N + Ax −
ΣFy = 0: By −
B x = 229 N
( 0.20 m ) ( 252
)
(
2 N − ( 0.08 m ) 90 2 N
− ( 0.20 m ) D = 0,
W
)
D = 216 2 N
D = 305 N
W
ΣFx = 0: 252 2 N − 162 2 N − C x = 0
Cx = 90 2 N,
C x = 127.3 N
W
ΣFy = 0: 90 2 N + C y − 216 2 N = 0
C y = 126 2 N,
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
C y = 178.2 N
W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 83.
FBD Frame:
ΣM B = 0:
( 0.2 m ) Ax − 60 N ⋅ m = 0
ΣFx = 0: 300 N − Bx = 0,
ΣFy = 0: By = 0,
A x = 300 N
W
B x = 300 N
W
By = 0 W
FBD member ABD:
ΣM C = 0:
ΣFy = 0:
( 0.20 m )( 300 N ) − ( 0.20 m ) Dy
C y − 300 N = 0
ΣFx = 0:
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
=0
D y = 300 N
W
C y = 300 N
W
Cx = 0 W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 84.
(a) FBD AC:
Note: CE is a two-force member
ΣM A = 0:
 FCE 
 FCE 
 + ( 2 in.) 
 − ( 6 in.)( 24 lb ) = 0
 2 
 2 
(8 in.) 
FCE = 14.4 2 lb,
so
E x = 14.40 lb
E y = 14.40 lb
ΣFx = 0:
− Ax + 14.4 lb = 0
A x = 14.40 lb
ΣFy = 0:
Ay − 24 lb + 14.40 lb = 0
A y = 9.60 lb
(b) FBD CE:
!
!
!
!
Note AC is a two-force member
ΣM E = 0:
1
 4

FAC +
FAC  − (1 in.)( 24 lb ) = 0
17
 17

( 3 in.) 
FAC = 1.6 17 lb,
A x = 6.40 lb
A y = 1.600 lb
(
)
(
)
ΣFx = 0: E x −
4
1.6 17 lb = 0,
17
ΣFy = 0:
1
1.6 17 lb − 24 lb = 0
17
Ey +
E x = 6.40 lb
E y = 22.4 lb
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
!
!
!
!
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 85.
(a) FBD AC:
Note: CE is a two-force member
ΣM A = 0:
1
 1

FCE +
FCE  − ( 0.3 m )( 320 lb )
2
 2

( 0.4 m ) 
E x = 120.0 N
FCE = 120 2 N,
E y = 120.0 N
(b) FBD CE:
(
)
1
120 2 N = 0,
2
ΣFx = 0:
− Ax +
ΣFy = 0:
Ay − 320 N +
(
A x = 120.0 N
)
1
120 2 N = 0
2
A y = 200 N
!
!
!
!
Note: AC is a two-force member
ΣM E = 0:
1
 1

FAC +
FAC  − ( 0.15 m )( 320 N ) = 0
2
 2

( 0.25 m ) 
FAC = 96 2 N,
A x = 96.0 N
A y = 96.0 N
ΣFx = 0: Ex − 96.0 N = 0
ΣFy = 0:
E y − 320 N + 96.0 N = 0
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
E x = 96.0 N
E y = 224 N
!
!
!
!
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 86.
(a) FBD AC:
Note: CE is a two-force member
ΣM A = 0:
 1

 1

− ( 8 in.) 
FCE  − ( 2 in.) 
FCE  + 192 lb ⋅ in. = 0
 2

 2

E x = 19.20 lb
FCE = 19.2 2 lb,
E y = 19.20 lb
(b) FBD CE:
ΣFx = 0:
Ax − 19.2 lb = 0,
ΣFy = 0:
Ay − 19.2 lb = 0,
A x = 19.20 lb
A y = 19.20 lb
!
!
!
!
Note: AC is a two-force member
ΣM E = 0:
4
 1

FAE +
FAE  + 192 lb ⋅ in. = 0
17
 17

( 3 in.) 
FAE = −12.8 17 lb,
Ax =
4
FAE ,
17
Ay =
1
FAE ,
17
ΣFx = 0: Ex − 51.2 lb = 0,
ΣFy = 0:
E y − 12.80 lb = 0,
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
A x = 51.2 lb
A y = 12.80 lb
E x = 51.2 lb
E y = 12.80 lb
!
!
!
!
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 87.
(a) FBD AC:
Note: CE is a two-force member
ΣM A = 0:
 2

FCE  = 0,
120 N ⋅ m − ( 0.4 m ) 
2


FCE = 150 2 N
E x = 150.0 N
E y = 150.0 N
(b) FBD CE:
ΣFx = 0:
Ax − 150.0 N = 0,
ΣFy = 0:
Ay − 150.0 N = 0,
A x = 150.0 N
A y = 150.0 N
!
!
!
!
Note: AC is a two-force member
 2

ΣM E = 0: − ( 0.25 m ) 
FAC  + 120 N ⋅ m = 0
 2

FAC = 240 2 N,
A x = 240 N
A y = 240 N
ΣFx = 0: 240 N − Ex = 0,
ΣFy = 0:
E y − 240 N = 0,
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
E x = 240 N
E y = 240 N
!
!
!
!
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 88.
FBD Frame:
Note: In analysis of entire frame, location of M is immaterial. Note also
that AB, BC, and FG are two force members.
ΣM H = 0:
( 6 ft ) I y
− 180 lb ⋅ ft = 0,
ΣFy = 0:
30 lb −
5
FAB = 0,
13
ΣM C = 0:
( 2.5 ft ) 
I y = 30.0 lb
!
(a) FBD AI:
12

78 lb − FFG  = 0,
 13

ΣFx = 0: FBC −
12
78 lb − 72 lb = 0,
13
FAB = 78.0 lb
22.6° !
FGH = 72.0 lb
!
FBC = 144.0 lb
!
(b) FBD AI:
FAB = 78.0 lb
As above, ΣFy = 0 yields
Then:
 12

ΣM C = 0: ( 2.5 ft )  78.0 lb − FFG  − 180 lb ⋅ ft = 0
 13

22.6° !
FFG = 0 !
ΣFx = 0:
FBC −
12
( 78.0 lb ) = 0,
13
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
FBC = 72.0 lb
!
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 89.
(a) FBD ACF:
Note: BC is a two-force member
ΣM F = 0:
 1

FBC  = 0,
 26

( 0.1 m )(120 N ) − ( 0.3 m ) 
B x = 200 N
FBC = 40 26 N,
B y = 40.0 N
ΣFx = 0:
(b) FBD BCE:
ΣFy = 0:
(
)
5
40 26 N − Fx = 0,
26
(
Fx = 200 N
)
1
40 26 N = 0,
26
Fy − 120 N −
Fy = 160.0 N
!
!
!
!
Note ACF is a two-force member
ΣM B = 0:
 1

FCF  = 0,
 5

( 0.4 m )(120 N ) − ( 0.3 m ) 
FCF = 160 5 N,
Fx = 320 N
Fy = 160.0 N
ΣFx = 0: Bx −
ΣFy = 0:
(
(
)
2
160 5 N = 0,
5
B x = 320 N
!
!
!
)
1
160 5 N − 120 N − B y = 0
5
B y = 40.0 N
!
(c) Moving the 120 N force from D to E does not affect the reactions. The answers are the same as in part (b). !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 90.
(a) FBD AF:
Note: BC is a two-force member
ΣM F = 0:
 1

FBC  − 48 N ⋅ m = 0,
 26

( 0.3 m ) 
FBC = 160 26 N
so
B x = 800 N
B y = 160.0 N
(b & c) FBD BCE:
ΣFx = 0:
Fx − 800 N = 0,
ΣFy = 0:
− Fy + 160 N = 0,
Fx = 800 N
Fy = 160.0 N
!
!
!
!
Note: ACF is a two-force member, and that the application point of the
couple is immaterial.
ΣM B = 0:
 1

FCF  − 48 N ⋅ m = 0
 5

( 0.3 m ) 
FCF = 160 5 N,
Fx = 320 N
Fy = 160.0 N
ΣFx = 0: 320 N − Bx = 0,
ΣFy = 0:
−160 N + By = 0,
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
B x = 320 N
B y = 160.0 N
!
!
!
!
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 91.
\
First note that, when a cable or cord passes over a frictionless, motionless pulley, the tension is unchanged.
ΣM C = 0:
rT1 − rT2 = 0
(a) Replace each force with an equivalent force-couple.
(b) Cut cable and replace forces on pulley with equivalent pair of forces at A as above.
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
T1 = T2
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 92.
FBD Pipe 2:
ΣFx′ = 0:
NG −
3
( 220 lb ) = 0,
5
NG = 132 lb
ΣFy′ = 0:
ND −
4
( 220 lb ) = 0,
5
N D = 176 lb
FBD Pipe 1:
θ = 90° − 2 tan −1
ΣFx′ = 0:
3
= 16.2602°
4
N F cos (16.2602° ) − 132 lb −
3
( 220 lb ) = 0
5
N F = 275.00 lb
ΣFy′ = 0:
N C + ( 275 lb ) cos (16.2602° ) −
4
( 220 lb ) = 0
5
NC = 99.00 lb
FBD Frame & Pipes:
ΣM A = 0:


4
5
( 48 in.) E y − 24 in. + 24 in. + (10 in.)  (220 lb) = 0


E y = 256
ΣFy = 0:
ΣFx = 0:
2
lb,
3
Ay − 2 ( 220 lb ) + 256
Ay = 183

1
lb
3
Ax − E x = 0,
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
E y = 257 lb
A y = 183.3 lb
2
lb = 0
3
Ax = E x
COSMOS: Complete Online Solutions Manual Organization System
FBD BE:
ΣM B = 0:
( 24 in.)  256

2 
 15 
lb  − (18 in.) E x +  in.  ( 275 lb ) = 0
3 
 4

E x = 399.5 lb,
E x = 400 lb
From above,
A x = 400 lb
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 93.
FBD Pipe 2:
FBD Pipe 1:
ΣFx' = 0:
NG −
3
( 220 lb ) = 0,
5
NG = 132 lb
ΣFy′ = 0:
ND −
4
( 220 lb ) = 0,
5
N D = 176 lb
ΣFx′ = 0:
ΣFy′ = 0:
ΣM B = 0:
FBD BE:
NF −
NC −
3
( 220 lb ) − 132 lb = 0,
5
4
( 220 lb ) = 0,
5
N F = 264 lb
NC = 176 lb
( 24 in.) E y − ( 32 in.) Ex + ( 5 in.) ( 264 lb ) = 0
4Ex − 3E y = 165
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
(1)
COSMOS: Complete Online Solutions Manual Organization System
FBD Frame with Pipes:
ΣM A = 0: ( 48 in.) E y − (14 in.) Ex + ( 5 in.) ( 264 lb )
− ( 35 in. + 45 in.) (176 lb ) = 0
(2)
solving with (1) above:
E y = 355
ΣFx = 0:
2
lb,
3
Ax − 308 lb −
E y = 356 lb
4
3
( 264 lb ) + ⋅ 2 (176 lb ) = 0
5
5
A x = 308 lb
ΣFy = 0:
Ay + 355
!
!
2
3
4
lb − ( 264 lb ) − ⋅ 2 (176 lb ) = 0
3
5
5
A y = 84.3 lb
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
!
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 94.
FBD AC:
ΣM c = 0:
( 0.6 m ) Ax − ( 0.06 m )(170 N ) = 0,
A x = 17.00 N
ΣFx = 0:
FBD CE:
ΣFx = 0:
−17 N +
133 N −
15
(170 N ) − Cx = 0,
17
Cx = 133 N
15
(170 N ) + Ex = 0,
17
E x = 17.00 N
ΣM C = 0:
W
W
( 0.48 m )(17.00 N ) + ( 0.90 m ) E y
− ( 0.45 m ) (170 N ) = 0,
E y = 75.9 N W
FBD Frame:
ΣFy = 0:
Ay − 170 N + 75.9 N = 0,
A y = 94.1 N W
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 95.
FBD Frame & pulley:
ΣM B = 0:
( 0.6 m ) Ay − ( 0.075 m )( 240 N ) = 0,
A y = 30.0 N W
ΣFy = 0:
− 30.0 N + By − 240 N = 0,
B y = 270 N W
ΣFx = 0:
FBD AD:
ΣM D = 0:
From above
− Ax + Bx = 0,
Ax = Bx
( 0.200 m ) Ax − ( 0.075 m ) ( 240 N )
+ ( 0.30 m ) ( 30 N ) = 0
Ax = Bx ,
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
A x = 45.0 N
W
B x = 45.0 N
W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 96.
(a)
FBD Entire machine:
ΣM A = 0:
( 95 in.) (16 kips) − ( 35 in.)(18 kips ) + (145 in.) ( 2B )
− (170 in.) ( 50 kips ) = 0,
B = 26.241 kips
ΣFy = 0:
B = 26.2 kips
W
−16 kips + 2 A − 18 kips + 2 ( 26.241 kips ) − 50 kips = 0,
A = 15.76 kips
W
(b)
FBD Motor unit:
ΣM D = 0:
( 30 in.) Cx + (85 in.)( 52.482 kips ) − (110 in.)( 50 kips ) = 0
Cx = 34.634 kips,
C x = 34.6 kips
W
ΣFx = 0:
Dx − 34.634 kips = 0,
D x = 34.6 kips
W
ΣFy = 0:
52.482 kips − 50 kips − Dy = 0,
D y = 2.48 kips
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 97.
(a)
FBD Entire Machine:
(145 in.) (2B) − ( 35 in.)(18 kips )
ΣM A = 0:
− (170 in.)( 50 kips ) = 0
2B = 62.966 kips,
ΣFy = 0:
B = 31.5 kips
!
2 A + 62.966 kips − 18 kips − 50 kips = 0
2 A = 5.034 kips,
A = 2.52 kips
!
where A and B are single
wheel forces
(b)
ΣM D = 0:
FBD Motor unit:
(85 in.)( 62.966 kips ) − (110 in.)( 50 kips )
+ (30 in.) Cx = 0,
Cx = 4.9297 kips
C x = 4.93 kips
ΣFy = 0:
− Dy − 50 kips + 62.966 kips = 0
D y = 12.97 kips
ΣFx = 0:
!
!
Dx − C x = 0, Dx = Cx ,
D x = 4.93 kips
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
!
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 98.
FBD Frame:
( 0.625 m ) F − ( 0.75 m )( 4 kN ) − (1.25 m )( 3 kN ) = 0
ΣM A = 0:
Fx = 10.8 kN
ΣFx = 0:
Ax − 10.8 kN = 0,
ΣFy = 0:
Ay − 4 kN − 3 kN = 0,
FBD BF: (I)
A x = 10.80 kN
W
A y = 7.00 kN W
FBD ABD: (II)
I:
ΣM C = 0:
( 0.375 m )(10.8 kN ) − ( 0.25 m ) Bx = 0,
II:
ΣM D = 0:
( 0.25 m )(10.8 kN + 16.2 kN ) + ( 0.5 m ) By − (1.00 m )( 7.0 kN ) = 0,
Bx = 16.2 kN,
By = 0.5 kN,
ΣFx = 0:
−10.8 kN − 16.20 kN + Dx = 0,
Dx = 27 kN,
ΣFy = 0:
7.0 kN − 0.5 kN − Dy = 0,
Dy = 6.5 kN,
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
B x = 16.20 kN
B y = 500 N
D x = 27.0 kN
W
W
W
D y = 6.50 kN W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 99.
FBD Frame:
( 0.3 m ) Dx = − ( 0.4 m ) ( 900 N ) = 0,
ΣM A = 0:
Dx = 1200 N
D x = 1.200 kN
ΣFx = 0: 1.200 kN − Ax = 0,
ΣFy = 0:
Ax = 1.200 kN,
A x = 1.200 kN
Ay − 900 N = 0
A y = 900 N
Note: AG is a two-force member.
( 0.2 m )  By − ( 0.4 m ) ( 900 N ) = 0,
ΣM A = 0:
FBD AC:
ΣFy = 0:
900 N + 1800 N − 900 N −
( FAGx = 2400 N,
FAG = 3000 N
ΣFx = 0:
−
ΣM E = 0:
ΣFx = 0:
FBD DF:
ΣFy = 0:
3
FAG = 0
5
FAGy = 1800 N
4
( 3000 N ) − 1200 N + Bx = 0,
5
On GBEH:
By = 1800 N
G x = 2.40 kN
,
B x = 3.60 kN
,
)
Bx = 3600 N
G y = 1.800 kN !
B y = 1.800 kN
3
FDH = 0,
5
1200 N − Ex = 0
− ( 0.2 m )
!
FDH = 0
Ex = 1200 N
Ey = 0
On GBEH:
E x = 1.200 kN
H x = 0,
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
,
Ey = 0 !
Hy = 0 !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 100.
FBD Frame:
( 6 in.) Dx − (13.5 in.) ( 48 lb ) − (16.5 in.) ( 20 lb ) = 0
ΣM A = 0:
Dx = 163 lb
ΣFx = 0:
− A x + 20 lb + 48 lb + 163 lb = 0
Ax = 231 lb
ΣFy = 0:
Ay = 0
FBD DE:
ΣM E = 0:
(19.5 in.)(163 lb ) − ( 6 in.) Bx = 0
Bx = 529.75 lb
ΣM C = 0:
( 4.5 in.) By − ( 7.5 in.)( 231 lb ) = 0,
B x = 530 lb
On ABC:
ΣFx = 0:
FBD ABC:
ΣFy = 0:
On ABC:
By = 385 lb
,
− 231 lb + 529.75 lb − Cx = 0,
C y − 385 lb = 0,
B y = 385 lb W
Cx = 298.75 lb
C y = 385 lb
C x = 299 lb
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
,
C y = 385 lb W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 101.
FBD Frame:
ΣM E = 0:
( 6 in.)(84 lb ) − ( 42 in.) Ay = 0
Ay = 12.00 lb
FBD CF: (I)
I:
II:
II:
ΣM D = 0:
ΣM B = 0:
ΣFx = 0:
ΣFy = 0:
FBD ABC: (II)
( 3.5 in.) Cx + ( 7 in.) C y − ( 5 in.) (84 lb ) = 0,
Cx + 2C y = 120 lb
(12 in.) C y − ( 3.5 in.) Cx − (12 in.) (12 lb ) = 0,
− Bx + 60.632 lb = 0,
12 lb + By + 29.684 lb = 0,
3.5 Cx − 12 C y = 144 lb
Solving:
Cx = 60.632 lb,
C y = 29.684 lb
On ABC:
C x = 60.6 lb,
C y = 29.7 lb W
Bx = 60.632 lb,
By = − 41.684 lb,
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
B x = 60.6 lb
W
B y = 41.7 lb W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 102.
FBD Stool:

w 1
N
= ( 56 kg )  9.81
 = 274.68 N
2 2
kg 

ΣM A = 0:
( 0.350 m ) By − ( 0.150 m )( 274.68 N ) = 0,
By = 117.72 N
FBD BE:
ΣM E = 0:
 1

FFG  = 0
 2

(1.25 m )(117.72 N ) − ( 0.15 m ) 
FFG = 98.1 2 N
ΣFx = 0:
ΣFy = 0:
Ex −
1
(98.1 2) N = 0,
2
− Ey +
E x = 98.1 N
W
1
(98.1 2 N) + 117.72 N = 0
2
E y = 215.82 N,
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
E y = 216 N
W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 103.
FBD BC:
Note that only one of the forces B y , B′y exists.
(
)
M C B y or B′y − ( 0.15 m )
ΣM C = 0:

2 
FFG = 
M B or B′y
 0.15 m  C y


(
(
So we seek a maximum of M C By or B′y
1
FFG = 0
2
)
(1)
)
( )
M C ( B′y ) = ( 0.175 m ) B′y
M C By = ( 0.125 m ) By
where
FBD stool:
w = ( 56 kg ) ( 9.81 N/kg ) = 549.36 N
consider contact at A′, A y = 0, either
ΣM A′ = ( 0.425 m ) By − ( 0.225 m )
w
=0
2
( )
M C By = ( 0.03309 m ) w
By = 0.26471 w,
or ( M A′ ) = ( 0.475 m ) B′y − ( 0.225 m )
(i)
w
=0
2
( )
B′y = ( 0.23684 ) w, M C B′y = ( 0.04145 m ) w
(ii)
consider contact at A, A′y = 0, either
ΣM A = 0:
( 0.35 m ) By − ( 0.15 m )
w
=0
2
( )
By = 0.21429 w, M C By = ( 0.02678 m ) w
or
ΣM A = 0:
B′y = 0.1875 w,
( 0.40 m )
B′y − ( 0.15 m )
(iii)
w
=0
2
( )
M C B′y = ( 0.03281 m ) w
(iv)
(a) The maximum is (ii), with contact at A′ and B′, and
!

2 
(b) FFG max = 
 ( 0.04145 m ) ( 549.36 N ) = 214.69 N,
 0.15 m 
FFG max = 215 N T !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 104.
Members FBDs:
I:
ΣM A = 0:
(12.8 ft ) Bx − ( 32 ft ) By − ( 20 ft )(14 kips ) = 0
II:
ΣM C = 0:
( 7.2 ft ) Bx + ( 24 ft ) By − (12 ft )( 21 kips ) = 0
Solving: Bx = 27.5 kips, By = 2.25 kips,
I:
ΣFx = 0:
A x = 27.5 kips
ΣFy = 0:
Ax − 27.5 kips = 0,
Ax = 27.5 kips,
(a)
!
Ay − 14 kips − 2.25 kips = 0, Ay = 16.25 kips,
A y = 16.25 kips
(b)
B x = 27.5 kips
B y = 2.25 kips
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
!
!
!
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 105.
Member FBDs:
II:
ΣM C = 0:
( 7.2 ft ) Bx − ( 24 ft ) By − (12 ft )(14 kips ) = 0
I:
ΣM A = 0:
(12.8 ft ) Bx + ( 32 ft ) By − ( 20 ft )( 21 kips ) = 0
Solving: Bx = 28.75 kips, By = 1.675 kips,
I:
ΣFx = 0:
A x = 28.8 kips
ΣFy = 0:
Ax − 28.75 kips = 0,
Ax = 28.75 kips,
(a)
!
Ay − 21 kips + 1.625 kips, Ay = 19.375 kips ,
A y = 19.38 kips
(b)
B x = 28.8 kips
B y = 1.625 kips
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
!
!
!
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 106.
\
FBD Frame:
ΣM A = 0:
(12 in.) Dx − ( 28 in.)( 411 lb ) = 0
D x = 959 lb
FBD DF:
Note that BF and CE are two-force members.
ΣFx = 0:
ΣM D = 0:
Solving:
959 lb +
4
15
FCE −
FBF = 0
5
17
(12 in.) 
3

 8

FCE  + ( 34.5 in.) 
FBF  = 0
5

 17

FBF = 357 lb,
ΣM E = 01:
( 22.5 in.)  ( 357 lb ) − (12 in.) Dy = 0
D y = 315 lb ,
8
17
so

(a)
(b)
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
D = 1009 lb
18.18°
FBF = 357 lb T
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 107.
FBD Frame:
(12 in.) Dx − ( 34.5 in.)( 274 lb ) = 0
ΣM A = 0:
Dx = 787.75 lb
FBD DF:
ΣM D = 0:
ΣFx = 0:
Solving:
ΣM E = 0:
(12 in.) 
3

 8

FCE  + ( 34.5 in.) 
FBF − 274 lb  = 0
5
17




787.75 lb +
4
15
FCE −
FBF = 0
5
17
FBF = 684.25 lb,
( 22.5 in.)  ( 684.25 lb ) − 274 lb  − (12 in.) Dy
Dy = 90 lb,
8
17

so
(a)
(b)
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
D = 793 lb
=0
6.52°
FBF = 684 lb T
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 108.
FBD AC:
Note that BE is a two force member
ΣFy = 0:
1
FBE − P = 0
5
FBE = 5 P
ΣΜ Α = 0:
 1
 5
( 0.4 m ) 
(
)

 2
5 P  + ( 0.1 m ) 

 5
(
)

5P 

− a P + ( 0.3 m ) C = 0
so
( a − 0.6 m ) P = ( 0.3 m ) C
since P > 0 and C ≥ 0, a – 0.6 m ≥ 0
a ≥ 0.6 m
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 109.
Member FBDs:
ΣΜ Α = 0:
( 0.6 m )
4
4

FCF − ( 0.9 m )  FDG  = 0
5
5


4
4
ΣΜ Ε = 0: − ( 0.3 m ) FCF + ( 0.6 m ) FDG − ( 0.9 m )( 800 N ) = 0
5
5
Solving:
FCF = 9000 N,
FDG = 6000 N
FCF = 9.00 kN C
FDG = 6.00 kN T
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 110.
Member FBDs:
4
FDG = 0
5
ΣM A = 0:
( 0.3 m ) FBF − ( 0.9 m )
ΣM E = 0:
4
− ( 0.3 m ) FBF + ( 0.6 m ) FDG − ( 0.9 m )( 0.8 kN ) = 0
5
Solving:
FBE = − 3 kN,
FBF = − 7.2 kN
so
FBF = 7.20 kN T
FDG = 3.00 kN C
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 111.
Member FBDs:
ΣM A = 0:
ΣM E = 0:
Solving:
( 0.6 m )
4
4
FCH − ( 0.3 m ) FBG = 0
5
5
4
4
− ( 0.9 m ) FCH + ( 0.6 m ) FBG − ( 0.9 m )( 0.8 kN ) = 0
5
5
FBG = 6 kN, FCH = 3 kN
so
FBG = 6.00 kN T
FCH = 3.00 kN C
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 112.
Member FBDs:
I:
ΣM J = 0:
2a C y + a Cx − a P = 0
Solving:
I:
II:
Cx =
II:
ΣM K = 0: 2a C y − a C x = 0
P
P
, Cy =
2
4
ΣFx = 0:
1
FBG − Cx = 0,
2
ΣFy = 0:
FAF +
ΣFx = 0:
Cx −
ΣFy = 0:
−
1  2 
P
P  − P + = 0,

4
2  2 
1
FDG = 0,
2
2
P,
2
FBG = 2C x =
FAF =
FDG = 2C x =
P
1  2 
P  + FEH = 0,
+

4
2  2 
FBG = 0.707 P T
P
,
4
FAF = 0.250 P T
2
P,
2
FDG = 0.707 P T
P
4
FEH = 0.250 P C
FEH = −
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 113.
Note that, if we assume P is applied to BF, each individual member FBD
looks like:
Fshort = 2Flong =
so
2
Fload
3
(by moment equations about S and L).
Labeling each interaction force with the letter corresponding to the joint
of application, we have:
F =
2(P + E)
3
= 2B
C
D
=
3
2
B
A
C =
=
3
2
E =
so
2(P + E)
= 2 B = 6C = 18E
3
P + E = 27 E
P
26
E=
so D = 2E =
A = 2C = 3E
F =
2
P
P +

3
26 
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
A=
F=
P
13
3P
13
9P
13
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 114.
Note that, if we assume P is applied to EG, each individual member FBD
looks like
so
2Fleft = 2 Fright = Fmiddle
Labeling each interaction force with the letter corresponding to the joint
of its application, we see that
B = 2 A = 2F
C = 2B = 2D
G = 2C = 2H
P + F = 2G ( = 4C = 8B = 16 F ) = 2E
From
P + F = 16 F , F =
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
P
15
so A =
P
15
D=
2P
15
H=
4P
15
E=
8P
15
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 115.
(a) Member FBDs:
FBD II:
ΣFy = 0:
FBD I:
ΣM A = 0:
By = 0
ΣM B = 0: aF2 = 0
aF2 − 2aP = 0
F2 = 0
but F2 = 0
so P = 0
not rigid for P ≠ 0 !
(b) Member FBDs:
(
)
Note: 7 unknowns Ax , Ay , Bx , By , F1, F2 , C but only 6 independent equations.
System is statically indeterminate !
System is, however, rigid !
(c) FBD whole:
FBD right:
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
FBD I:
ΣM A = 0: 5aBy − 2aP = 0
ΣFy = 0: Ay − P +
FBD II:
FBD I:
ΣM c = 0:
2
P=0
5
By =
Ay =
a
5a
Bx −
By = 0
2
2
ΣFx = 0: Ax + Bx = 0
2
P
5
3
P
5
Bx = 5By
Ax = − Bx
B x = 2P
A x = 2P
A = 2.09P
16.70° !
B = 2.04P
11.31° !
System is rigid !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 116.
Note: In all three cases, the right member has only three forces acting, two of which are parallel. Thus the
third force, at B, must be parallel to the link forces.
(a) FBD whole:
ΣM A = 0:
− 2aP −
a 4
1
B + 5a
B=0
4 17
17
4
B=0
17
ΣFx = 0:
Ax −
ΣFy = 0:
Ay − P +
1
B=0
17
B = 2.06 P
B = 2.06P
14.04° !
A = 2.06P
14.04° !
A x = 2P
Ay =
P
2
rigid !
(b) FBD whole:
Since B passes through A,
ΣM A = 2aP = 0 only if P = 0
∴ no equilibrium if P ≠ 0
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
not rigid !
COSMOS: Complete Online Solutions Manual Organization System
(c) FBD whole:
ΣM A = 0: 5a
1
3a 4
B+
B − 2aP = 0
4 17
17
ΣFx = 0: Ax +
4
B=0
17
ΣFy = 0: Ay − P +
1
B=0
17
B=
17
P
4
B = 1.031P
14.04° !
A = 1.250P
36.9° !
Ax = − P
Ay = P −
P 3P
=
4
4
System is rigid !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 117.
(a) Member FBDs:
FBD I:
ΣM A = 0: aF1 − 2aP = 0
FBD II:
ΣM B = 0: − aF2 = 0
F1 = 2P;
ΣFy = 0: Ay − P = 0
F2 = 0
ΣFx = 0: Bx + F1 = 0, Bx = − F1 = −2P
B x = 2P
ΣFy = 0: By = 0
FBD I:
Ay = P
ΣFx = 0: − Ax − F1 + F2 = 0
Ax = F2 − F1 = 0 − 2P
so B = 2P
!
A x = 2P
so A = 2.24P
26.6° !
Frame is rigid !
(b) FBD left:
FBD I:
FBD II:
FBD whole:
ΣM E = 0:
a
a
5a
P + Ax −
Ay = 0
2
2
2
ΣM B = 0: 3aP + aAx − 5aAy = 0
Ax − 5 Ay = − P
Ax − 5 Ay = −3P
This is impossible unless P = 0 ∴ not rigid !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
(c) Member FBDs:
FBD I: ΣFy = 0: A − P = 0
FBD II:
ΣM D = 0: aF1 − 2aA = 0
F1 = 2 P
ΣFx = 0: F2 − F1 = 0
F2 = 2 P
ΣM B = 0: 2aC − aF1 = 0
C =
ΣFx = 0: F1 − F2 + Bx = 0
ΣFx = 0: By + C = 0
F1
= P
2
A= P
!
C= P
!
Bx = P − P = 0
By = −C = − P
B= P !
Frame is rigid !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 118.
FBD ABC:
ΣM C = 0: 0.045 m + ( 0.30 m ) sin 30° ( 400 N ) sin 30° 
+ 0.030 m + ( 0.30 m ) cos 30°  ( 400 N ) cos 30°
 12

 5

− ( 0.03 m )  FBD  − ( 0.045 m )  FBD  = 0
13
13




FBD = 3097.64 N
ΣFx = 0:
5
( 3097.64 N ) − ( 400 N ) sin 30° − Cx = 0
13
C x = 991.39 N
ΣFy = 0:
12
( 3097.64 N ) − ( 400 N ) cos 30° − C y = 0
13
C y = 2512.9 N
FBD Blade:
(a)
Vertical component at D =
12
( 3097.64 N )
13
= 2.86 kN
(b)
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
C = 2.70 kN
68.5°
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 119.
FBD Blade:
ΣFy = 0:
FBD ABC:
ΣM C = 0:
3 kN −
12
FBD = 0: FBD = 3.25 kN C
13
0.045 m + ( 0.3 m ) sin 30°  P sin 30°
+ 0.030 m + ( 0.3 m ) cos 30° P cos 30°
− ( 0.045 m )
− ( 0.03 m )
5
( 3.25 kN )
13
12
( 3.25 kN ) = 0
13
P = 0.41968 kN
so
ΣFx = 0:
P = 420 N
(a)
− ( 0.41968 kN ) sin 30° +
5
( 3.25 kN ) − Cx = 0
13
C x = 1.04016 kN
ΣFy = 0:
− ( 0.41968 kN ) cos30° +
12
( 3.25 kN ) − C y = 0
13
C y = 2.6365 kN
so
(b)
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
C = 2.83 kN
68.5°
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 120.
FBD Stamp D:
ΣFy = 0:
E − FBD cos 20° = 0,
ΣM A = 0:
FBD ABC:
E = FBD cos 20°
( 0.2 m )( sin 30°)( FBD cos 20°)
+ ( 0.2 m )( cos 30° )( FBD sin 20° )
− ( 0.2 m ) sin 30° + ( 0.4 m ) cos15°  ( 250 N ) = 0
FBD = 793.64 N C
and, from above,
E = ( 793.64 N ) cos 20°
E = 746 N
(a)
ΣFx = 0:
!
Ax − ( 793.64 N ) sin 20° = 0
A x = 271.44 N
ΣFy = 0:
Ay + ( 793.64 N ) cos 20° − 250 N = 0
A y = 495.78 N
so
(b )
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
A = 565 N
61.3° !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 121.
\
FBD Stamp D:
ΣFy = 0:
900 N − FBD cos 20° = 0,
FBD = 957.76 N C
(a)
FBD ABC:
ΣM A = 0:
( 0.2 m )( sin 30° )  ( 957.76 N ) cos 20°
+ ( 0.2 m )( cos 30° )  ( 957.76 N ) sin 20°
− ( 0.2 m ) sin 30° + ( 0.4 m ) cos15° P = 0
P = 301.70 N,
P = 302 N
(b)
ΣFx = 0:
Ax − ( 957.76 N ) sin 20° = 0
A x = 327.57 N
ΣFy = 0:
− Ay + ( 957.76 N ) cos 20° − 301.70 N = 0
A y = 598.30 N
so
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
A = 682 N
61.3°
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 122.
FBD ABC:
( 45 mm ) sin 25° − 10 mm 
θ = sin −1 
ΣM C = 0:
100 mm
= 5.1739°
( 70 mm )(110 N ) − ( 45 mm ) sin 25° FBD cos 5.1739°
+ ( 45 mm ) cos 25° FBD sin 5.1739° = 0
FBD = 504.50 N
ΣFx = 0:
( 504.50 N ) cos 5.1739° − (110 N ) sin 25° − Cx = 0,
Cx = 455.96 N
FBD CE:
ΣFx = 0:
455.96 N − Q = 0
Q = 455.96 N,
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
(a)
Q = 456 N
(b)
FBD = 540 N T
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 123.
Member FBDs:
θ = tan −1
= tan −1
( BC + CD ) sin 60°
AF + ( AB + BC − CD ) cos 60°
( 2.5 in. + 1 in.) sin 60°
4.5 in. + (1.5 in. + 2.5 in. − 1 in.) cos 60°
θ = 26.802°
From FBD CDE:
ΣM C = 0:
( 7.5 in.)( 20 lb ) − (1 in.) FDF cos ( 30° − 26.802°) = 0,
FDF = 150.234 lb C
ΣFx = 0:
(150.234 lb ) cos ( 26.802°) − ( 20 lb ) sin 60° − Cx = 0,
Cx = 116.774 lb
ΣFy = 0:
(150.234 lb ) sin 26.802° − ( 20 lb ) cos 60° − C y
=0
C y = 57.742 lb
From FBD ABC:
ΣM A = 0:
(1.5 in.) sin 60° Bx + (1.5 in.) cos 60° By − ( 4 in.) sin 60°  (116.774 lb )
+ ( 4 in.) cos 60° ( 57.742 lb ) = 0,
3 Bx + By = 385.37 lb
continued
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
(1)
COSMOS: Complete Online Solutions Manual Organization System
From FBD BFG:
ΣM G = 0:
− (1.5 in.) (150.234 lb ) sin 26.802° + (1.5 in.) cos 30°  (150.234 lb ) cos 26.802°
− (1.5 in.) cos 30° Bx − 6 in. + (1.5 in.) sin 30°  B y = 0,
Bx = 243.32 lb,
Solving (1) and (2):
ΣFx = 0:
ΣFy = 0:
3Bx − 9 B y = 96.775 lb
By = − 36.075 lb
243.32 lb − (150.234 lb ) cos 26.802° − Gx = 0,
G x = 109.226 lb
G y = 31.667 lb
G y − (150.234 lb ) sin 26.802° + 36.075 lb = 0,
On G,
G = 113.7 lb
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
16.17° !
(2)
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 124.
Member FBDs:
θ = tan −1
= tan −1
( BC + CD ) sin 60°
AF + AB cos 75° + ( BC − CD ) cos 60°
( 2.5 in. + 1 in.) sin 60°
4.5 in. + (1.5 in.) cos 75° + ( 2.5 in. − 1 in.)  cos 60°
θ = 28.262°
From FBD CDE:
ΣM C = 0:
( 7.5 in.)( 20 lb ) − (1 in.) FDF cos ( 30° − 28.262°) = 0
FDF = 150.069 lb
ΣFx = 0:
(150.069 lb ) cos 28.262° − ( 20 lb ) sin 60° − Cx = 0
Cx = 114.859 lb
ΣFy = 0:
(150.069 lb ) sin 28.262° − ( 20 lb ) cos 60° − C y
=0
C y = 61.058 lb
From FBD ABC:
ΣM A = 0:
(1.5 in.) sin 75° Bx − (1.5 in.) cos 75° By − (1.5 in.) sin 75° + ( 2.5 in.) sin 60° (114.859 lb )
+ (1.5 in.) cos 75° + ( 2.5 in.) cos 60° ( 61.058 lb ) = 0
or 1.44889 Bx − 0.38823 B y = 315.07 lb
(1)
continued
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
From FBD BFG:
ΣM G = 0:
− (1.5 in.) (150.069 lb ) sin 28.262° + (1.5 in.) cos15° (150.069 lb ) cos 28.262° 
− (1.5 in.) cos15° Bx + ( 6 in.) + (1.5 in.) sin15°  B y = 0
or 1.44889 Bx − 6.3882 B y = 84.926 lb
Solving (1) and (2):
Bx = 227.74 lb,
(2)
By = 38.358 lb
ΣFx = 0:
227.74 lb − (150.069 lb ) cos 28.262° − Gx = 0,
G x = 95.560 lb
ΣFy = 0:
G y − (150.069 lb ) sin 28.262° + 38.358 lb = 0,
G y = 32.700 lb
So force on G is
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
G = 101.8 lb
18.89° !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 125.
FBD Piston:
ΣFy = 0:
180 lb − FCE cos 20° = 0,
FCE = 191.552 lb C
FBD Joint C:
FBC
191.552°
=
sin 40°
sin 40°
FBC = 191.552 lb
FBD AB:
ΣM A = 0:
( 6 in.)(191.552 lb ) cos 30° − M
=0
M = 995 lb ⋅ in.
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 126.
FBD AB:
ΣM A = 0:
( 6 in.) FBC cos 30° − 660 lb ⋅ in. = 0,
FBC = 127.017 lb T
FBD Piston:
180 lb
C
cos φ
ΣFy = 0:
180 lb − FCE cos φ = 0,
FCE =
ΣFy = 0:
180 lb
cos φ + (127.017 lb ) sin 30°
cos φ
− FCD cos φ = 0,
FBD Joint C:
FCD cos φ = 243.51 lb
ΣFx = 0:
(1)
(127.017 lb ) cos30° −
180 lb
sin φ − FCD sin φ = 0
cos φ
FCD sin φ = 110.000 lb − (180 lb ) tan φ
divide (2) by (1)
(2)
tan φ = 0.45173 − 0.73919 tan φ
tan φ = 0.25974,
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
φ = 14.56°
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 127.
(a) FBD E:
ΣFx = 0:
FDE cos α −160 N = 0,
where α = tan −1
FBD Joint D:
β = tan −1
FDE =
160 N
cosα
38.8 mm
= 22.823°, so FDE = 173.59 N C
92.2 mm
37.5 mm − 18.8 mm
= 25.732°
38.8 mm
FCD cos ( 45° − 25.732° )
ΣFx = 0:
− (173.59 N ) cos ( 25.732° + 22.823° ) = 0
FCD = 121.717 N C
FBD BC:
ΣM B = 0:
( 0.045 m )(121.717 N ) sin 45° − M
M = 3.873 N ⋅ m,
= 0,
M = 3.87 N ⋅ m
(b) FBD E:
ΣFx = 0:
FDE cos α −160 N = 0,
where α = tan −1
FDE =
160 N
cos α
20 mm + 35.4 mm − 26.4 mm
= 16.8584°
92.2 mm + 27.8 mm − 24.3 mm
so FDE = 167.185 N C
continued
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
!
COSMOS: Complete Online Solutions Manual Organization System
FBD Joint D:
β = tan −1
27.8 mm − 24.3 mm
= 7.5520°
26.4 mm
γ = tan −1
20 mm + 35.4 mm − 26.4 mm
= 42.363°
24.3 mm + 7.5 mm
ΣFy′ = 0:
(167.185 N ) sin (16.8584° + 42.363°)
− FCD sin ( 90° − 7.5520° − 42.363° ) = 0,
FCD = 223.07 N C
FBD BC:
δ = tan −1
35.4 mm
= 51.857°
27.8 mm
ΣM B = 0: M − ( 0.045 m )( 223.07 N ) sin ( 90° − 51.857° − 7.5520° ) = 0
M = 5.12 N ⋅ m !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 128.
(a) FBD BC:
( 0.045 m ) FCD sin 45° − 6.00 N ⋅ m = 0
ΣM B = 0:
FCD = 188.562 N C
FBD Joint D:
α = tan −1
38.8 mm
= 22.823°
92.2 mm
β = tan −1
37.5 mm − 18.8 mm
= 25.732°
38.8 mm
ΣFx′ = 0:
(188.562 N ) cos ( 45° − 25.732°)
− FDE cos ( 22.823° + 25.732° ) = 0, FDE = 268.92 N C
FBD E:
( 268.92 N ) cos ( 22.823°) − P = 0,
ΣFx = 0:
P = 248 N
(b) FBD BC:
δ = tan −1
35.4 mm
= 51.857°
27.8 mm
β = tan −1
27.8 mm − 24.3 mm
= 7.5520°
26.4 mm
ΣM B = 0:
!
( 0.045 m ) FCD sin ( 90° − 51.857° − 7.5510°) − 6.00 N ⋅ m = 0
FCD = 262.00 N C
continued
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
FBD Joint D:
β = 7.5520°,
α = tan −1
γ = tan −1
20 mm + 35.4 mm − 26.4 mm
= 42.363°
24.3 mm + 7.5 mm
20 mm + 35.4 mm − 26.4 mm
= 16.8584°
92.2 mm + 27.8 mm − 24.3 mm
ΣFy′ = 0:
FDE sin (16.8584° + 42.363° )
− ( 262.00 N ) sin ( 90° − 7.5520° − 42.363° ) = 0,
FDE = 196.366 N C
FBD E:
ΣFx = 0:
(196.366 N )( cos16.8584°) − P = 0,
P = 187.9 N
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
!
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 129.
Member FBDs:
BCD:
AB:
ΣFy = 0:
By = 0
ΣM C = 0:
( 0.128 m ) Bx − ( 0.10 m )(80 N ) = 0,
ΣM A = 0:
(0.15 m) ( 62.5 N ) − M = 0,
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
Bx = 62.5 N
M = 9.38 N ⋅ m
!
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 130.
FBD BCD:
d = 0.24 m
ΣFy = 0:
0.100 m
= 0.1875 m
0.128 m
(80 N ) sin 45° − By
=0
By = 40 2 N
ΣM C = 0:
( 0.100 m )(80 N ) cos 45°
+ ( 0.1875 m )( 80 N ) sin 45°
(
+ ( 0.24 m ) 40 2 N
)
− ( 0.128 m ) Bx = 0
Bx = 233.13 N
FBD AB:
ΣM A = 0:
( 0.2 m ) ( 40
)
2 N − ( 0.15 m )( 233.13 N ) = 0
M = 46.3 N ⋅ m.
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
!
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 131.
FBD BD:
θ = 90° − 25° = 65°
ΣM B = 0:
 2 ( 0.2 m ) cos 25° D sin 65° −15 N ⋅ m = 0
D = 45.654 N
FBD AC:
ΣM A = 0:
M A = ( 0.2 m )( 45.654 N ) = 0
M A = 9.13 N ⋅ m
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
!
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 132.
FBD BD:
ΣM B = 0:
 2 ( 0.2 m ) cos 25° C −15 N ⋅ m = 0
C = 41.377 N
FBD AC:
ΣM A = 0:
M A − ( 0.2 m )( 41.377 N ) sin 65° = 0
M A = 7.50 N ⋅ m
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
!
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 133.
FBD Member CD:
Noting that B is Perpendicular to CD,
ΣFx′ = 0:
Dy sin 30° − ( 80 lb ) cos 30° = 0,
D y = 138.564 lb
FBD Machine:
ΣM A = 0:
10 in.
(138.564 lb ) − M = 0,
sin 30°
M = 2771.3 lb ⋅ in.
M = 2.77 kip ⋅ in.
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
!
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 134.
FBD CD:
Since B is ⊥ to CD
ΣFx′ = 0:
Dy sin 30° − ( 80 lb ) cos 30° = 0
D y = 138.564 lb
FBD Whole:
a = 10 in. − ( 8 in.) cos 30° = 3.07180 in.
b=
a
= 5.32051 in.
tan 30°
d = b − 4 in. = 1.32051 in.
ΣM A = 0:
(10 in.)(80 lb ) + (1.32051 in.)(138.564 lb ) − M
=0
M = 983 lb ⋅ in.
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
!
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 135.
FBD ABD:
First note, by inspection, that P = W = 250 lb.
Note: that BC is a two-force member
ΣM D = 0:
( 5 in.)
3
1
FBC + ( 3 in.)
FBC − ( 9 in.)( 250 lb ) = 0
10
10
FBC = 125 10 lb T
ΣFx = 0:
Dx −
(
)
3
125 10 lb = 0
10
D x = 375 lb
ΣFy = 0:
250 lb −
(
)
1
125 10 lb − Dy = 0
10
D y = 125 lb
(note, that Dy = 125 lb is evident by symmetry of the barrel)
so D = 395 lb
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
18.43°
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 136.
\
FBD whole:
By symmetry A = B = 22.5 kN
FBD ADF:
ΣM F = 0:
( 75 mm ) CD − (100 mm )( 22.5 kN ) = 0
CD = 30 kN
ΣFx = 0:
ΣFy = 0:
Fx − CD = 0
22.5 kN − Fy = 0
W
Fx = CD = 30 kN
Fy = 22.5 kN
so F = 37.5 kN
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
36.9° W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 137.
FBD Joint A:
TB TC 2.1 kN
=
=
17 17
30
TB = TC = 1.19 kN
FBD BF and CF w/cables to A:
By symmetry
Ex = Fx = 1.05 kN
and E y = Fy
FBD Machine:
By symmetry
H x = I x = 1.05 kN
and H y = I y
ΣM D = 0:
FBD BF:
( 0.075 m ) Fy − ( 0.08 m ) (1.05 kN )
15
(1.19 kN )
17
8
− ( 0.16 m )
(1.19 kN ) = 0
16
Fy = 4.48 kN
− ( 0.15 m )
ΣM G = 0:
( 0.075 m ) ( 4.48 kN ) + ( 0.08 m ) (1.05 kN )
+ ( 0.02 m ) (1.05 kN ) − ( 0.15 m ) H y = 0
FBD FGH:
H y = 2.94 kN
ΣFy = 0:
2.94 kN − G y + 4.48 kN = 0 ,
G y = 7.42 kN
ΣFx = 0:
−1.05 kN + Gx + 1.05 kN = 0 ,
Gx = 0
so
H = 3.12 kN
70.3° !
G = 7.42 kN !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 138.
FBD Machine:
By inspection, P = W = 110 lb,
By symmetry:
ΣFy = 0:
H x = Ix,
110 lb − 2H y = 0 ,
ΣFy = 0:
H y = I y = 55 lb,
FCE = FBD
By symmetry:
FBD ABC:
Hy = Iy
4

110 lb − 2  FCE  = 0 ,
5


FCE = FBD =
550
lb C
8
 4  550  
 3  550  
lb   + (17 in.)  
lb  



5  8
5  8
( 2.5 in.) 
ΣM H = 0:
− (4 in.) FFG = 0
FBD DFH:
FFG = 209.69 lb,
ΣFx = 0:
209.69 lb −
FFG = 210 lb
W
3  550 
lb  − H x = 0

5 8

H x = 168.440 lb
so
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
H = 177.2 lb
18.08° W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 139.
FBD top handle:
Note CD and DE are two-force members
ΣM A = 0:
( 4 in.)
6
5
FCD − (1.5 in.)
FCD − (13.2 in.) ( 90 lb ) = 0
61
61
FCD = 72 61 lb
FDE = FCD = 72 61 lb
By symmetry:
FBD Joint D:
ΣFx = 0:
D−2
(
)
5
72 61 lb = 0,
61
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
D = 720 lb W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 140.
FBD top handle:
By symmetry the horizontal force at F must be zero
ΣFx = 0: Dx = 0
ΣM F = 0:
( 0.015 m ) Fy − ( 0.185 m )(135 N ) = 0
Dy = 1665 N
FBD ABD:
ΣM B = 0:
( 40 mm ) (1665 N ) − ( 30 mm ) A = 0
A= 2220 N
A= 2.22 kN W
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 141.
FBD cutter AC:
ΣM C = 0:
( 32 mm )1.5 KN − ( 28 mm ) Ay − (10 mm ) Ax
=0
 11 
10 Ax + 28  Ax  = 48 kN
 13 
Ax = 1.42466 kN
Ay = 1.20548 kN
FBD handle AD:
ΣM D = 0:
(15 mm )(1.20548 kN ) − ( 5 mm )(1.42466 kN )
− (70 mm) P = 0
P = 0.1566 kN = 156.6 N W
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 142.
FBD CD:
θ = tan − 1
ΣM D = 0:
0.85 in. − 0.3 in.
= 22.932°
2.6 in. − 1.3 in.
( 3.4 in.)( 30 lb ) − ( 0.8 in.) FAC sin 22.932°
+ ( 0.3 in.) FAC cos 22.932° = 0
FAC = 2879.6 lb
ΣFx = 0: Dx − ( 2879.6 lb ) cos 22.932° = 0,
Dx = 2652.0 lb
ΣFy = 0: Dy − ( 2879.6 lb ) sin 22.932° + 30 lb = 0,
Dy = 1092.00 lb
FBD BD:
ΣM B = 0:
( 2.6 in.) F + ( 0.2 in.)(1092.00 lb )
− ( 0.85 in.)( 2652.0 lb ) = 0
F = 783 lb W
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 143.
FBD Joint B:
α = tan − 1
1
= 5.7106°
10
ΣFx′ = 0: FBC cos ( 30° + 5.7106° ) − ( 54 lb ) cos 30° = 0
FBC = 57.595 lb T
FCD = 57.595 lb
By symmetry:
FBD Joint C:
ΣFx′′ = 0: 2 ( 57.595 lb ) sin ( 5.7106° ) − P = 0
P = 11.4618 lb
so
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
P = 11.46 lb
30.0° W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 144.
FBD BC:
ΣM B = 0:
(1 in.) Cx − ( 2.65 in.) ( 270 lb ) = 0
Cx = 715.5 lb
FBD CF:
( 3.5 in.) P − ( 0.75 in.) FDE cos 40°
+ ( 0.65 in.) FDE sin 40° = 0
ΣM C = 0:
FDE = 22.333 P
ΣFx = 0: C x − ( 22.333 P ) cos 40° + P sin 30° = 0
P = 43. 081 lb
P = 43.1 lb
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
60.0˚ W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 145.
FBD Handles:
ΣM C = 0: − aBx + b (100 N − 100 N ) = 0
Bx = 0
FBD top handle:
ΣM A = 0:
( 28 mm ) By − (110 mm ) (100 N ) = 0
By = 392.86 N
FBD top blade:
ΣM D = 0:
( 40 mm ) ( 392.86 N ) − ( 30 mm ) E = 0
E = 523.81 N
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
E = 524 N W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 146.
FBD BE:
ΣM B = 0:
(
)
746 mm C − ( 640 mm ) ( 0.25 kN ) = 0
C=
160
kN
746
FBD Blade:
ΣM A = 0:
 2  160

kN  


 5  746
( 38 mm ) F − ( 47 mm ) 
 1  160

kN   = 0
− ( 23 mm ) 


 5  746
F = 8.82 kN W
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 147.
FBD Boom with bucket and man:
θ = tan − 1
3 ft + ( 6 ft ) sin 35°
= 62.070°
( 6 ft ) cos 35° − 1.5 ft
ΣM A = 0: ( 6 ft ) cos 35° ( FBD sin 62.070° )
− ( 6 ft ) sin 35° ( FBD cos 62.070° )
− ( 9ft ) cos 35°  (1400 lb ) − ( 20 ft ) cos 35°  ( 450 lb ) = 0
FBD = 17693 lb,
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
FBD = 6.48 kips
62.1° W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 148.
FBD AED:
ΣM A = 0:
(a)
( 36 in.) (1500 lb ) − ( 25 in.) F = 0,
F = 2160 lb
ΣFy = 0: Ay − 1500 lb = 0,
ΣFy = 0: Ax − 2160 lb = 0,
FBD Entire machine:
ΣM B = 0:
Ay = 1500 lb
Ax = 2160 lb
(84 in.) (1500 lb ) − ( 25 in.) Gx = 0,
Gx = 5040 lb
ΣM C = 0:
FBD AG:
( 24 in.) ( Gy + 1500 lb )
− (12.5 in.) ( 5040 lb − 2160 lb ) = 0,
Gy = 0 W
(b)
ΣFy = 0: C y − 1500 lb + 0 = 0,
C y = 1500 lb
ΣFx = 0:
Cx − 2160 lb − 5040 lb = 0,
Cx = 7200 lb
so on BCF,
C x = 7.20 kips
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
,
C y = 1.500 kips
W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 149.
FBD boom:
θ = tan −1
Note:
( 3.2sin 24° − 1) m
( 3.2cos 24° − 0.6 ) m
θ = 44.73°
(a)
ΣM A = 0:
( 6.4 m ) cos 24° ( 2.3544 kN )
− ( 3.2 m ) cos 24° B sin 44.73°
+ ( 3.2 m ) sin 24° B cos 44.73° = 0
B = 12.153 kN
B = 12.15 kN
44.7° W
ΣFx = 0: Ax − (12.153 kN ) cos 44.73° = 0
A x = 8.633 kN
(b)
ΣFy = 0:
− 2.3544 kN + (12.153 kN ) sin 44.73° − Ay = 0
A y = 6.198 kN
On boom:
On carriage:
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
A = 10.63 kN
A = 10.63 kN
35.7°
35.7° W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 150.
FBD Bucket (one side):
(a)
ΣM D = 0:
( 0.8 m ) ( 2.4525 kN ) − ( 0.5 m ) FAB = 0
FAB = 3.924 kN
w = ( 250 kg ) ( 9.81 N/kg ) = 2452.5 N = 2.4525 kN
FBD link BE:
ΣM E = 0:
( 0.68 m )( 3.924 kN ) − ( 0.54 m ) 
80

FCD 
 89

 39

+ ( 0.35 m )  FCD  = 0,
 89

FCD = 7.682 kN
FCD = 7.68 kN C W
FBD top blad
(b)
FBD Entire linkage:
ΣM G = 0:
( 2.5 m ) ( 2.4525 kN )
1
1
FGH − ( 0.6 m )
FGH = 0,
2
2
FFH = 21.677 kN,
FFH = 21.7 kN C W
+ ( 0.2 m )
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 151.
\
FBD Central Gear:
By symmetry: F1 = F2 = F3 = F
ΣM A = 0:
3 ( rA F ) − 15 N ⋅ m = 0,
ΣM C = 0:
rB ( F − F4 ) = 0,
F =
5
N⋅m
rA
FBD Gear C:
ΣFx′ = 0:
C x′ = 0
ΣFy′ = 0:
C y′ − 2 F = 0,
F4 = F
C y′ = 2 F
Gears B and D are analogous, each having a central force of 2F
FBD Spider:
ΣM A = 0:
75 N ⋅ m − 3 ( rA + rB ) 2 F = 0
25 N ⋅ m − ( rA + rB )
10
N⋅m = 0
rA
rA + rB
r
= 2.5 = 1 + B ,
rA
rA
FBD Outer gear:
Since
rA = 24 mm,
ΣM A = 0:
rB = 1.5 rA
rB = 36.0 mm !
(a)
3 ( rA + 2rB ) F − M E = 0
3 ( 24 mm + 72 mm )
(b)
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
5 N⋅m
−ME = 0
24 mm
M E = 60.0 N ⋅ m
!
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 152.
FBD Gear A: looking from C
(a)
M A = 15 lb ⋅ ft
ΣM A = 0:
rA = 4 in.
M A − P rA = 0
P=
M A 180 lb ⋅ in.
=
rA
4 in.
P = 45 lb
FBD Gear B: looking from F
ΣM B = 0:
M 0 − rB P = 0
M 0 = rB P = ( 2.5 in.)( 45 lb ) = 112.5 lb ⋅ in.
M 0 = 112.5 lb ⋅ in. i !
FBD ABC: looking down
(b)
( 2 in.)( 45 lb ) − ( 5 in.) C
ΣM B = 0:
ΣFz = 0:
FBD BEG:
=0
45 lb − B + 18 lb = 0
By analogy, using FBD DEF
ΣFz = 0:
E = 63 lb k
C = 18 lb k
B = −63 lb k
F = 18 lb k
Gz + 63 lb − 63 lb = 0
Gz = 0
ΣFy = 0
Gy = 0
ΣM G = 0
M G − ( 6.5 in.)( 63 lb ) = 0
M G = ( 410 lb ⋅ in.) i !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
FBD CFH:
ΣF = 0:
Hz = H y = 0
ΣM H = 0
M H = − ( 6.5 in.)(18 lb )
= −117 lb ⋅ in.
M G = − (117.0 lb ⋅ in.) i !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 153.
Note: The couples exerted by the two yokes on the crosspiece must be equal and opposite. Since neither yoke
can exert a couple along the arm of the crosspiece it contacts, these equal and opposite couples must be
normal to the plane of the crosspiece.
If the crosspiece arm attached to shaft CF is horizontal, the plane of the crosspiece is normal to shaft AC, so
couple M C is along AC.
FBDs shafts with yokes:
(a)
ΣM x = 0:
FBD CDE:
M C cos30° − 50 N ⋅ m = 0
M C = 57.735 N ⋅ m
FBD BC: ΣM x′ = 0: M A − M C = 0
(b)
M A = 57.7 N ⋅ m W
ΣM C = 0: M A i′ + ( 0.5 m ) Bz j′ − ( 0.5 m ) By′ k = 0
ΣF = 0: B + C = 0
so
B = 0W
C=0
FBD CDF : ΣM Dy = 0: − ( 0.6 m ) Ez + ( 57.735 N ⋅ m ) sin 30° = 0
E z = 48.1 N k
ΣFx = 0: Ex = 0
ΣM Dz = 0: ( 0.6 m ) E y = 0
Ey = 0
0
ΣF = 0: C + D + E = 0
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
so E = ( 48.1 N ) k W
D = −E = − ( 48.1 N ) k W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 154.
Note: The couples exerted by the two yokes on the crosspiece must be equal and opposite. Since neither yoke
can exert a couple along the arm of the crosspiece it contacts, these equal and opposite couples must be
normal to the plane of the crosspiece.
If the crosspiece arm attached to CF is vertical, the plane of the crosspiece is normal to CF, so the couple M C
is along CF.
(a)
FBD CDE:
FBD BC:
ΣM x = 0: M C − 50 N ⋅ m = 0
ΣM x′ = 0: M A − M C cos 30° = 0
M C = 50 N ⋅ m
M A = ( 50 N ⋅ m ) cos 30°
M A = 43.3 N ⋅ m W
(b)
ΣM Cy′ = 0: M C sin 30° + ( 0.5 m ) Bz = 0
ΣM Cz = 0: − ( 0.5 m ) By = 0
ΣF = 0: B + C = 0
FBD CDE:
Bz = −
( 50 N ⋅ m )( 0.5)
0.5 m
= −50 N
so B = − ( 50.0 N ) k W
By = 0
C = −B
so
C = ( 50 N ) k on BC
ΣM Dy = 0: − ( 0.4 m ) C z − ( 0.6 m ) Ez = 0
4
Ez = − ( 50 N )   = −33.3 N
6
ΣM Dz = 0: E y = 0
ΣFx = 0: Ex = 0
ΣF = 0: C + D + E = 0
so E = − ( 33.3 N ) k W
− ( 50 N ) k + D − ( 33.3 N ) k = 0
D = ( 83.3 N ) k W
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 155.
FBD Joint A:
By inspection of FBD of entire device, FA = w = 1500 lb
By symmetry, FAB = FAC
 1

FAC  = 0,
1500 lb − 2 
 5

ΣFy = 0:
FBD CDF:
ΣFx = 0:
(
FAC = 750 5 lb
)
2
750 5 lb = 0,
5
Dx − Fx −
Dx − Fx = 1500 lb
ΣFy = 0:
(
(1)
)
1
750 5 lb = 0,
5
Fy − D y +
Dy − Fy = 750 lb
ΣM D = 0:
FBD EF:
(2)
 1

 2

(4.6 ft) 
750 5 lb  + ( 0.6 ft ) 
750 5 lb 
 5

 5

(
)
(
)
− (1 ft ) Fy − ( 3.6 ft ) Fx = 0,
3.6 Fx + Fy = 4350 lb
By symmetry:
ΣM F = 0:
Ex = Dx ,
E y = Dy
Hy = Jy =
w
= 750 lb,
2
(4), (5)
H y = 750 lb
!
( 3 ft ) E y − ( 3.6 ft ) Ex − (1 ft ) H y + ( 0.6 ft )( 750 lb ) = 0
3.6Ex − 3E y + H y = 450 lb
ΣFx = 0:
(3)
Dx + Fx − H x = 0,
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
(6)
(7)
COSMOS: Complete Online Solutions Manual Organization System
Solving Eqns, (1)–(7), on EFH
Fx = 540 lb
Fy = 2410 lb
E x = 2040 lb
E y = 3160 lb
H x = 2580 lb
H y = 750 lb
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
!
!
!
!
!
!
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 156.
FBD Truss:
ΣM E = 0: ( 9 ft ) Fy − ( 6.75 ft )( 4 kips ) − (13.5 ft )( 4 kips ) = 0
Fy = 9 kips
ΣFy = 0: − E y + 9 kips = 0
E y = 9 kips
ΣFx = 0: − Ex + 4 kips + 4 kips = 0
E x = 8 kips
By inspection of joint E:
FEF = 8.00 kips T !
Joint FBDs:
Joint F:
FEC = 9.00 kips T !
By inspection of joint B:
FAB = 0 !
FBD = 0 !
ΣFx = 0:
Joint C:
4
FCF − 8 kips = 0
5
ΣFy = 0: FDF −
3
(10 kips) = 0
5
ΣFx = 0: 4 kips −
FCF = 10.00 kips C !
FDF = 6.00 kips T !
4
(10 kips ) + FCD = 0
5
FCD = 4.00 kips T !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
ΣFy = 0: FAC − 9 kips +
3
(10 kips ) = 0
5
FAC = 3.00 kips T !
Joint A:
ΣFx = 0: 4 kips −
4
FAD = 0
5
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
FAD = 5.00 kips C !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 157.
FBD Truss:
ΣM A = 0: ( 6 ft )  6Ly − 3 ( 0.5 kip ) − 2 (1 kip ) − 1(1 kip )  = 0
L y = 0.75 kip
FJK = 0 !
Inspection of joints K , J , and I , in order, shows that
FIJ = 0 !
FHI = 0 !
Joint FBDs:
FIK = FKL ; FHJ = FJL and FGI = FIK
and that
0.75
F
F
= JL = KL
1
8
5
FJL = 2.1213 kips
FKL = 1.6771 kips
FJL = 2.12 kips C !
FKL = 1.677 kips T !
FHJ = 2.12 kips C !
and, from above:
FGI = FIK = 1.677 kips T !
ΣFx = 0:
2
1
( FFH + FGH ) − ( 2.1213 kips ) = 0
5
2
ΣFy = 0:
1
1
( FGH + FFH ) + ( 2.1213 kips ) = 0
5
2
Solving:
FFH = 2.516 kips
FGH = − 0.8383 kips
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
FFH = 2.52 kips C !
FGH = 0.838 kips T !
COSMOS: Complete Online Solutions Manual Organization System
ΣFx = 0:
2
( FDF − 2.516 kips ) = 0
5
ΣFy = 0: FFG − 0.5 kip +
FDF = 2.52 kips C
1
( 2 )( 2.516 kips ) = 0
5
FFG = 1.750 kips T !
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 158.
FBD Truss:
ΣM F = 0: (10 m ) G y − ( 7.5 m )( 80 kN ) − ( 8 m )( 30 kN ) = 0
G y = 84 kN
ΣFx = 0: − Fx + 30 kN = 0
Fx = 30 kN
ΣFy = 0: Fy + 84 kN − 80 kN = 0
Fy = 4 kN
FEG = 0 !
By inspection of joint G:
FCG = 84 kN C !
84 kN
F
F
= CE = AC = 10.5 kN
8
61
29
Joint FBDs:
FCE = 82.0 kN T !
FAC = 56.5 kN C !
ΣFy = 0:
2
6
FAE +
(82.0 kN ) − 80 kN = 0
5
61
FAE = 19.01312
ΣFx = 0: − FDE −
1
(19.013 kN ) +
5
FDE = 43.99 kN
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
FAE = 19.01 kN T !
5
(82.0 kN ) = 0
61
FDE = 44.0 kN T !
COSMOS: Complete Online Solutions Manual Organization System
ΣFx = 0:
1
FDF − 30 kN = 0
5
FDF = 67.082 kN
ΣFy = 0:
2
( 67.082 kN ) − FBF − 4 kN = 0
5
FBF = 56.00 kN
ΣFx = 0: 30 kN +
ΣFy = 0: 56 kN −
Solving:
ΣFy = 0:
FDF = 67.1 kN T !
5
FBD −
61
6
FBD −
61
FBF = 56.0 kN C !
5
FAB = 0
29
3
FAB = 0
29
FBD = 42.956 kN
FBD = 43.0 kN T !
FAB = 61.929 kN
FAB = 61.9 kN C !
6
2
( 42.956 N ) + ( FAD − 67.082 N ) = 0
61
5
FAD = 30.157 kN
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
FAD = 30.2 N T !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 159.
FBD Truss:
ΣFx = 0: A x = 0
By load symmetry, A y = I y = 8 kips
FBD Section:
ΣM D = 0:
( 7 ft )( 2 kips − 8 kips ) + ( 3 ft ) ( FCE ) = 0
FCE = 14 kips
ΣM E = 0:
FCE = 14.00 kips T W
( 7 ft ) 1( 4 kips ) + 2 ( 2 kips − 8 kips )
( 4.5 ft )
FDF =
7
FDF = 0
51.25
8 51.25
kips
4.5
ΣFy = 0: 8 kips − 2 kips − 4 kips +
FDF = 12.73 kips C W
1.5 8 51.25
kips −
51.25 4.5
FDE = −1.692 kips
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
3
FDE = 0
58
FDE = 1.692 kips C W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 160.
FBD Section:
p = 8.4 ( 2.7 m ) = 1.89 m
Note: BG
12.0
ΣM A = 0:
( 2.7 m ) FFG − ( 3.6 m )(8 kN ) − ( 7.8 m )(8 kN ) − (12 m )( 4 kN ) = 0
FFG = 51.56 kN
ΣM G = 0:
FFG = 51.6 kN C W
(1.89 m ) 
12

FAB  − ( 4.2 m )( 8 kN ) − ( 8.4 m )( 4 kN ) = 0
 12.3

FAB = 36.44 kN
ΣM D = 0:
FAB = 36.4 kN T W
( 4.2 m )(8 kN ) + (8.4 m )(8 kN ) − ( 8.4 m ) 
3

FAG  = 0
5


Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
FAG = 20.0 kN T W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 161.
FBD Truss:
ΣM A = 0:
( 24 ft ) K x − (18 ft )( 2 kips ) − ( 36 ft )( 2 kips ) = 0
K x = 4.5 kips
FBD Section:
ΣM F = 0:
(12 ft )( 4.5 kips ) − (18 ft )( 2 kips ) − ( 36 ft )( 2 kips ) + ( 36 ft ) FEJ = 0
FEJ = 1.500 kips T W
ΣM J = 0:
(18 ft )( 2 kips ) + (12 ft )( 4.5 kips ) − ( 36 ft ) FAF
=0
FAF = 2.50 kips T W
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 162.
FBD Frame:
ΣM F = 0:
(10.8 ft ) Ay − (12 ft )( 4.5 kips ) = 0
A y = 5.00 kips
ΣFx = 0:
W
− Ax + 4.5 kips = 0
A x = 4.50 kips
FBD member ABC:
W
Note: BE is a two-force member
ΣM C = 0:
ΣFx = 0:
ΣFy = 0:
(12 ft ) FBE + (10.8 ft )( 5 kips ) − (18 ft )( 4.5 kips ) = 0
FBE = 2.25 kips
W
C x = 2.25 kips
W
Cx + 2.25 kips − 4.5 kips = 0
C y − 5 kips = 0
C y = 5.00 kips
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 163.
FBD Frame:
ΣM E = 0:
( 3.75 in.) Bx + (8.75 in.)( 75 lb ) = 0
Bx = 175 lb
FBD member ACE:
ΣFx = 0:
Ex − Bx = 0
ΣFy = 0:
E y + By − 75 lb = 0
B x = 175.0 lb
W
E x = 175.0 lb
W
By = 75 lb − E y
ΣM C = 0:
Thus
− (1.25 in.)( 75 lb ) + ( 3.75 in.)(175 lb ) − ( 4.5 in.) E y = 0
E y = 125.0 lb
W
B y = 50.0 lb
W
By = 75 lb − 125 lb = −50 lb
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 164.
FBD Frame:
Q = 65 N
P = 15 N
ΣM D = 0:
( 0.25 m )( P + Q ) − (.15 m )( P + Q ) − ( 0.08 m ) Ex
Ex = 1.2 ( P + Q ) = 100 N
ΣFx = 0:
(b)
Dx − Ex = 0 = Dx − 100 N
(a)
ΣFy = 0:
E x = 100.0 N
=0
W
Dx = 100 N
D x = 100.0 N
W
E y − Dy − 2P − 2Q = 0
E y = D y + 2 ( P + Q ) = Dy + 160 N
FBD member BF:
ΣM C = 0:
( 0.15 m )( 65 N ) − ( 0.1 m ) Dy − ( 0.04 m )(100 N )
− ( 0.25 m )(15 N ) = 0
Dy = 20 N
From above
D y = 20.0 N
W
E y = 20 N + 160 N = 180 N (b) E y = 180.0 N
W
ΣFx = 0:
(a)
− C x + 100 N = 0
(a)
ΣFy = 0:
C x = 100.0 N
W
− 65 N + C y − 20 N − 15 N = 0
(a)
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
C y = 100.0 N
W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 165.
Member FBDs:
FBD I:
ΣM C = 0:
R ( FBD sin 30° )
−  R (1 − cos 30° )  ( 200 N ) − R ( 100 N ) = 0
FBD = 253.6 N
ΣFx = 0:
(b)
FBD = 254 N T W
− C x + ( 253.6 N ) cos 30° = 0
C x = 219.6 N
ΣFy = 0:
C y + ( 253.6 N ) sin 30° − 200 N − 100 N = 0
C y = 173.2 N
(c) so C = 280 N
ΣM A = 0:
aP − a ( 253.6 N ) cos 30°  = 0
FBD II:
(a)
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
38.3° W
P = 220 N
W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 166.
FBD jaw AB:
ΣFx = 0: Bx = 0
ΣM B = 0:
( 0.01 m ) G − ( 0.03 m ) A = 0
A=
ΣFy = 0:
G
3
A + G − By = 0
By = A + G =
4G
3
FBD handle ACE:
By symmetry and FBD jaw DE: D = A =
E y = By =
ΣM C = 0:
G
, Ex = Bx = 0,
3
4G
3
( 0.105 m )( 240 N ) + ( 0.015 m )
G
4G
− ( 0.015 m )
=0
3
3
G = 1680 N W
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 167.
Note: By symmetry the vertical components of pin forces C and D are
zero.
FBD handle ACF: (not to scale)
ΣFy = 0:
ΣM C = 0:
ΣFx = 0:
Ay = 0
(13.5 in.)( 20 lb ) − (1.5 in.) Ax
= 0 Ax = 180 lb
C − Ax − 20 lb = 0 C = (180 + 20 ) lb = 200 lb
FBD blade DE:
ΣM D = 0:
( 9 in.) E − ( 3 in.)(180 lb ) = 0
E = 60.0 lb W
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.