Statics Problems: Free-Body Diagrams & Equilibrium

5–1.
Draw the free-body diagram of the dumpster D of the
truck, which has a weight of 5000 lb and a center of gravity
at G. It is supported by a pin at A and a pin-connected
hydraulic cylinder BC (short link). Explain the significance
of each force on the diagram. (See Fig. 5–7b.)
1m
3m
A
SOLUTION
The Significance of Each Force:
W is the effect of gravity (weight) on the dumpster.
Ay and Ax are the pin A reactions on the dumpster.
FBC is the hydraulic cylinder BC reaction on the dumpster.
1.5 m
G
D
B
20
30
C
5–2.
Draw the free-body diagram of member ABC which is
supported by a smooth collar at A, rocker at B, and short link
CD. Explain the significance of each force acting on the
diagram. (See Fig. 5–7b.)
SOLUTION
The Significance of Each Force:
NA is the smooth collar reaction on member ABC.
NB is the rocker support B reaction on member ABC.
FCD is the short link reaction on member ABC.
2.5 kN is the effect of external applied force on member ABC.
4 kN # m is the effect of external applied couple moment on member ABC.
3m
60
A
4 kN m
B
45
4m
D
C
2.5 kN
6m
5–3.
Draw the free-body diagram of the beam which supports the
80-kg load and is supported by the pin at A and a cable which
wraps around the pulley at D. Explain the significance of
each force on the diagram. (See Fig. 5–7b.)
D
5
4
3
A
B
E
SOLUTION
T force of cable on beam.
Ax, Ay force of pin on beam.
80(9.81)N force of load on beam.
C
2m
2m
1.5 m
5– 4
Draw the free-body diagram of the hand punch, which is pinned at A and bears down on the
smooth surface at B.
Given:
F 40 N
a 0.45 m
b 0.06 m
c 0.6 m
Solution:
5–5.
Draw the free-body diagram of the uniform bar, which has a
mass of 100 kg and a center of mass at G. The supports A, B,
and C are smooth.
0.5 m
1.25 m
C
G
1.75 m
SOLUTION
A
B
0.1 m
30⬚
0.2 m
5– 6.
Draw the free-body diagram of the jib crane AB, which is pin-connected at A and supported by
member (link) BC.
Units Used:
3
kN
10 N
Given:
F
20 kN
a
3.5 m
b
4.5 m
c
0.4 m
d
3
e
4
Solution:
5–7.
Draw the free-body diagram of the beam, which is pin
connected at A and rocker-supported at B.
500 N
SOLUTION
800 N⭈m
B
5m
A
8m
4m
5–9.
Draw the free-body diagram of the jib crane AB, which is pin
connected at A and supported by member (link) BC.
C
5
3
4
SOLUTION
B
0.4 m
A
4m
3m
8 kN
5–10.
4 kN
Determine the horizontal and vertical components of
reaction at the pin A and the reaction of the rocker B on
the beam.
B
A
30⬚
SOLUTION
6m
Equations of Equilibrium: From the free-body diagram of the beam, Fig. a, NB can
be obtained by writing the moment equation of equilibrium about point A.
a
+ ©MA = 0;
NB cos 30°(8) - 4(6) = 0
NB = 3.464 kN = 3.46 kN
Ans.
Using this result and writing the force equations of equilibrium along the x and
y axes, we have
+ ©F = 0;
:
x
A x - 3.464 sin 30° = 0
A x = 1.73 kN
+ c ©Fy = 0;
Ans.
A y + 3.464 cos 30° - 4 = 0
A y = 1.00 kN
Ans.
2m
5–11.
Determine the magnitude of the reactions on the beam at A
and B. Neglect the thickness of the beam.
600 N
3
5
4
B
A
4m
SOLUTION
a + ©MA = 0;
By (12) - (400 cos 15°)(12) - 600(4) = 0
By = 586.37 = 586 N
+ ©F = 0;
:
x
Ans.
Ax - 400 sin 15° = 0
Ax = 103.528 N
+ c ©Fy = 0;
Ay - 600 - 400 cos 15° + 586.37 = 0
Ay = 400 N
FA = 2(103.528)2 + (400)2 = 413 N
Ans.
15⬚
400 N
8m
5–12.
Determine the components of the support reactions at the
fixed support A on the cantilevered beam.
6 kN
30⬚
SOLUTION
30⬚
Equations of Equilibrium: From the free-body diagram of the cantilever beam, Fig. a,
Ax, Ay, and MA can be obtained by writing the moment equation of equilibrium about
point A.
+ ©F = 0;
:
x
4 cos 30° - A x = 0
A x = 3.46 kN
+ c ©Fy = 0;
Ans.
A y - 6 - 4 sin 30° = 0
A y = 8 kN
Ans.
a+ ©MA = 0;MA - 6(1.5) - 4 cos 30° (1.5 sin 30°) - 4 sin 30°(3 + 1.5 cos 30°) = 0
MA = 20.2 kN # m
Ans.
1.5 m
A
1.5 m
1.5 m
4 kN
5–13.
The 75-kg gate has a center of mass located at G. If A
supports only a horizontal force and B can be assumed as a
pin, determine the components of reaction at these supports.
A
G
1m
B
SOLUTION
Equations of Equilibrium: From the free-body diagram of the gate, Fig. a, By and
Ax can be obtained by writing the force equation of equilibrium along the y axis and
the moment equation of equilibrium about point B.
+ c ©Fy = 0;
By - 75(9.81) = 0
By = 735.75 N = 736 N
a+©MB = 0;
Ans.
A x(1) - 75(9.81)(1.25) = 0
A x = 919.69 N = 920 N
Ans.
Using the result Ax = 919.69 N and writing the force equation of equilibrium along
the x axis, we have
+ ©F = 0;
:
x
Bx - 919.69 = 0
Bx = 919.69 N = 920 N
1.25 m
Ans.
5–14.
The overhanging beam is supported by a pin at A and
the two-force strut BC. Determine the horizontal and
vertical components of reaction at A and the reaction at B
on the beam.
1m
A
SOLUTION
1.5 m
Equations of Equilibrium: Since line BC is a two-force member, it will exert a force FBC
directed along its axis on the beam as shown on the free-body diagram, Fig. a. From the
free-body diagram, FBC can be obtained by writing the moment equation of equilibrium
about point A.
a+ ©MA = 0;
3
FBC a b(2) - 600(1) - 800(4) - 900 = 0
5
FBC = 3916.67 N = 3.92 kN
Ans.
Using this result and writing the force equations of equilibrium along the x and
y axes, we have
+
: ©Fx = 0;
4
3916.67a b - A x = 0
5
A x = 3133.33 N = 3.13 kN
+ c ©Fy = 0;
Ans.
3
-A y - 600 - 800 + 3916.67a b = 0
5
A y = 950 N
Ans.
C
600 N
1m
800 N
2m
B
900 N⭈m
.
.
.
5–16.
Determine the components of reaction at the supports A and
B on the rod.
P
L
––
2
SOLUTION
Equations of Equilibrium: Since the roller at A offers no resistance to vertical
movement, the vertical component of reaction at support A is equal to zero. From
the free-body diagram, Ax, By, and MA can be obtained by writing the force
equations of equilibrium along the x and y axes and the moment equation of
equilibrium about point B, respectively.
+ ©F = 0;
:
x
Ax = 0
+ c ©Fy = 0;
By - P = 0
By = P
a + ©MB = 0;
Pa
Ans.
Ans.
L
b - MA = 0
2
MA =
PL
2
Ans.
A
L
––
2
B
5–17.
If the wheelbarrow and its contents have a mass of 60 kg and
center of mass at G, determine the magnitude of the resultant
force which the man must exert on each of the two handles in
order to hold the wheelbarrow in equilibrium.
B
G
0.6 m
0.5 m
A
SOLUTION
a +©MB = 0;
0.5 m
- Ay (1.4) + 60(9.81)(0.9) = 0
Ay = 378.39 N
+ c ©Fy = 0;
378.39 - 60(9.81) + 2By = 0
By = 105.11 N
+ ©F = 0;
:
x
Bx = 0
FB = 105 N
Ans.
0.9 m
5 –18.
Determine the reactions at the roller A and pin B.
Given:
M 800kN˜ m
c 3m
F 390kN
d 5
a 8m
e 12
b 4m
T 30deg
Solution:
Guesses
RA 1kN
Bx 1kN
By 1kN
Given
RA˜ sin ( T ) Bx d
2
e d
RA˜ cos ( T ) By 2
e
2
e d
2
˜F = 0
˜F = 0
M RA˜ cos ( T ) ˜ ( a b) Bx˜ c = 0
§ RA ·
¨ ¸
¨ Bx ¸ Find RA Bx By
¨ ¸
© By ¹
RA
105 kN
§¨ Bx ¸·
¨ By ¸
© ¹
§ 97.4 ·
¨
¸ kN
© 269 ¹
Ans.
5–19.
The shelf supports the electric motor which has a mass of
15 kg and mass center at Gm. The platform upon which it
rests has a mass of 4 kg and mass center at Gp. Assuming that
a single bolt B holds the shelf up and the bracket bears
against the smooth wall at A, determine this normal force
at A and the horizontal and vertical components of reaction
of the bolt on the bracket.
Bx (60) - 4(9.81)(200) - 15(9.81)(350) = 0
40 mm
B
A
Bx = 989.18 = 989 N
Ans.
+ ©F = 0;
:
x
Ax = 989.18 = 989 N
Ans.
+ c ©Fy = 0;
By = 4(9.81) + 15(9.81)
By = 186.39 = 186 N
150 mm
50 mm
60 mm
SOLUTION
a + ©MA = 0;
200 mm
Ans.
Gp
Gm
.
.
.
.
.
5–22. Spring CD remains in the horizontal position at all
times due to the roller at D. If the spring is unstretched
when u = 0° and the bracket achieves its equilibrium
position when u = 30°, determine the stiffness k of the
spring and the horizontal and vertical components of
reaction at pin A.
D k
C
0.45 m
0.6 m
B
u
F ⫽ 300 N
A
5–23. Spring CD remains in the horizontal position at all
times due to the roller at D. If the spring is unstretched
when u = 0° and the stiffness is k = 1.5 kN>m, determine
the smallest angle u for equilibrium and the horizontal and
vertical components of reaction at pin A.
D k
C
0.45 m
0.6 m
B
u
F ⫽ 300 N
A
5–24.
The platform assembly has weight W1 and center of gravity at G1. If it is intended to support a
maximum load W2 placed at point G2,,determine the smallest counterweight W that should be
placed at B in order to prevent the platform from tipping over.
Given:
W1 1000N
W2 1600N
a 0.5m
b 3m
c 0.5m
d 4m
e 3m
f 1m
Solution:
When tipping occurs, R c = 0
6MD = 0;
W2˜ f W1˜ c WB˜ ( b c) = 0
WB WB
W2˜ f W1˜ c
bc
314 N
Ans.
5–25.
Determine the reactions on the bent rod which is supported by a smooth surface at B and by a
collar at A, which is fixed to the rod and is free to slide over the fixed inclined rod.
Given:
F 100N
M 20N˜ m
a 0.3m
b 0.3m
c 0.2m
d 3
e 4
f 12
g 5
Solution:
Initial Guesses:
NA 20N
NB 10N
MA 30N˜ m
Given
6MA = 0; MA F˜ a M NB˜
f
2
2
˜ ( a b) NB˜
f g
6Fx = 0;
NA˜
e
2
e d
6Fy = 0;
§ NA
NA˜
2
d
2
e d
2
NB˜
g
2
2
g
2
2
˜c = 0
f g
= 0
f g
NB˜
·
¨
¸
¨ NB ¸ Find NA NB MA
¨
¸
© MA ¹
f
2
2
F= 0
f g
§¨ NA ¸·
¨ NB ¸
© ¹
§ 39.7 ·
¨
¸N
© 82.5 ¹
MA
10.6 N˜ m
Ans.
.
.
5–27.
The sports car has a mass of 1.5 Mg and mass center at G. If
the front two springs each have a stiffness of kA = 58 kN>m
and the rear two springs each have a stiffness of
kB = 65 kN>m, determine their compression when the car
is parked on the 30° incline. Also, what friction force FB
must be applied to each of the rear wheels to hold the car in
equilibrium? Hint: First determine the normal force at A
and B, then determine the compression in the springs.
A
0.4 m
0.8 m
14 715 cos 30°11.22
- 14 715 sin 30°10.42 - 2NA 122 = 0
NA = 3087.32 N
2FB - 14 715 sin 30° = 0
FB = 3678.75 N = 3.68 kN
Q+ ©Fy¿ = 0;
FB
30°
Equations of Equilibrium: The normal reaction NA can be obtained directly by
summing moments about point B.
a+ ©Fx¿ = 0;
B
1.2 m
SOLUTION
a + ©MB = 0;
G
Ans.
2NB + 213087.322 - 14 715 cos 30° = 0
NB = 3284.46 N
Spring Force Formula: The compression of the sping can be determined using the
Fsp
.
spring formula x =
k
xA =
3087.32
= 0.05323 m = 53.2 mm
5811032
Ans.
xB =
3284.46
= 0.05053 m = 50.5 mm
65 103
Ans.
5–28.
Determine the magnitude and direction T of the minimum force P needed to pull the roller of
mass M over the smooth step.
Given:
a
0.6 m
b
0.1 m
T1
20 deg
M
50 kg
g
9.81
m
2
s
Solution:
For Pmin, NA tends to 0
a b·
I
acos §¨
I
33.56 deg
¸
© a ¹
6MB = 0 M g sin T 1 ( a b) M g cos T 1 a sin I ª¬P cos T ( a b)º¼ P sin T a sin I
P
0
M gª¬sin T 1 ( a b) cos T 1 a sin I º¼
cos T ( a b) a sin I sin T
For Pmin :
M gª¬sin T 1 ( a b) cos T 1 a sin I º¼
dP
dT
ª¬cos T ( a b) a sin I sin T º¼
which gives,
P
T
atan §¨ sin I ˜
©
a
2
ª¬a sin I cos T ( a b)sin T º¼
·
¸
a b¹
M gª¬sin T 1 ( a b) cos T 1 a sin I º¼
cos T ( a b) a sin I sin T
0
T
33.6 deg
Ans.
P
395 N
Ans.
.
.
5–31. The jib crane is supported by a pin at C and rod AB.
If the load has a mass of 2 Mg with its center of mass located
at G, determine the horizontal and vertical components of
reaction at the pin C and the force developed in rod AB on
the crane when x = 5 m.
4m
A
3.2 m
C
0.2 m
B
D
x
G
5–32. The jib crane is supported by a pin at C and rod AB.
The rod can withstand a maximum tension of 40 kN. If the
load has a mass of 2 Mg, with its center of mass located at G,
determine its maximum allowable distance x and the
corresponding horizontal and vertical components of
reaction at C.
4m
A
3.2 m
C
0.2 m
B
D
x
G
5–33.
The woman exercises on the rowing machine. If she exerts a
holding force of F = 200 N on handle ABC, determine the
horizontal and vertical components of reaction at pin C and
the force developed along the hydraulic cylinder BD on the
handle.
F ⫽ 200 N
30⬚
0.25 m
A
B
0.25 m
C
D
SOLUTION
0.75 m
Equations of Equilibrium: Since the hydraulic cylinder is pinned at both ends, it can
be considered as a two-force member and therefore exerts a force FBD directed
along its axis on the handle, as shown on the free-body diagram in Fig. a. From the
free-body diagram, FBD can be obtained by writing the moment equation of
equilibrium about point C.
a+ ©MC = 0;
FBD cos 15.52°(250) + FBD sin 15.52°(150) - 200 cos 30°(250 + 250)
-200 sin 30°(750 + 150) = 0
FBD = 628.42 N = 628 N
Ans.
Using the above result and writing the force equations of equilibrium along the x and
y axes, we have
+
: ©Fx = 0;
Cx + 200 cos 30° - 628.42 cos 15.52° = 0
Cx = 432.29 N = 432 N
+ c ©Fy = 0;
0.15 m
Ans.
200 sin 30° - 628.42 sin 15.52° + Cy = 0
Cy = 68.19 N = 68.2 N
Ans.
0.15 m
5–34.
A uniform glass rod having a length L is placed in the smooth hemispherical bowl having a
radius r. Determine the angle of inclination T for equilibrium.
Solution:
By Observation I = T.
Equilibirium :
6MA = 0; NB2 r cos T W
NB
L
2
cos T
WL
4r
6F x = 0; NA cos T W sin T
6F y = 0; W tan T sin T sin T
2 cos T
cos T
0
2
2
cos T
L
4r
L
WL
4r
2
NA
0
W cos T
1 2 cos T
cos T 1
2
0
2
L
4r
cos T
0
2
L 128 r
16 r
W tan T
T
§ L L2 128 r2 ·
¸
16 r
©
¹
acos ¨
Ans.
5–35.
The toggle switch consists of a cocking lever that is pinned to
a fixed frame at A and held in place by the spring which has
an unstretched length of 250 mm. Determine the magnitude
of the resultant force at A and the normal force on the peg
at B when the lever is in the position shown.
300 mm
30
100 mm
SOLUTION
B
l = 2(0.3)2 + (0.4)2 - 2(0.3)(0.4) cos 150° = 0.67664 m
sin 150°
sin u
=
;
0.3
0.67664
A
u = 12.808°
300 mm
Fs = ks = 5(0.67664 - 0.25) = 2.1332 N
a + ©MA = 0;
-(2.1332 sin 12.808°)(0.4) + NB (0.1) = 0
NB = 1.89159 N = 1.89 N
Q+ ©Fx = 0;
Ans.
Ax - 2.1332 cos 12.808° = 0
Ax = 2.0801 N
+a©Fy = 0;
Ay + 1.89159 - 2.1332 sin 12.808° = 0
Ay = -1.4187 N
FA = 2(2.0801) 2 + (-1.4187)2 = 2.52 N
Ans.
k 5 N/m
5–36.
The drainpipe of mass M is held in the tines of the fork lift. Determine the normal forces at A
and B as functions of the blade angle Tand plot the results of force (ordinate) versus T (abscissa)
for 0 d T d 90 deg.
Units used:
3
Mg
10 kg
Given:
M
1.4 Mg
a
0.4 m
g
9.81
m
2
s
Solution:
T
0 90
NA T
M g sin T deg
Ans.
3
10
NB T
M g cos T deg
Ans.
3
10
Force in kN
15
NA T 10
NB T
5
0
0
20
40
60
T
Angle in Degrees
80
100
5–37.
The boom supports the two vertical loads. Neglect the size
of the collars at D and B and the thickness of the boom,
and compute the horizontal and vertical components of
force at the pin A and the force in cable CB. Set
F1 = 800 N and F2 = 350 N.
C
3
5
4
SOLUTION
a + ©MA = 0;
1m
-800(1.5 cos 30°) - 350(2.5 cos 30°)
+
3
4
F CB (2.5 sin 30°) + FCB(2.5 cos 30°) = 0
5
5
1.5 m
B
D
F2
FCB = 781.6 = 782 N
+ ©F = 0;
:
x
Ax -
4
(781.6) = 0
5
Ay - 800 - 350 +
Ay = 681 N
30
A
Ax = 625 N
+ c ©Fy = 0;
Ans.
Ans.
3
(781.6) = 0
5
Ans.
F1
5–38.
The boom is intended to support two vertical loads, F1 and F2.
If the cable CB can sustain a maximum load of 1500 N before
it fails, determine the critical loads if F1 = 2F2. Also, what is
the magnitude of the maximum reaction at pin A?
C
3
5
4
SOLUTION
a + ©MA = 0;
1m
- 2F2(1.5 cos 30°) - F2(2.5 cos 30°)
+
4
3
(1500)(2.5 sin 30°) + (1500)(2.5 cos 30°) = 0
5
5
1.5 m
B
D
F2
F2 = 724 N
Ans.
F1 = 2F2 = 1448 N
A
F1 = 1.45 kN
+ ©F = 0;
:
x
Ax -
Ans.
4
(1500) = 0
5
Ax = 1200 N
+ c ©Fy = 0;
Ay - 724 - 1448 +
30
3
(1500) = 0
5
Ay = 1272 N
FA = 2(1200)2 + (1272)2 = 1749 N = 1.75 kN
Ans.
F1
5–39. The airstroke actuator at D is used to apply a force of
F = 200 N on the member at B. Determine the horizontal
and vertical components of reaction at the pin A and the
force of the smooth shaft at C on the member.
C
15⬚
600 mm
B
A
60⬚
200 mm
600 mm
F
D
5–40. The airstroke actuator at D is used to apply a force
of F on the member at B. The normal reaction of the
smooth shaft at C on the member is 300 N. Determine the
magnitude of F and the horizontal and vertical components
of reaction at pin A.
C
15⬚
600 mm
B
A
60⬚
200 mm
600 mm
F
D
2l
2l
 2l 
2l − W cosθ   = 0
2
.
l
.
.
l
5– 42.
The rigid metal strip of negligible weight is used as part of an electromagnetic switch. If the
stiffness of the springs at A and B is k, and the strip is originally horizontal when the springs are
unstretched, determine the smallest force needed to close the contact gap at C.
Units Used:
mN
10
3
N
Given:
a
50 mm
b
50 mm
c
10 mm
k
5
N
m
Solution:
Initial Guesses:
F
0.5 N
yA
yB
1 mm
1 mm
Given
c yA
yB yA
ab
a
§ yA ·
¨ ¸
¨ yB ¸
¨F¸
© ¹
Find yA yB F
k yA k yB F
§ yA ·
¨ ¸
© yB ¹
0
§ 2 ·
¨ ¸ mm
©4¹
k yB a F( a b)
F
10 mN
0
Ans.
5– 43.
The rigid metal strip of negligible weight is used as part of an electromagnetic switch. Determine
the maximum stiffness k of the springs at A and B so that the contact at C closes when the
vertical force developed there is F . Originally the strip is horizontal as shown.
Units Used:
mN
10
3
N
Given:
a
50 mm
b
50 mm
c
10 mm
F
0.5 N
Solution:
Initial Guesses: k
1
N
m
yA
yB
1 mm
1 mm
Given
c yA
yB yA
ab
a
§ yA ·
¨ ¸
¨ yB ¸
¨k ¸
© ¹
Find yA yB k
k yA k yB F
§ yA ·
¨ ¸
© yB ¹
k yB a F( a b)
0
§ 2 ·
¨ ¸ mm
©4¹
k
250
N
m
0
Ans.
5–44.
The upper portion of the crane boom consists of the jib AB,
which is supported by the pin at A, the guy line BC, and the
backstay CD, each cable being separately attached to the
mast at C. If the 5-kN load is supported by the hoist line,
which passes over the pulley at B, determine the magnitude
of the resultant force the pin exerts on the jib at A for
equilibrium, the tension in the guy line BC, and the tension
T in the hoist line. Neglect the weight of the jib. The pulley
at B has a radius of 0.1 m.
SOLUTION
C
1.5 m
0.1 m
T
5 kN
D
T(0.1) - 5(0.1) = 0;
T = 5 kN
Ans.
From the jib,
a + ©MA = 0;
-5(5) + TBC a
1.6
227.56
b (5) = 0
TBC = 16.4055 = 16.4 kN
+ c ©Fy = 0;
-Ay + (16.4055)a
1.6
227.56
Ans.
b - 5 = 0
Ay = 0
+ ©F = 0;
:
x
Ax - 16.4055a
5
227.56
b - 5 = 0
FA = Fx = 20.6 kN
r 0.1 m
5m
From pulley, tension in the hoist line is
a + ©MB = 0;
B
A
Ans.
5–45.
The device is used to hold an elevator door open. If the spring
has a stiffness of k = 40 N>m and it is compressed 0.2 m,
determine the horizontal and vertical components of reaction
at the pin A and the resultant force at the wheel bearing B.
150 mm
125 mm
k
A
100 mm
SOLUTION
B
Fs = ks = (40)(0.2) = 8 N
a + ©MA = 0;
- (8)(150) + FB(cos 30°)(275) - FB(sin 30°)(100) = 0
FB = 6.37765 N = 6.38 N
+ ©F = 0;
:
x
Ax - 6.37765 sin 30° = 0
Ax = 3.19 N
+ c ©Fy = 0;
Ans.
Ans.
Ay - 8 + 6.37765 cos 30° = 0
Ay = 2.48 N
Ans.
30
5–46.
Three uniform books, each having a weight W and length a,
are stacked as shown. Determine the maximum distance d
that the top book can extend out from the bottom one so
the stack does not topple over.
a
SOLUTION
Equilibrium: For top two books, the upper book will topple when the center of
gravity of this book is to the right of point A. Therefore, the maximum distance from
the right edge of this book to point A is a/2.
Equation of Equilibrium: For the entire three books, the top two books will topple
about point B.
a + ©MB = 0;
a
W(a- d) -W ad - b = 0
2
d =
3a
4
Ans.
d
5–47.
The horizontal beam is supported by springs at its ends.
Each spring has a stiffness of k = 5 kN>m and is originally
unstretched when the beam is in the horizontal position.
Determine the angle of tilt of the beam if a load of 800 N is
applied at point C as shown.
800 N
kB
kA
C
A
1m
SOLUTION
Equations of Equilibrium: The spring force at A and B can be obtained directly by
summing moment about points B and A, respectively.
a + ©MB = 0;
800(2) - FA(3) = 0
FA = 533.33 N
a + ©MA = 0;
FB(3) - 800(1) = 0
FB = 266.67 N
Spring Formula: Applying ¢ =
F
, we have
k
¢A =
533.33
= 0.1067 m
5(103)
¢B =
266.67
= 0.05333 m
5(103)
Geometry: The angle of tilt a is
a = tan - 1 a
0.05333
b = 1.02°
3
Ans.
B
2m
5–48.
The horizontal beam is supported by springs at its ends. If
the stiffness of the spring at A is kA = 5 kN> m , determine
the required stiffness of the spring at B so that if the beam is
loaded with the 800-N force it remains in the horizontal
position. The springs are originally constructed so that the
beam is in the horizontal position when it is unloaded.
800 N
kA
kB
C
A
1m
SOLUTION
Equations of Equilibrium: The spring forces at A and B can be obtained directly by
summing moments about points B and A, respectively.
a + ©MB = 0;
800(3) - FA(3) = 0
FA = 533.33 N
a + ©MA = 0;
FB(3) - 800(1) = 0
FB = 266.67 N
Spring Formula: Applying ¢ =
¢A =
F
, we have
k
533.33
= 0.1067 m
5(103)
¢B =
266.67
kB
Geometry: Requires, ¢ B = ¢ A. Then
266.67
= 0.1067
kB
kB = 2500 N>m = 2.50 kN>m
Ans.
B
2m
5–49.
The wheelbarrow and its contents have a mass of
m = 60 kg with a center of mass at G. Determine
the normal reaction on the tire and the vertical
force on each hand to hold it at u = 30°. Take
a = 0.3 m , b = 0.45 m , c = 0.75 m and d = 0.1 m .
G
r
u
b
SOLUTION
Equations of Equilibrium: Referring to the FBD of the wheelbarrow shown in Fig. a,
we notice that force P can be obtained directly by writing the moment equation of
equilibrium about A.
a+ ©MA = 0;
60(9.81) sin 30°(0.3) - 60(9.81) cos 30°(0.45)
+ 2P cos 30°(1.2) - 2P sin 30°(0.1) = 0
P = 71.315 N = 71.3 N
Ans.
Using this result to write the force equation of equilibrium along vertical,
+ c ©Fy = 0;
N + 2(71.315) - 60(9.81) = 0
N = 445.97 N = 446 N
d
a
Ans.
c
5–50.
The wheelbarrow and its contents have a mass m and center
of mass at G. Determine the greatest angle of tilt u without
causing the wheelbarrow to tip over.
G
r
d
a
u
b
SOLUTION
Require point G to be over the wheel axle for tipping. Thus
b cos u = a sin u
u = tan - 1
b
a
Ans.
c
5–51.
The rigid beam of negligible weight is supported
horizontally by two springs and a pin. If the springs are
uncompressed when the load is removed, determine the
force in each spring when the load P is applied. Also,
compute the vertical deflection of end C. Assume the spring
stiffness k is large enough so that only small deflections
occur. Hint: The beam rotates about A so the deflections in
the springs can be related.
P
A
C
B
k
L
SOLUTION
a + ©MA = 0;
3
FB(L) + FC(2L) - Pa Lb = 0
2
FB + 2FC = 1.5P
L
2L
=
¢B
¢C
¢ C = 2¢ B
FC
2FB
=
k
k
FC = 2FB
5FB = 1.5P
Deflection,
FB = 0.3P
Ans.
FC = 0.6P
Ans.
0.6P
k
Ans.
xC =
k
L/2
L/2
5–52.
A boy stands out at the end of the diving board, which is
supported by two springs A and B, each having a stiffness of
k = 15kN>m. In the position shown the board is horizontal.
If the boy has a mass of 40 kg, determine the angle of tilt
which the board makes with the horizontal after he jumps
off. Neglect the weight of the board and assume it is rigid.
1m
A
SOLUTION
Equations of Equilibrium: The spring force at A and B can be obtained directly by
summing moments about points B and A, respectively.
a + ©MB = 0;
FA (1) - 392.4(3) = 0
FA = 1177.2 N
a + ©MA = 0;
FB (1) - 392.4(4) = 0
FB = 1569.6 N
Spring Formula: Applying ¢ =
¢A =
F
, we have
k
1177.2
= 0.07848 m
15(103)
¢B =
1569.6
= 0.10464 m
15(103)
Geometry: The angle of tilt a is
a = tan - 1 a
0.10464 + 0.07848
b = 10.4°
1
Ans.
3m
B
5–53.
The uniform beam has a weight W and length l and is
supported by a pin at A and a cable BC. Determine the
horizontal and vertical components of reaction at A and
the tension in the cable necessary to hold the beam in the
position shown.
C
l
B
A
SOLUTION
Equations of Equilibrium: The tension in the cable can be obtained directly by
summing moments about point A.
a + ©MA = 0;
l
T sin 1f - u2l - W cos u a b = 0
2
T =
Using the result T =
+ ©F = 0;
:
x
a
Ans.
W cos u
2 sin 1f - u2
W cos u
bcos f - Ax = 0
2 sin 1f - u2
Ax =
+ c ©Fy = 0;
W cos u
2 sin 1f - u2
Ay + a
Ay =
W cos f cos u
2 sin 1f - u2
Ans.
W cos u
b sin f - W = 0
2 sin 1f - u2
W1sin f cos u - 2 cos f sin u2
2 sin f - u
Ans.
5–54.
Determine the distance d for placement of the load P for
equilibrium of the smooth bar in the position u as shown.
Neglect the weight of the bar.
d
u
a
SOLUTION
+ c ©Fy = 0;
R cos u - P = 0
a + ©MA = 0;
- P(d cos u) + Ra
Rd cos2 u = Ra
d =
a
b = 0
cos u
a
b
cos u
a
cos3 u
Ans.
Also;
Require forces to be concurrent at point O.
AO = d cos u =
a>cos u
cos u
Thus,
d =
a
cos3 u
Ans.
P
5–55.
If d = 1 m, and u = 30°, determine the normal reaction at
the smooth supports and the required distance a for the
placement of the roller if P = 600 N. Neglect the weight of
the bar.
d
SOLUTION
u
a
Equations of Equilibrium: Referring to the FBD of the rod shown in Fig. a,
a+ ©MA = 0;
a+ ©Fy¿ = 0;
+Q ©F = 0;
x¿
a
b - 600 cos 30°(1) = 0
cos 30°
450
NB =
a
NB = a
NB - NA sin 30° - 600 cos 30° = 0
NB - 0.5NA = 600 cos 30°
NA cos 30° - 600 sin 30° = 0
NA = 346.41 N = 346 N
Substitute this result into Eq (2),
NB - 0.5(346.41) = 600 cos 30°
NB = 692.82
N = 693 N
Ans.
Substitute this result into Eq (1),
450
a
a = 0.6495 m
692.82 =
a = 0.650 m
Ans.
P
5–56.
The disk B has a mass of 20 kg and is supported on the
smooth cylindrical surface by a spring having a stiffness of
k = 400 N>m and unstretched length of l0 = 1 m. The spring
remains in the horizontal position since its end A is attached
to the small roller guide which has negligible weight.
Determine the angle u for equilibrium of the roller.
0.2 m
A
r
u
SOLUTION
+ c ©Fy = 0;
R sin u - 20(9.81) = 0
+ ©F = 0;
:
x
R cos u - F = 0
tan u =
Since cos u =
1.0 +
20(9.81)
F
F
400
2.2
2.2 cos u = 1.0 +
20(9.81)
400 tan u
880 sin u = 400 tan u + 20(9.81)
Solving,
u = 27.1°
and u = 50.2°
B
k
Ans.
2m
5–57. The winch cable on a tow truck is subjected to a
force of T = 6 kN when the cable is directed at u = 60°.
Determine the magnitudes of the total brake frictional
force F for the rear set of wheels B and the total normal
forces at both front wheels A and both rear wheels B for
equilibrium. The truck has a total mass of 4 Mg and mass
center at G.
.
.
.
u
G
1.25 m
A
2m
B
2.5 m
F
1.5 m
3m
T
5–58. Determine the minimum cable force T and critical
angle u which will cause the tow truck to start tipping, i.e., for
the normal reaction at A to be zero. Assume that the truck is
braked and will not slip at B. The truck has a total mass of
4 Mg and mass center at G.x
1.25 m
A
2m
.
.
u
G
B
2.5 m
F
1.5 m
3m
T
5–59.
The thin rod of length l is supported by the smooth tube.
Determine the distance a needed for equilibrium if the
applied load is P.
a
A
2r
B
l
SOLUTION
2r
+ ©F = 0;
:
x
24 r2 + a 2
a + ©MA = 0;
-P¢
NB - P = 0
2r
24 r 2 + a 2
≤ l + NB 24r2 + a2 = 0
4 r2 l
- 24 r 2 + a2 = 0
4 r2 + a2
4 r2 l = A 4 r2 + a2 B 2
3
A 4 r2 l B 3 = 4 r2 + a2
2
2
a = 2(4 r2 l)3 - 4 r2
Ans.
P
5–60.
The 30-N uniform rod has a length of l = 1 m. If s = 1.5 m,
determine the distance h of placement at the end A along the
smooth wall for equilibrium.
C
SOLUTION
h
Equations of Equilibrium: Referring to the FBD of the rod shown in Fig. a, write
the moment equation of equilibrium about point A.
a + ©MA = 0;
A
T sin f(1) - 3 sin u(0.5) = 0
T =
s
1.5 sin u
sin f
l
B
Using this result to write the force equation of equilibrium along y axis,
a
+ c ©Fy = 0;
15 sin u
b cos (u - f) - 3 = 0
sin f
sin u cos (u - f) - 2 sin f = 0
(1)
Geometry: Applying the sine law with sin (180° - u) = sin u by referring to Fig. b,
sin f
sin u
=
;
h
1.5
sin u = a
h
b sin u
1.5
(2)
Substituting Eq. (2) into (1) yields
sin u[cos (u - f) -
4
h] = 0
3
since sin u Z 0, then
cos (u - f) - (4>3)h
cos (u - f) = (4>3)h
(3)
Again, applying law of cosine by referring to Fig. b,
l2 = h2 + 1.52 - 2(h)(1.5) cos (u - f)
cos (u - f) =
h2 + 1.25
3h
(4)
Equating Eqs. (3) and (4) yields
h2 + 1.25
4
h =
3
3h
3h2 = 1.25
h = 0.645 m
Ans.
5–61.
The uniform rod has a length l and weight W. It is supported
at one end A by a smooth wall and the other end by a cord
of length s which is attached to the wall as shown. Determine
the placement h for equilibrium.
C
h
s
A
SOLUTION
Equations of Equilibrium: The tension in the cable can be obtained directly by
summing moments about point A.
l
B
a + ©MA = 0;
l
T sin f(l) - W sin ua b = 0
2
T =
Using the result T =
+ c ©Fy = 0;
W sin u
2 sin f
W sin u
,
2 sin f
W sin u
cos (u - f) - W = 0
2 sin f
sin u cos (u - f) - 2 sin f = 0
(1)
Geometry: Applying the sine law with sin (180° - u) = sin u, we have
sin f
sin u
=
s
h
sin f =
h
sin u
s
(2)
Substituting Eq. (2) into (1) yields
cos (u - f) =
2h
s
(3)
Using the cosine law,
l2 = h2 + s2 - 2hs cos (u - f)
cos (u - f) =
h2 + s2 - l2
2hs
(4)
Equating Eqs. (3) and (4) yields
h2 + s 2 - l2
2h
=
s
2hs
h =
s2 - l 2
A 3
Ans.
5–62.
The uniform load has a mass of 600 kg and is lifted using a
uniform 30-kg strongback beam and four wire ropes as
shown. Determine the tension in each segment of rope and
the force that must be applied to the sling at A.
F
1.25 m
A
B
2m
SOLUTION
Equations of Equilibrium: Due to symmetry, all wires are subjected to the same
tension. This condition statisfies moment equilibrium about the x and y axes and
force equilibrium along y axis.
©Fz = 0;
4
4Ta b - 5886 = 0
5
T = 1839.375 N = 1.84 kN
Ans.
The force F applied to the sling A must support the weight of the load and
strongback beam. Hence
©Fz = 0;
F - 60019.812 - 3019.812 = 0
F = 6180.3 N = 6.18 kN
1.25 m
Ans.
1.5 m
1.5 m
C
5– 63.
Due to an unequal distribution of fuel in the wing tanks, the centers of gravity for the airplane
fuselage A and wings B and C are located as shown. If these components have weights WA,
WB and W C, determine the normal reactions of the wheels D, E, and F on the ground.
Units Used:
3
kN 10 N
Given:
WA 225kN
WB 40kN
WC 30kN
a 2.4m
e 6m
b 1.8m f 1.2m
c 2.4m
g 0.9m
d 1.8m
Solution:
Initial guesses:
RD 1kN
RE 1kN
RF 1kN
Given
6Mx = 0;
WB˜ b RD˜ ( a b) WC˜ c RE˜ ( c d) = 0
6My = 0;
WB˜ f WA˜ ( g f) WC f RF ˜ ( e g f) = 0
6Fz = 0;
RD RE RF WA WB WC = 0
§ RD ·
¨ ¸
¨ RE ¸ Find RD RE RF
¨ ¸
© RF ¹
§ RD ·
¨ ¸
¨ RE ¸
¨ ¸
© RF ¹
§ 113.1 ·
¨ 113.1 ¸ kN
¨
¸
© 68.7 ¹
Ans.
5–64.
z
The wing of the jet aircraft is subjected to a thrust of
T = 8 kN from its engine and the resultant lift force
L = 45 kN. If the mass of the wing is 2.1 Mg and the mass
center is at G, determine the x, y, z components of reaction
where the wing is fixed to the fuselage at A.
5m
A
G
3m
7m
x
©Fx = 0;
- Ax + 8000 = 0
T
Ax = 8.00 kN
Ans.
©Fy = 0;
Ay = 0
Ans.
©Fz = 0;
- Az - 20 601 + 45 000 = 0
Az = 24.4 kN
©My = 0;
Ans.
45 000(15) - 20 601(5) - Mx = 0
Mx = 572 kN # m
©Mz = 0;
Ans.
My - 2.5(8000) = 0
My = 20.0 kN # m
©Mx = 0;
y
2.5 m
SOLUTION
Ans.
Mz - 8000(8) = 0
Mz = 64.0 kN # m
Ans.
8 kN
L
45 kN
.
.
.
.
.
.
.
.
.
.
.
5–68.
z
Determine the force components acting on the ball-andsocket at A, the reaction at the roller B and the tension on the
cord CD needed for equilibrium of the quarter circular plate.
D
350 N
200 N
2m
1m
A
x
SOLUTION
Equations of Equilibrium: The normal reactions NB and Az can be obtained directly
by summing moments about the x and y axes, respectively.
©Mx = 0;
NB(3) - 200(3) - 200(3 sin 60°) = 0
NB = 373.21 N = 373 N
©My = 0;
350(2) + 200(3 cos 60°) - Az(3) = 0
A z = 333.33 N = 333 N
©Fz = 0;
Ans.
Ans.
TCD + 373.21 + 333.33 - 350 - 200 - 200 = 0
TCD = 43.5 N
Ans.
©Fx = 0;
Ax = 0
Ans.
©Fy = 0;
Ay = 0
Ans.
C
60
200 N
3m
B
y
z
5–69. Determine the components of reaction acting at the
smooth journal bearings A, B, and C.
450 N
C
0.6 m
300 N m 45⬚
A
0.4 m
x
B
0.8 m
0.4 m
y
5–70. The circular plate has a weight W and center of
gravity at its center. If it is supported by three vertical cords
tied to its edge, determine the largest distance d from the
center to where any vertical force P can be applied so as not
to cause the force in any one of the cables to become zero.
B
P
120⬚
120⬚
r
A
.
120⬚
d
C
5–71.
z
Determine the support reactions at the smooth collar A and
the normal reaction at the roller support B.
A
800 N
600 N
B
SOLUTION
x
Equations of Equilibrium: From the free-body diagram, Fig. a, NB, (MA)z, and A y
can be obtained by writing the moment equations of equilibrium about the x and
z axes and the force equation of equilibrium along the y axis.
©Mx = 0;
NB(0.8 + 0.8) - 800(0.8) - 600(0.8 + 0.8) = 0
NB = 1000 N
©Mz = 0;
(MA)z = 0
Ans.
Ans.
©Fy = 0;
Ay = 0
Ans.
Using the result NB = 1000 N and writing the moment equation of equilibrium
about the y axis and the force equation of equilibrium along the z axis, we have
©My = 0;
©Fz = 0;
(MA)y - 600(0.4) + 1000(0.8) = 0
(MA)y = -560 N # m
Ans.
A z + 1000 - 800 - 600 = 0
A z = 400 N
Ans.
The negative sign indicates that (M A)y acts in the opposite sense to that shown on
the free-body diagram. If we write the force equation of equilibrium along the
x axis, ©Fx = 0, and so equilibrium is satisfied.
0.8 m
0.8 m
0.4 m
0.4 m
y
5–72.
z
The pole is subjected to the two forces shown. Determine
the components of reaction of A assuming it to be a balland-socket joint. Also, compute the tension in each of the
guy wires, BC and ED.
30°
20°
2 m F = 450 N
2
45°
E
D
F1 = 860 N
SOLUTION
B
3m
6m
Force Vector and Position Vectors:
6m
FA = Ax i + Ay j + Az k
C
F1 = 8605cos 45°i - sin 45°k6 N = 5608.11i - 608.11k6 N
4m
4.5 m
F2 = 4505-cos 20° cos 30°i + cos 20° sin 30°k - sin 20°k6 N
x
1-6 - 02i + 1-3 - 02j + 10 - 62k
21-6 - 022 + 1 -3 - 022 + 10 - 622
R
2
1
2
= - FEDi - FEDj - FEDk
3
3
3
FBC = FBC B
=
16 - 02i + 1-4.5 - 02j + 10 - 42k
216 - 022 + 1-4.5 - 022 + 10 - 422
R
9
8
12
F i F j F k
17 BC
17 BC
17 BC
r1 = 54k6 m
r2 = 58k6 m
r3 = 56k6 m
Equations of Equilibrium: Force equilibrium requires
©F = 0;
FA + F1 + F2 + FED + FBC = 0
a Ax + 608.11 - 366.21 -
2
12
+
F
F bi
3 ED
17 BC
+ aAy + 211.43 -
1
9
FED FBC bj
3
17
+ aAz - 608.11 - 153.91 -
2
8
F
F bk
3 ED
17 BC
0
Equating i, j and k components, we have
©Fx = 0;
Ax + 608.11 - 366.21 -
©Fy = 0;
Ay + 211.43 -
©Fz = 0;
Az - 608.11 - 153.91 -
2
12
F
+
F = 0
3 ED
17 BC
1
9
FED F = 0
3
17 BC
2
8
F
F = 0
3 ED
17 BC
6m
y
= 5-366.21i + 211.43j - 153.91k6 N
FED = FED B
A
(1)
(2)
(3)
5–72. (continued)
Moment equilibrium requires
©MA = 0;
4k * a
r1 * FBC + r2 * 1F1 + F22 + r3 * FED = 0
9
8
12
F i F j F kb
17 BC
17 BC
17 BC
+ 8k * 1241.90i + 211.43j - 762.02k2
2
1
2
+ 6k * a - FEDi - FEDj - FEDkb = 0
3
3
3
Equating i, j and k components, we have
©Mx = 0;
36
FBC + 2FED - 1691.45 = 0
17
(4)
©My = 0;
48
F - 4FED + 1935.22 = 0
17 BC
(5)
Solving Eqs. (4) and (5) yields
FBC = 205.09 N = 205 N
FED = 628.57 N = 629 N
Ans.
Substituting the results into Eqs. (1), (2) and (3) yields
Ax = 32.4 N
Ay = 107 N
Az = 1277.58 N = 1.28 kN
Ans.
5–73. If P = 6 kN, x = 0.75 m and y = 1 m, determine
the tension developed in cables AB, CD, and EF. Neglect
the weight of the plate.
z
B
F
P
x
D
A
y
E
x
C
2m
2m
y
z
5–74. Determine the location x and y of the point of
application of force P so that the tension developed in
cables AB, CD, and EF is the same. Neglect the weight of
the plate.
B
F
P
x
D
A
y
E
x
C
2m
2m
y
5–75.
z
If the pulleys are fixed to the shaft, determine the magnitude
of tension T and the x, y, z components of reaction at the
smooth thrust bearing A and smooth journal bearing B.
1m
1m
A
1m
0.2 m
0.3 m
x
Equations of Equilibrium: From the free-body diagram of the shaft, Fig. a, A y, T,
and Bx can be obtained by writing the force equation of equilibrium along the y axis
and the moment equations of equilibrium about the y and z axes, respectively.
©Fy = 0;
Ay = 0
©My = 0;
400(0.2) - 900(0.2) - 900(0.3) + T(0.3) = 0
©Mz = 0;
-Bx(3) - 400(1) - 900(1) = 0
Ans.
T = 1233.33 N = 1.23 kN
Bx = -433.33 N = -433N
Ans.
Ans.
Using the above results and writing the moment equation of equilibrium about the
x axis and the force equation of equilibrium along the x axis, we have
©Mx = 0;
Bz(3) - 900(2) - 1233.33(2) = 0
Bz = 1422.22 N = 1.42 kN
©Fx = 0;
Ans.
400 + 900 - 433.33 - A x = 0
A x = 866.67 N = 867 N
Ans.
Finally, writing the force equation of equilibrium along the z axis, yields
©Fz = 0;
y
{400 i} N
{900 i} N
SOLUTION
A z - 1233.33 - 900 + 1422.22 = 0
A z = 711.11 N = 711 N
Ans.
B
{⫺900 k} N
T
5 –76.
A ball of mass M rests between the grooves A and B of the incline and against a vertical wall at
C. If all three surfaces of contact are smooth, determine the reactions of the surfaces on the
ball. Hint: Use the x, y, z axes, with origin at the center of the ball, and the z axis inclined as
shown.
Given:
M
2 kg
T1
10 deg
T2
45 deg
Solution:
6F x = 0; F c cos T 1 M g sin T 1
Fc
M g˜ tan T 1
Fc
0.32 kg˜ m
6F y = 0; NA cos T 2 NB cos T 2
NA
0
Ans.
0
NB
6F z = 0; 2 NA sin T 2 M g cos T 1 F c sin T 1
NA
NA
N
1 M g˜ cos T 1 F c˜ sin T 1
˜
2
sin T 2
1.3 kg˜ m
NA
NB
Ans.
0
5–77. A uniform square table having a weight W and sides
a is supported by three vertical legs. Determine the smallest
vertical force P that can be applied to its top that will cause
it to tip over.
a/2
a/2
a
.
z
5–78. The shaft is supported by three smooth journal
bearings at A, B, and C. Determine the components of
reaction at these bearings.
900 N
600 N
450 N
C
0.6 m
A
0.9 m
0.6 m
x
0.9 m
500 N
0.9 m
B
0.9 m
0.9 m
y
5 – 79.
The platform has mass M and center of mass located at G. If it is lifted using the three cables,
determine the force in each of these cables.
Units Used:
3
3
Mg 10 kg
g 9.81
kN 10 N
m
2
s
Given:
M 3Mg
a 4m
b 3m
c 3m
d 4m
e 2m
Solution:
The initial guesses are:
FAC 10N
FBC 10N
FDE 10N
Given
b˜ FAC
2
a b
2
c˜ FBC
2
2
= 0
a c
M˜ g˜ e FAC˜ a˜
de
2
a b
a
2
2
2
F BC˜
a˜ ( d e)
2
2
= 0
a c
˜ FBC( b c) M˜ g˜ b FDE˜ b = 0
a c
§ FAC ·
¨
¸
F
¨ BC ¸ Find FAC FBC FDE
¨F ¸
© DE ¹
§ FAC ·
¨
¸
F
¨ BC ¸
¨
¸
© FDE ¹
§ 6.13 ·
¨ 6.13 ¸ kN
¨
¸
© 19.62 ¹
Ans.
5– 80.
The platform has a mass of M and center of mass located at G. If it is lifted using the three
cables, determine the force in each of the cables. Solve for each force by using a single
moment equation of equilibrium.
Units Used:
Mg 1000kg
3
kN 10 N
g 9.81
m
2
s
Given:
M 2Mg
c 3m
a 4m
d 4m
b 3m
e 2m
Solution:
§¨ 0 ¸·
rBC ¨ c ¸
¨a ¸
© ¹
§¨ 0 ¸·
rAC ¨ b ¸
¨a ¸
© ¹
§¨ e d ¸·
rAD ¨ b ¸
¨ 0 ¸
©
¹
§¨ d e ¸·
rBD ¨ c ¸
¨ 0 ¸
©
¹
First find FDE.
6My' = 0;
Next find FBC.
FDE˜ ( d e) M˜ g˜ d = 0
Guess
FDE M˜ g˜ d
de
FDE
13.1 kN
FBC 1kN
ª«§¨ e ¸· §¨ 0 ·¸ §¨ e d ¸·
§
Given «¨ 0 ¸ u ¨ 0 ¸ ¨ c ¸ u ¨ FBC˜
«¨ 0 ¸ ¨ M˜ g ¸ ¨ 0 ¸ ©
¬© ¹ ©
¹ ©
¹
º
·»
¸» ˜ r = 0
rBC » AD
¹¼
rBC
FBC Find F BC
FBC
4.09 kN
Ans.
Now find FAC.
Guess
FAC 1kN
ª«§¨ e ·¸ §¨ 0 ·¸ §¨ e d ·¸
§
Given «¨ 0 ¸ u ¨ 0 ¸ ¨ b ¸ u ¨ FAC˜
«¨ 0 ¸ ¨ M˜ g ¸ ¨ 0 ¸ ©
¬© ¹ ©
¹ ©
¹
º
·»
¸» ˜ rBD = 0
rAC ¹»
¼
rAC
FAC Find FAC
FAC
4.09 kN
Ans.
5–81.
z
The sign has a mass of 100 kg with center of mass at G.
Determine the x, y, z components of reaction at the ball-andsocket joint A and the tension in wires BC and BD.
1m
D
2m
C
SOLUTION
1m
2m
Equations of Equilibrium: Expressing the forces indicated on the free-body
diagram, Fig. a, in Cartesian vector form, we have
A
FA = A xi + A yj + A zk
x
B
W = {-100(9.81)k} N = {-981k} N
FBD = FBDuBD = FBD ≥
FBC = FBCuBC = FBC ≥
G
(-2 - 0)i + (0 - 2)j + (1 - 0)k
2(-2 - 0) + (0 - 2) + (1 - 0)
2
2
2
¥ = a-
2
2
1
F i - FBDj + FBDkb
3 BD
3
3
1m
(1 - 0)i + (0 - 2)j + (2 - 0)k
1
2
2
¥ = a FBCi - FBCj + FBCkb
3
3
3
2(1 - 0) + (0 - 2) + (2 - 0)
2
2
2
Applying the forces equation of equilibrium, we have
©F = 0;
FA + FBD + FBC + W = 0
2
2
1
1
2
2
(A xi + A yj + A zk) + a - FBDi - FBDj + FBDkb + a FBCi - FBCj + FBCkb + ( - 981 k) = 0
3
3
3
3
3
3
a Ax -
2
1
2
2
1
2
F + FBC b i + a A y - FBD - FBC bj + aA z + FBD + FBC - 981 bk = 0
3 BD
3
3
3
3
3
Equating i, j, and k components, we have
Ax -
2
1
F + FBC = 0
3 BD
3
(1)
Ay -
2
2
F - FBC = 0
3 BD
3
(2)
Az +
1
2
F + FBC - 981 = 0
3 BD
3
(3)
In order to write the moment equation of equilibrium about point A, the position
vectors rAG and rAB must be determined first.
rAG = {1j} m
rAB = {2j} m
Thus,
©M A = 0; rAB * (FBC + FBD) + (rAG * W) = 0
2
2
2
2
1
1
(2j) * c a FBC - FBD bi - a FBC + FBD bj + a FBC + FBD bk d + (1j) * ( - 981k) = 0
3
3
3
3
3
3
2
4
2
4
a FBC + FBD - 981 b i + a FBD - FBC b k = 0
3
3
3
3
Equating i, j, and k components we have
4
2
F + FBC - 981 = 0
3 BC
3
(4)
4
2
F - FBC = 0
3 BC
3
(5)
1m
y
5–81. (continued)
Soving Eqs. (1) through (5), yields
FBD = 294.3 N = 294 N
Ans.
FBC = 588.6 N = 589 N
Ans.
Ax = 0
Ans.
Ay = 588.6 N = 589 N
Ans.
Az = 490.5 N
Ans.
5–82.
Determine the tension in cables BD and CD and the x, y, z components of reaction at the
ball-and-socket joint at A.
Given:
F
300 N
a
3m
b
1m
c
0.5 m
d
1.5 m
Solution:
rBD
§ b ·
¨d¸
¨ ¸
©a¹
rCD
§ b ·
¨ d ¸
¨ ¸
©a¹
Initial Guesses:
TBD
1N
TCD
1N
Ax
1N
Ay
Az
1N
Given
§ Ax ·
§ 0 ·
¨ ¸
rBD
rCD
TCD
¨ 0 ¸
¨ Ay ¸ TBD
¨ ¸
rBD
rCD
¨A ¸
© F ¹
© z¹
0
§d·
§d·
§d c· § 0 ·
rBD · ¨ ¸ §
rCD · ¨
¨ d ¸ u § T
d u T
0 ¸u¨ 0 ¸
¨ ¸ ¨© BD rBD ¸¹ ¨ ¸ ¨© CD rCD ¸¹ ¨
¸ ¨ ¸
©0¹
©0¹
© 0 ¹ © F ¹
0
1N
§ TBD ·
¨
¸
¨ TCD ¸
¨ Ax ¸
¨
¸
¨ Ay ¸
¨ A ¸
© z ¹
Find TBD TCD A x A y A z
§ TBD ·
¨
¸
© TCD ¹
§ Ax ·
¨ ¸
¨ Ay ¸
¨A ¸
© z¹
§ 116.7 ·
¨
¸N
© 116.7 ¹
§ 66.7 ·
¨ 0 ¸N
¨
¸
© 100 ¹
Ans.
Ans.
5–83.
Both pulleys are fixed to the shaft and as the shaft turns
with constant angular velocity, the power of pulley A is
transmitted to pulley B. Determine the horizontal tension T
in the belt on pulley B and the x, y, z components of
reaction at the journal bearing C and thrust bearing D if
u = 0°. The bearings are in proper alignment and exert only
force reactions on the shaft.
z
200 mm
©Fx = 0;
©Fy = 0;
80 N
6510.082 - 8010.082 + T10.152 - 5010.152 = 0
Ans.
165 + 80210.452 - Cz 10.752 = 0
Ans.
150 + 58.0210.22 - Cy 10.752 = 0
Cy = 28.8 N
Ans.
Dx = 0
Ans.
Dy + 28.8 - 50 - 58.0 = 0
Dy = 79.2 N
©Fz = 0;
y
80 mm
A
65 N
Cz = 87.0 N
©Mz = 0;
150 mm
B
T
T = 58.0 N
©My = 0;
θ
C
Equations of Equilibrium:
©Mx = 0;
D
300 mm
x
SOLUTION
50 N
250 mm
Ans.
Dz + 87.0 - 80 - 65 = 0
Dz = 58.0 N
Ans.
5–84.
Both pulleys are fixed to the shaft and as the shaft turns
with constant angular velocity, the power of pulley A is
transmitted to pulley B. Determine the horizontal tension T
in the belt on pulley B and the x, y, z components of
reaction at the journal bearing C and thrust bearing D if
u = 45°. The bearings are in proper alignment and exert
only force reactions on the shaft.
SOLUTION
z
200 mm
©Fx = 0;
©Fy = 0;
B
y
80 mm
A
65 N
80 N
6510.082 - 8010.082 + T10.152 - 5010.152 = 0
Ans.
165 + 80210.452 - 50 sin 45°10.22 - Cz 10.752 = 0
Ans.
58.010.22 + 50 cos 45°10.22 - Cy 10.752 = 0
Cy = 24.89 N = 24.9 N
Ans.
Dx = 0
Ans.
Dy + 24.89 - 50 cos 45° - 58.0 = 0
Dy = 68.5 N
©Fz = 0;
150 mm
C
T
Cz = 77.57 N = 77.6 N
©Mz = 0;
θ
x
T = 58.0 N
©My = 0;
D
300 mm
Equations of Equilibrium:
©Mx = 0;
50 N
250 mm
Ans.
Dz + 77.57 + 50 sin 45° - 80 - 65 = 0
Dz = 32.1 N
Ans.
5–85.
If the roller at B can sustain a maximum load of
3 kN, determine the largest magnitude of each of the three
forces F that can be supported by the truss.
A
2m
45
2m
2m
2m
SOLUTION
Equations of Equilibrium: The unknowns Ax and Ay can be eliminated by summing
moments about point A.
a + ©MA = 0;
F(6) + F(4) + F(2) - 3 cos 45°(2) = 0
F = 0.3536 kN = 354 N
Ans.
F
F
F
B
5–86.
Determine the normal reaction at the roller A and horizontal
and vertical components at pin B for equilibrium of the
member.
10 kN
0.6 m
0.6 m
A
SOLUTION
Equations of Equilibrium: The normal reaction NA can be obtained directly by
summing moments about point B.
a + ©MA = 0;
6 kN
0.8 m
10(0.6 + 1.2 cos 60°) + 6 (0.4) - NA (1.2 + 1.2 cos 60°) = 0
NA = 8.00 kN
+ ©F = 0;
:
x
Bx - 6 cos 30° = 0
+ c ©Fy = 0;
By + 8.00 - 6 sin 30° - 10 = 0
Bx = 5.20 kN
By = 5.00 kN
60
0.4 m
Ans.
Ans.
Ans.
B
5–87.
The symmetrical shelf is subjected to a uniform load of 4 kPa.
Support is provided by a bolt (or pin) located at each end A
and A¿ and by the symmetrical brace arms, which bear
against the smooth wall on both sides at B and B¿. Determine
the force resisted by each bolt at the wall and the normal
force at B for equilibrium.
A¿
B¿
A
1.5 m
0.15 m
SOLUTION
B
Equations of Equilibrium: Each shelf’s post at its end supports half of the applied
load, ie, 4000 (0.2) (0.75) = 600 N. The normal reaction NB can be obtained directly
by summing moments about point A.
a + ©MA = 0;
NB (0.15) - 600(0.1) = 0
NB = 400 N
Ans.
FA = 2A2x + A2y = 24002 + 6002 = 721 N
Ans.
+ ©F = 0;
:
x
400 - Ax = 0
Ax = 400 N
+ c ©Fy = 0;
Ay - 600 = 0
Ay = 600 N
The force resisted by the bolt at A is
0.2 m
4 kPa
5–88.
z
Determine the x and z components of reaction at the
journal bearing A and the tension in cords BC and BD
necessary for equilibrium of the rod.
C
3m
A
3m
2m
D
SOLUTION
x
F1 = 5-800k6 N
6m
F2 = 5350j6 N
FBC = FBC
1-3j + 4k2
F2
B
5
= 5-0.6FBCj + 0.8FBCk6 N
FBD = FBD
F1
13j + 4k2
5
= 50.6FBDj + 0.8FBDk6 N
©Fx = 0;
Ax = 0
©Fy = 0;
350 - 0.6FBC + 0.6FBD = 0
©Fz = 0;
Az - 800 + 0.8FBC + 0.8FBD = 0
©Mx = 0;
©My = 0;
©Mz = 0;
Ans.
(MA)x + 0.8FBD162 + 0.8FBC162 - 800162 = 0
800122 - 0.8FBC122 - 0.8FBD122 = 0
(MA)z - 0.6FBC122 + 0.6FBD122 = 0
FBD = 208 N
Ans.
FBC = 792 N
Ans.
Az = 0
Ans.
(MA)x = 0
Ans.
(MA)z = 700 N # m
Ans.
{ 800 k} N
4m
{350j} N
y
5–89.
The uniform rod of length L and weight W is supported on
the smooth planes. Determine its position u for equilibrium.
Neglect the thickness of the rod.
L
u
c
SOLUTION
a + ©MB = 0;
-Wa
L
cos u b + NA cos f (L cos u) + NA sin f (L sin u) = 0
2
NA =
W cos u
2 cos (f - u)
+ ©F = 0;
:
x
NB sin c - NA sin f = 0
+ c ©Fy = 0;
NB cos c + NA cos f - W = 0
NB =
W - NA cos f
cos c
(1)
(2)
(3)
Substituting Eqs. (1) and (3) into Eq. (2):
aW -
W cos u sin f
W cos u cos f
b tan c = 0
2 cos (f - u)
2 cos (f - u)
2 cos (f - u) tan c - cos u tan c cos f - cos u sin f = 0
sin u (2 sin f tan c) - cos u (sin f - cos f tan c) = 0
tan u =
sin f - cos f tan c
2 sin f tan c
u = tan - 1 a
1
1
cot c - cot f b
2
2
Ans.
f
5–90. Determine the horizontal and vertical components
of force at the pin A and the reaction at the rocker B of the
curved beam.
500 N
200 N
10⬚ 15⬚
2m
A
.
.
.
B
5–91.
z
Determine the x, y, z components of reaction at the fixed
wall A. The 150-N force is parallel to the z axis and
the 200-N force is parallel to the y axis.
150 N
A
x
2m
y
1m
SOLUTION
2.5 m
©Fx = 0;
Ax = 0
©Fy = 0;
Ay + 200 = 0
Ay = -200 N
©Fz = 0;
2m
Ans.
Az - 150 = 0
Az = 150 N
©Mx = 0;
Ans.
200 N
Ans.
-150(2) + 200(2) - (MA)x = 0
(MA)x = 100 N # m
Ans.
©My = 0;
(MA)y = 0
Ans.
©Mz = 0;
200(2.5) - (MA)z = 0
(MA)z = 500 N # m
Ans.
5–92.
Determine the reactions at the supports A and B for
equilibrium of the beam.
400 N/m
200 N/m
A
4m
SOLUTION
Equations of Equilibrium: The normal reaction of NB can be obtained directly by
summing moments about point A.
+ ©MA = 0;
NB(7) - 1400(3.5) - 300(6) = 0
NB = 957.14 N = 957 N
Ag - 1400 - 300 + 957 = 0
+
: ©Fx = 0;
B
Az = 0
Ans.
Ag = 743 N
Ans.
3m