Chapter 4
PROBLEM 4.1
The boom on a 4300-kg truck is used to unload a pallet of shingles of
mass 1600 kg. Determine the reaction at each of the two (a) rear
wheels B, (b) front wheels C.
SOLUTION
WA
mA g
1600 kg 9.81 m/s 2 15696 N
or
WA
15.696 kN
WG
mG g
4300 kg 9.81 m/s2 42 183 N
WG
or
42.183 kN
(a) From f.b.d. of truck with boom
6M C
0:
15.696 kN ª¬ 0.5 0.4 6 cos15q m º¼ 2FB ª¬ 0.5 0.4 4.3 m º¼
42.183 kN 0.5 m ? 2 FB
0
126.185
5.2
24.266 kN
or FB
12.13 kN W
or FC
16.81 kN W
(b) From f.b.d. of truck with boom
6M B
0:
15.696 kN ª¬ 6 cos15q 4.3 m º¼ 42.183 kN ª¬ 4.3 0.4 m º¼
2 FC ª¬ 4.3 0.9 m º¼
? 2 FC
Check:
6Fy
0:
0
174.786
5.2
33.613 kN
33.613 42.183 24.266 15.696 kN
57.879 57.879 kN
0 ok
0?
PROBLEM 4.14
For the beam and loading shown, determine the range of values of the
distance a for which the reaction at B does not exceed 225 N
downward or 450 N upward.
675 N
675 N
75 mm
112.5 N
100 mm
150 mm
50 mm
SOLUTION
675 N
To determine amax the two 675 N forces need to be as close to B without
having the vertical upward force at B exceed 450 N
675 N
From f.b.d. of beam with B = 450 N
ÂMD = 0:
112.5 N
– (675 N) (amax – 100 mm) – (675 N) (amax – 25 mm)
– (112.5 N) (50 mm) + (450 N) (200 mm) = 0
450 N
amax = 125 mm
or
675 N
675 N
225 N
To determine amin the two 675 N forces need to be as close to A without
having the vertical downward force at B exceed 225 N.
From f.b.d. of beam with B = 225 N
ÂMD = 0:
112.5 N
(675 N) (100 mm – amin) – (675 N) (amin – 25 mm)
– (112.5 N) (50 mm) – (225 N) (200 mm) = 0
or
Therefore,
amin = 25 mm
or 25 mm £ a £ 125 mm J
40 mm
D
A follower ABCD is held against a circular cam by a stretched spring,
which exerts a force of 21 N for the position shown. Knowing that the
tension in rod BE is 14 N, determine (a) the force exerted on the roller
at A, (b) the reaction at bearing C.
50 mm
C
B
E
20 mm
PROBLEM 4.15
20 mm
A
60°
SOLUTION
Note: From f.b.d. of ABCD
A x = A cos 60 q
A
2
Ay
A
A sin 60q
3
2
(a) From f.b.d. of ABCD
6M C
§ A·
0: ¨ ¸ 40 mm 21 N 40 mm ©2¹
14 N 20 mm ? A
0
28 N
or A
28.0 N
60q W
(b) From f.b.d. of ABCD
6Fx
? Cx
6Fy
? Cy
Then
and
0: C x 14 N 28 N cos 60q
28 N
0:
Cx
or
0
28.0 N
C y 21 N 28 N sin 60q
3.2487 N
C
C x2 C y2
T
§ Cy ·
tan 1 ¨
¸
© Cx ¹
or
Cy
3.25 N
282 3.2487 2
§ 3.2487 ·
tan 1 ¨
¸
© 28 ¹
or C
0
28.188 N
6.6182q
28.2 N
6.62q W
B
A
PROBLEM 4.24
90 N
90 N
300 mm
300 mm
B
90 Nm 150 mm
9 Nm 150 mm
45 N 45°
45 N
50 mm
50 mm
A 225 mm
225 mm
(a)
(b)
SOLUTION
A steel rod is bent to form a mounting bracket. For each of the
mounting brackets and loadings shown, determine the reactions at
A and B.
(a) From f.b.d. of mounting bracket
 MA = 0:
(a)
– B (200 mm) – (90 N) (300 mm)
90 N
+ (45 N) (50 mm) – 9000 N.mm = 0
\ B = – 168.75 N
J
or B = 168.8 N
9000 N.mm
45 N
 Fx = 0:
– 168.75 N – 45 N + Ax = 0
\ Ax = 213.75 N
Ax = 213.8 N
or
 Fy = 0:
– 90 N + Ay = 0
\ Ay = 90 N
Ay = 90 N
or
b213.75g + b90g = 231.92 N
FG A IJ = tan FG 90 IJ = 22.834°
HA K
H 213.75K
2
Ax2 + Ay2 =
Then
A =
and
q = tan –1
y
2
–1
x
or A = 231.9
22.8° J
(b) From f.b.d. of mounting bracket
(b)
 MA = 0:
90 N
– (B cos 45°) (200 mm) – (90 N) (300 mm)
– 9000 N.mm + (45 N) (50 mm) = 0
\ B = – 238.65 N
9000 N.mm
45 N
or B = 238.65
 Fx = 0:
Ax + (– 238.65) cos 45° – 45 N = 0
\ Ax = 213.75 N
or
Ax = 213.8 N
45° J
PROBLEM 4.24 CONTINUED
 Fy = 0:
Ay – (238.65 N) sin 45° – 90 N = 0
\ Ay = – 78.75 N
or
Ay = 78.8 N
b213.75g + b78.75g = 227.795 N
F A I = tan FG - 78.75IJ = – 20.225°
GH A JK
H 213.75K
2
Ax2 + Ay2 =
Then
A=
and
q = tan –1
y
2
–1
x
or A = 227.8 N
20.2°J
PROBLEM 4.31
Neglecting friction, determine the tension in cable ABD and the reaction
at support C.
625 mm
135 N
250 mm 250 mm
SOLUTION
From f.b.d. of inverted T-member
 MC = 0: T (625 mm) – T (250 mm) – (135 N) (250 mm) = 0
\ T = 90 N
or T = 90 N J
ÂFx = 0: Cx – 90 N = 0
\ Cx = 90 N
135 N
Cx= 90 N
 Fy = 0:
Cy + 90 N – 135 N = 0
\ Cy = 45 N
or
Cy = 45 N
C x2 + C y2 =
Then
C =
and
q = tan –1
b90g + b45g
2
2
= 100.6 N
FG C IJ = tan FG 45IJ = 26.565°
H 90 K
HC K
y
–1
x
or
C = 100.6 N
26.6° J
PROBLEM 4.39
C
B
q
A
45°
25 N
60°
O
Rod ABCD is bent in the shape of a circular arc of radius 80 mm and
rests against frictionless surfaces at A and D. Knowing that the collar
at B can move freely on the rod, determine (a) the value of T for
which the tension in cord OB is as small as possible, (b) the
corresponding value of the tension, (c) the reactions at A and D.
D
45°
SOLUTION
(a) From f.b.d. of rod ABCD
6M E
0:
25 N cos 60q dOE T cosT dOE 12.5 N
cosT
T
or
0
(1)
? T is minimum when cosT is maximum,
or T
0q W
(b) From Equation (1)
T
12.5 N
cos 0
12.5 N
or Tmin
6Fx
(c)
0: N A cos 45q N D cos 45q 12.5 N
25 N cos 60q
0
? ND N A
0
ND
or
6Fy
12.50 N W
NA
(2)
0: N A sin 45q N D sin 45q 25 N sin 60q
? ND N A
30.619 N
0
(3)
Substituting Equation (2) into Equation (3),
2N A
NA
30.619
15.3095 N
or N A
15.31 N
45.0q W
and N D
15.31 N
45.0q W
PROBLEM 4.45
0.45 m
0.45 m
B
A
A
(a)
90 N
0.45 m
B
B
A
(b)
90 N
(c)
A 90 N weight can be supported in the three different ways shown.
Knowing that the pulleys have a 100 mm radius, determine the reaction
at A in each case.
90 N
SOLUTION
(a) From f.b.d. of AB
6Fx
90 N
0: Ax
0
 Fy = 0: Ay – 90 N = 0
or
Ay = 90 N
J
and A = 90 N
 MA = 0: MA – (90 N) (0.45 m) = 0
\ MA = 33.75 N.m
or
MA = 33.8 N.m
J
From f.b.d. of AB
90 N
ÂFx = 0:
Ax – 90 N = 0
or
Ax = 90 N
90 N
 Fy = 0:
or
Then
Ay – 90 N = 0
Ay = 90 N
A =
Ax2 + Ay2 =
b90g + b90g
2
2
= 127.279 N
\ A = 127.3 N
 MA = 0:
45° J
MA + (90 N) (0.1 m)
– (90 N) (0.45 m + 0.1 m) = 0
\ MA = 33.75 N.m
or MA = 33.8 N.m
J
PROBLEM 4.45 CONTINUED
(c) From f.b.d. of AB
6Fx
0: Ax
 Fy = 0:
0
Ay – 90 N – 90 N = 0
Ay = 180 N
or
90 N 90 N
and A = 180 N
 MA = 0:
J
MA – (90 N) (0.45 m – 0.1 m)
– (90 N) (0.45 m + 0.1 m) = 0
\ MA = 67.5 N.m
or MA = 67.5 N.m J
PROBLEM 4.54
A slender rod AB, of weight W, is attached to blocks A and B, which
move freely in the guides shown. The blocks are connected by an elastic
cord which passes over a pulley at C. (a) Express the tension in the cord
in terms of W and T . (b) Determine the value of T for which the tension
in the cord is equal to 3W.
SOLUTION
(a) From f.b.d. of rod AB
6M C
ª§ l ·
º
0: T l sin T W «¨ ¸ cosT » T l cosT ¬© 2 ¹
¼
?T
0
W cosT
2 cosT sin T Dividing both numerator and denominator by cosT ,
T
W§
1
·
¨
¸
2 © 1 tan T ¹
or T
(b) For T
3W ,
3W
§W ·
¨ ¸
© 2¹
1 tan T ? 1 tan T
or
§W ·
¨ ¸
© 2 ¹ W
1 tan T T
§5·
tan 1 ¨ ¸
©6¹
1
6
39.806q
or T
39.8q W
PROBLEM 4.63
Horizontal and vertical links are hinged to a wheel, and forces are applied
to the links as shown. Knowing that a 75 mm, determine the value of P
and the reaction at A.
100 mm 100 mm
100 mm
95 N
SOLUTION
As shown on the f.b.d., the wheel is a three-force body. Let point D be
the intersection of the three forces.
From force triangle
75 mm
A
P
95
=
=
125 100
75
\ P=
100
(95 N) = 126.67 N
75
or P = 126.7 N J
95 N
and
95 N
75
A =
125
100
q = tan –1
125
(95 N) = 158.33 N
75
FG 75 IJ = 36.870°
H 100K
\ A = 158.3 N
36.9° J
PROBLEM 4.77
A small hoist is mounted on the back of a pickup truck and is used to lift
a 120-kg crate. Determine (a) the force exerted on the hoist by the
hydraulic cylinder BC, (b) the reaction at A.
D
1.2 m
30°
C
75°
B
A
0.4 m
SOLUTION
W
First note
mg
120 kg 9.81 m/s2 1177.2 N
From the geometry of the three forces acting on the small hoist
and
Then
x AD
1.2 m cos 30q
1.03923 m
y AD
1.2 m sin 30q
0.6 m
yBE
x AD tan 75q
1.03923 m tan 75q
D
§ y 0.4 m ·
tan 1 ¨ BE
¸
x AD
©
¹
E
75q D
T
180q 15q E
§ 3.4785 ·
tan 1 ¨
¸
© 1.03923 ¹
75q 73.366q
3.8785 m
73.366q
1.63412q
165q 1.63412q
163.366q
Applying the law of sines to the force triangle,
W
sin E
or
(a)
B
sin T
1177.2 N
sin1.63412q
B
A
sin15q
B
sin163.366q
11 816.9 N
or B
(b)
A
A
sin15q
11.82 kN
75.0q W
10.68 kN
73.4q W
10 684.2 N
or A
PROBLEM 4.79
A modified peavey is used to lift a 0.2-m-diameter log of mass 36 kg.
Knowing that T 45q and that the force exerted at C by the worker is
perpendicular to the handle of the peavey, determine (a) the force exerted
at C, (b) the reaction at A.
SOLUTION
First note
W
mg
36 kg 9.81 m/s2 353.16 N
From the geometry of the three forces acting on the modified peavey
E
1.1 m
§
·
tan 1 ¨
¸
1.1
m
0.2
m
©
¹
D
45q E
40.236q
45q 40.236q
4.7636q
Applying the law of sines to the force triangle,
W
sin E
or
(a)
(b)
C
sin D
353.16 N
sin 40.236q
C
A
A
sin135q
C
sin 4.7636
A
sin135q
45.404 N
or C
45.4 N
45.0q W
or A
387 N
85.2q W
386.60 N
PROBLEM 4.90
A 10-kg slender rod of length L is attached to collars which can slide
freely along the guides shown. Knowing that the rod is in equilibrium and
that E 25q, determine (a) the angle T that the rod forms with the
vertical, (b) the reactions at A and B.
SOLUTION
(a) As shown in the f.b.d. of the slender rod AB, the three forces
intersect at C. From the geometry of the forces
xCB
yBC
tan E
where
and
xCB
1
L sin T
2
yBC
L cosT
1
tan T
2
? tan E
tan T
or
E
For
2 tan E
25q
tan T
2 tan 25q
? T
43.003q
0.93262
or T
(b)
W
mg
10 kg 9.81 m/s 2 43.0q W
98.1 N
From force triangle
A
W tan E
98.1 N tan 25q
45.745 N
or A
and
B
W
cos E
98.1 N
cos 25q
45.7 N
W
108.241 N
or B
108.2 N
65.0q W
PROBLEM 4.99
y
150 mm
z
150 N
C
100 mm
50 mm
A
E
25 mm
D
T
90 mm
x
B
For the portion of a machine shown, the 100 mm-diameter pulley A and
wheel B are fixed to a shaft supported by bearings at C and D. The spring
of constant 360 N/m is unstretched when T 0, and the bearing at C does
not exert any axial force. Knowing that T 180° and that the machine is
at rest and in equilibrium, determine (a) the tension T, (b) the reactions at
C and D. Neglect the weights of the shaft, pulley, and wheel.
300 mm
F
25 mm
SOLUTION
25 mm
150 N
50 mm
90 mm
50 mm
First, determine the spring force, FE, at q = 180°
FE = k s x
where
ks = 360 N/m
x = (yE)final – (yE)initial = (300 mm + 90 mm) – (300 mm – 90 mm) = 180 mm
\ FE = (360 N/m) (180/1000 m) = 64.8 N
(a) From f.b.d. of machine part
 Mx = 0:
(150 N) (50 mm) – T (50 mm) = 0
\ T = 150 N
(b)
 MD (z-axis) = 0:
or
T = 150 N J
– Cy (250 mm) – FE (50 mm + 25 mm) = 0
– Cy (250 mm) – 64.8 N (75 mm) = 0
\ Cy = – 19.44 N
 MD (y-axis) = 0:
or
Cy = – (19.44 N) j
Cz (250 mm) + (150 N) + (100 mm) (150 mm) (100 mm) = 0
\ Cz = – 120 N
or Cz = – (120 N) k
and
C = – (19.4 N) j – (120 N) k J
PROBLEM 4.99 CONTINUED
 Fx = 0:
Dx = 0
 MC (z-axis) = 0:
Dy (250 mm) – FE (300 mm + 25 mm) = 0
or
Dy (250 mm) – 64.8 (325 mm) = 0
\ Dy = 84.24 N
ÂMC (y-axis) = 0:
or Dy = (84.2 N) j
– 2 (150 N) (150 mm) – Dz (250 mm) = 0
\ Dz = –180 N
or
Dz = – (180 N) k
and
D = (84.2 N) j – (180 N) k J
PROBLEM 4.103
The 200 u 200-mm square plate shown has a mass of 25 kg and is
supported by three vertical wires. Determine the mass and location of the
lightest block which should be placed on the plate if the tensions in the
three cables are to be equal.
SOLUTION
WG
First note
W1
25 kg 9.81 m/s2 m p1g
mg
m 9.81 m/s 2
245.25 N
9.81m N
From f.b.d. of plate
6Fy
6M x
0: 3T WG W1
0 (1)
0: WG 100 mm W1 z T 100 mm T 200 mm or 300T 100WG W1z
6M z
0
(2)
0: 2T 160 mm WG 100 mm W1 x or 320T 100WG W1x
0
0
0
(3)
Eliminate T by forming 100 u ª¬ Eq. 1 Eq. 2 º¼
100W1 W1z
? z
100 mm
0
0 d z d 200 mm, ? okay
Now, 3 u ª¬ Eq. 3 º¼ 320 u ª¬ Eq. 1º¼ yields
3 320T 3 100 WG 3W1x 320 3T 320WG 320W1
0
PROBLEM 4.103 CONTINUED
20WG 320 3x W1
or
W1
WG
or
The smallest value of
20
3
x
320 W1
will result in the smallest value of W1 since WG is given.
WG
? Use x
W1
WG
and then
WG
14
? W1
and
0
m
W1
g
xmax
200 mm
20
3 200 320
245.25 N
14
17.5179 N
9.81 m/s 2
1
14
17.5179 N minimum 1.78571 kg
at x
or m
1.786 kg W
200 mm, z
100 mm W
PROBLEM 4.114
An 2.4 m-long boom is held by a ball-and-socket joint at C and by two
cables AD and BE. Determine the tension in each cable and the reaction
at C.
0.9 m
1.2 m
0.6 m
1.8 m
0.3 m
2.4 m
880 N
SOLUTION
From f.b.d. of boom
 MCE = 0:
l CE =
where
lCE ◊ (rA/C ¥ TAD) + lCE ◊ (rA/C ¥ FA) = 0
b0.6 mg j - b0.9 mg k =
b0.6g + b0.9g m
2
2
1
117
.
(0.6 j – 0.9 k)
rA/C = (2.4 m) i
TAD = lAD TAD =
=
b
g b g b g
b2.4g + b0.3g + b1.2g m
- 2.4 m i + 0.3 m j + 1.2 m k
2
FG 1 IJ T
H 2.7 K
AD
2
2
TAD
(– 2.4 i + 0.3 j + 1.2 k)
FA = – (880 N) j
\
F
GH
0
0.6 - 0.9
TAD
2.4
0
0
2.7 117
.
- 2.4 0.3 1.2
I+
JK
0 0.6 - 0.9
2.4 0
0
0 -1
0
F 880 I = 0
GH 117
J
. K
PROBLEM 4.114 CONTINUED
(–1.728 – 0.648)
TAD
2.7 117
.
880
+ (2.16)
=0
117
.
\ TAD = 2160 N
or
TAD = 2160 N J
 MCD = 0: lCD ◊ (rB/C ¥ TBE) + lCD ◊ (rA/C ¥ FA)
where
lCD =
b0.3 mg j + b1.2 mg k
1.53 m
=
1
153
.
(0.3 j + 1.2 k)
rB/C = (1.8 m) i
TBE = lBETBE =
b g b g b g
b18. g + b0.6g + b0.9g m
. m i + 0.6 m j - 0.9 k
- 18
2
2
2
TBE =
FG 1 IJ T
H 2.1K
BE
(– 1.8 i + 0.6 j – 0.9 k)
0
0.3 1.2
0 0.3 1.2
880
TBE
+ 2.4 0
=0
\ 18
0
0
0
.
2.1 153
153
.
.
0 -1 0
. 0.6 - 0.9
- 18
(0.486 + 1.296)
TBE
+ (– 2.88) 880 = 0
2.1
\ TBE = 2986.67 N
or
TBE = 2986.7 N J
 Fx = 0: Cx – (TAD)x – (TBE)x = 0
Cx –
FG 2.4 IJ 2160 – FG 1.8 IJ 2986.67 = 0
H 2.7 K
H 2.1K
\ Cx = 4480 N
 Fy = 0:
Cy + (TAD)y + (TBE)y – 880 N = 0
Cy +
FG 0.3 IJ 2160 + FG 0.6IJ 2986.67 – 880 N = 0
H 2.7 K
H 2.1K
\ Cy = – 213.33 N
 Fz = 0: Cz + (TAD)z – (TBE)z = 0
Cz +
FG 1.2 IJ 2160 – FG 0.9 IJ 2986.67 = 0
H 2.7 K
H 2.1K
\ Cz = 320 N
or
C = (4480 N) i – (213.3 N) j + (320 N) k J
PROBLEM 4.122
y
The rectangular plate shown has a mass of 15 kg and is held in the
position shown by hinges A and B and cable EF. Assuming that the hinge
at B does not exert any axial thrust, determine (a) the tension in the cable,
(b) the reactions at A and B.
z
x
SOLUTION
W
First note
TEF
O EFTEF
mg
15 kg 9.81 m/s2 ª 0.08 m i 0.25 m j 0.2 m k º
«
» TEF
2
2
2
0.08 0.25 0.2 m ¼»
¬«
147.15 N
TEF
0.08i 0.25j 0.2k 0.33
From f.b.d. of rectangular plate
6M x
or
0:
147.15 N 0.1 m TEF y 0.2 m 0
ª§ 0.25 ·
º
14.715 N ˜ m «¨
¸ TEF » 0.2 m ¬© 0.33 ¹
¼
0
TEF
or
97.119 N
or TEF
6Fx
0: Ax TEF x
0
§ 0.08 ·
Ax ¨
¸ 97.119 N © 0.33 ¹
? Ax
23.544 N
0
97.1 N W
PROBLEM 4.122 CONTINUED
6M B z -axis 0: Ay 0.3 m TEF y 0.04 m W 0.15 m 0
ª§ 0.25 ·
º
Ay 0.3 m Ǭ
¸ 97.119 N » 0.04 m 147.15 N 0.15 m ¬© 0.33 ¹
¼
or
? Ay
6M B y -axis 63.765 N
0: Az 0.3 m TEF x 0.2 m TEF z 0.04 m 0
ª§ 0.08 ·
º
ª§ 0.2 ·
º
Az 0.3 m Ǭ
¸ TEF » 0.2 m «¨
¸ TEF » 0.04 m ¬© 0.33 ¹
¼
¬© 0.33 ¹
¼
? Az
0: Ay W TEF y By
23.5 N i 63.8 N j 7.85 N k W
0
§ 0.25 ·
63.765 N 147.15 N ¨
¸ 97.119 N By
© 0.33 ¹
? By
6Fz
0: Az TEF z Bz
0
7.848 N
and A
6Fy
0
0
9.81 N
0
§ 0.2 ·
7.848 N ¨
¸ 97.119 N Bz
© 0.33 ¹
? Bz
0
66.708 N
and B
9.81 N j 66.7 N k W