ALL CHAPTERS 2 - 22 COVERED
,Solutions Manual
SOLUTIONS TO CHAPTER 2 PROBLEMS
S.2.1 By inspection:
a. Translation parallel to BA.
b. Translation parallel to BD, clockwise rotation.
c. No translation, possible clockwise rotation.
d. The force F at C may be resolved into two components as shown in Fig. S.2.1(a). The force
system is then equivalent to a force parallel to AD of 2F Fcos 45° ¼ 1.293F and a force of
0.707F parallel to AB both acting at the centre of the block together with an anticlockwise
torque as shown in Fig. S.2.1(b). The resultant of the two forces then acts at an angle α to the
direction of AD given by
0:707F
tan α ¼ ¼ 0:547
1:293F
which gives
α ¼ 28:7°
2F
A B A B
0.707F
F
α
45º 45º
Fsin 45º
D C D C
1.293F
(a) Fcos 45º (b)
FIGURE S.2.1
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S.2.2 a. Vectors representing the 10 and 15 kN forces are drawn to a suitable scale as shown in
Fig. S.2.2. Parallel vectors AC and BC are then drawn to intersect at C. The resultant is the
vector OC which is 21.8 kN at an angle of 23.4° to the 15 kN force.
B C
10 kN
R
60°
q
O 15 kN A
FIGURE S.2.2
b. From Eq. (2.1) and Fig. S.2.2
R 2 ¼ 152 þ 102 þ 2 15 10 cos 60°
which gives
R ¼ 21:8 kN
Also, from Eq. (2.2)
10 sin 60°
tan θ ¼
15 þ 10 cos 60°
so that
θ ¼ 23:4°:
S.2.3 a. The vectors do not have to be drawn in any particular order. Fig. S.2.3 shows the vector diagram
with the vector representing the 10 kN force drawn first.
The resultants R is then equal to 8.6 kN and makes an angle of 23.9° to the negative
direction of the 10 kN force.
b. Resolving forces in the positive x direction
Fx ¼ 10 þ 8 cos 60° 12 cos 30° 20 cos 55° ¼ 7:9 kN
Then, resolving forces in the positive y direction
Fy ¼ 8 cos 30° þ 12 cos 60° 20 cos 35° ¼ 3:5 kN
The resultant R is given by
R 2 ¼ ð 7:9Þ2 þ ð 3:5Þ2
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, Solutions to Chapter 2 Problems e3
12 kN
8 kN
q
20 kN R 10 kN
FIGURE S.2.3
so that
R ¼ 8:6 kN
Also
3:5
tan θ ¼
7:9
which gives
θ ¼ 23:9°:
S.2.4 Since the joints at A, B, and C are hinged and the load is applied at joint B, the forces in AB and BC
will be purely axial. Assume that the force in AB is tension and in BC is compression (if a wrong
assumption is made the answer will be negative).
1.5 m
A FAB B
10 kN
3m
FCB
α
C
FIGURE S.2.4(a)
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,Solutions Manual
SOLUTIONS TO CHAPTER 2 PROBLEMS
S.2.1 By inspection:
a. Translation parallel to BA.
b. Translation parallel to BD, clockwise rotation.
c. No translation, possible clockwise rotation.
d. The force F at C may be resolved into two components as shown in Fig. S.2.1(a). The force
system is then equivalent to a force parallel to AD of 2F Fcos 45° ¼ 1.293F and a force of
0.707F parallel to AB both acting at the centre of the block together with an anticlockwise
torque as shown in Fig. S.2.1(b). The resultant of the two forces then acts at an angle α to the
direction of AD given by
0:707F
tan α ¼ ¼ 0:547
1:293F
which gives
α ¼ 28:7°
2F
A B A B
0.707F
F
α
45º 45º
Fsin 45º
D C D C
1.293F
(a) Fcos 45º (b)
FIGURE S.2.1
e1
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S.2.2 a. Vectors representing the 10 and 15 kN forces are drawn to a suitable scale as shown in
Fig. S.2.2. Parallel vectors AC and BC are then drawn to intersect at C. The resultant is the
vector OC which is 21.8 kN at an angle of 23.4° to the 15 kN force.
B C
10 kN
R
60°
q
O 15 kN A
FIGURE S.2.2
b. From Eq. (2.1) and Fig. S.2.2
R 2 ¼ 152 þ 102 þ 2 15 10 cos 60°
which gives
R ¼ 21:8 kN
Also, from Eq. (2.2)
10 sin 60°
tan θ ¼
15 þ 10 cos 60°
so that
θ ¼ 23:4°:
S.2.3 a. The vectors do not have to be drawn in any particular order. Fig. S.2.3 shows the vector diagram
with the vector representing the 10 kN force drawn first.
The resultants R is then equal to 8.6 kN and makes an angle of 23.9° to the negative
direction of the 10 kN force.
b. Resolving forces in the positive x direction
Fx ¼ 10 þ 8 cos 60° 12 cos 30° 20 cos 55° ¼ 7:9 kN
Then, resolving forces in the positive y direction
Fy ¼ 8 cos 30° þ 12 cos 60° 20 cos 35° ¼ 3:5 kN
The resultant R is given by
R 2 ¼ ð 7:9Þ2 þ ð 3:5Þ2
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, Solutions to Chapter 2 Problems e3
12 kN
8 kN
q
20 kN R 10 kN
FIGURE S.2.3
so that
R ¼ 8:6 kN
Also
3:5
tan θ ¼
7:9
which gives
θ ¼ 23:9°:
S.2.4 Since the joints at A, B, and C are hinged and the load is applied at joint B, the forces in AB and BC
will be purely axial. Assume that the force in AB is tension and in BC is compression (if a wrong
assumption is made the answer will be negative).
1.5 m
A FAB B
10 kN
3m
FCB
α
C
FIGURE S.2.4(a)
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