(^282) „„„„„ A Textbook of Engineering Mechanics
(a) Tension in the cable
Let T= Tension in the cable and
a= Length of each side of the equilateral triangle.
Taking moments about the joint 5 and equating the same,
(T cos 60°) × 2a= (1 × 1.5 a) + (1 × 0.5 a)
(T × 0.5) 2a=2a
∴ T= 2 kN Ans.
(b) Nature and magnitude of the force in each bar
Fig. 13.59.
First of all consider the joint 1. We have already found out that tension in the cable is 2 kN as
shown in the figure. Let the directions of P1–2 and P1–4 be assumed as shown in Fig. 13.59 (a).
Resolving the forces vertically and equating the same,
P1–2 sin 60° = 2 sin 30°
∴ 1– 2
2sin 30 2 0.5
1.154 kN (Tension)
sin 60 0.866
P
°×

°
and now resolving the forces horizontally and equating the same,
P1–4= 2 cos 30° + 1.154 cos 60° kN
= (2 × 0.866) + (1.154 × 0.5) = 2.309 kN (Compression)
Now consider the joint 2. We have already found out that the force in member 1-2 (i.e. P1–2) is
1.54 kN (Tension). Let the directions of the forces P2–4 and P2–3 be assumed as shown in Fig 13.59
(b). Resolving the forces vertically and equating the same,
P2–4 sin 60° = 1 – 1.154 sin 60° = 1 – (1.154 × 0.866) = 0
∴ P2–4=0
and now resolving the forces horizontally and equating the same,
P2–3= 1.154 cos 60° = 1.154 × 0.5 = 0.577 kN (Tension)
Now consider the joint 4. A little consideration will show that the force P3–4 will be zero. This
will happen as the force P2–4 is zero and the vertical components of the forces P1–4 and P4–5 are also
zero. Moreover, the force P4–5 will be equal to the force P1–4 i.e., 2.309 kN (Compression). This will
happen as the forces P2–4 and P2–5 (being zero) will have their vertical components as zero.