Engineering Mechanics

Chapter 5 : Equilibrium of Forces „„„„„ 59

Again applying Lami’s equation at joint C,

1000

sin 120 sin 120 sin 120

TTBC ==CD

°°°

∴

1000 sin 120

1000 N.

CD sin 120

T

°

==

°

Ans

Example 5.3. A light string ABCDE whose extremity A is fixed, has weights W 1 and W 2
attached to it at B and C. It passes round a small smooth peg at D carrying a weight of 300 N at the
free end E as shown in Fig. 5.7.

Fig. 5.7.
If in the equilibrium position, BC is horizontal and AB and CD make 150° and 120° with BC,
find (i) Tensions in the portion AB, BC and CD of the string and (ii) Magnitudes of W 1 and W 2.

Solution. Given : Weight at E = 300 N
For the sake of convenience, let us split up the string ABCD into two parts. The system of
forces at joints B and C is shown in Fig. 5.8. (a) and (b).

Fig. 5.8.

(i) Tensions is the portion AB, BC and CD of the string

Let TAB= Tension in the portion AB, and

TBC = Tension in the portion BC,

We know that tension in the portion CD of the string.

TCD = TDE = 300 N Ans.

Applying Lami’s equation at C,

2 300

sin 150 sin 120 sin 90

TBC ==W

°°°