(^60) „„„„„ A Textbook of Engineering Mechanics
2 300
sin 30 sin 60 1
TBC ==W
°° ...[Q sin (180° – θ) = sin θ]
∴ TBC = 300 sin 30° = 300 × 0.5 = 150 N Ans.
and W 2 = 300 sin 60° = 300 × 0.866 = 259.8 N
Again applying Lami’s equation at B,
1
sin90sin150sin120
TWAB ==TBC
°°°
1 150
1sin30sin60
TWAB ==
°°
...[Q sin (180° – θ) = sin θ]
∴
150 150
173.2 N
AB sin 60 0.866
T ===
°
Ans.
and 1
150 sin 30 150 0.5
86.6 N
sin 60 0.866
W
°×

°
(ii) Magnitudes of W 1 and W 2
From the above calculations, we find that the magnitudes of W 1 and W 2 are 86.6 N and
259.8 N respectively. Ans.
EXERCISE 5.1

1. Two men carry a weight of 2 kN by means of two ropes fixed to the weight. One rope is
   inclined at 45° and the other at 30° with their vertices. Find the tension in each rope.
   [Ans. 1.04 kN ; 1.46 kN]


2. Three forces acting on a particle are in equilibrium. The angles between the first and
   second is 90° and that between the second and third is 120°. Find the ratio of the forces.
   [Ans. 1.73 : 1 : 2]


3. A smooth sphere of weight W is supported by a string fastened to a point A on the smooth
   vertical wall, the other end is in contact with point B on the wall as shown in Fig. 5.9

Fig. 5.9. Fig. 5.10.

If length of the string AC is equal to radius of the sphere, find tension (T) in the string and

reaction of the wall. [Ans. 1.155 W ; 0.577 W]

Hint. Since AO = 2 OB, therefore ∠ AOB = 60°