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Question: Block A of weight 100 N rests on a frictionless inclined plane of slope angle \(30 ^ { \circ }\) as...

Block A of weight 100 N rests on a frictionless inclined plane of slope angle 30∘30 ^ { \circ } as shown in the figure. A flexible cord attached to A passes over a frictionless pulley and is connected to block B of weight W :

Find the weight W for which the system is in equilibrium :

A

25 N

B

50 N

C

75 N

D

100 N

Answer

50 N

Explanation

Solution

As the system is in equilibrium

∴T=W\therefore T = W …. (i)

and T=mgsin⁡30∘T = m g \sin 30 ^ { \circ } ……(ii)

From (i) and (ii) we get

W=mgsin⁡30∘=(100 N)×12=50 NW = m g \sin 30 ^ { \circ } = ( 100 \mathrm {~N} ) \times \frac { 1 } { 2 } = 50 \mathrm {~N}