Block `A` of weight `100N` rests on a frictionless inclined plane of slope angle `30^(@)`(Fig. 5.7). `A` flexible cord attached to `A` passes over a frictonless pulled and is connected to block `B` of weight `W`. Find the weight `W` for which the system in equilibrium.

Options

A. `80N`
B. `50N`
C. `40N`
D. `100N`

Answer

B

Explanation

In equilbrium the force mg sin `theta` acting on block A parallel to the plane should be balanced by the tension in the string
`mg sin theta = T =F---(i)`
`w =T =F---(ii)`
and for block `B` where w is the weight of block `B` From Eqs (i) and (ii) we get
`w =mg sin theta`
`= 100 xx sin 30^(@)`
`= 100 xx (1)/(2) N = 50N` .