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.

Explanation

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