Question

Determine the least weight w necessary to present download motion of 1100N block.

Answer 1

In equilbrium the force mg sin θ acting on block A parallel to the plane should be balanced by the tension in the string

mgsinθ=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=mgsinθ

=100×sin30∘

=100×12N=50N .