Question

A triangular block of mass M with angles 30°, 60°, and 90° rests with its 30°–90° side on a horizontal table. A cubical block of mass m rests on the 60°–30° side. The acceleration which M must have relative to the table to keep m stationary relative to the triangular block assuming frictionless contact is(a) g (b) $\frac{g}{\sqrt{2}}$(c) $\frac{g}{\sqrt{3}}$(d) $\frac{g}{\sqrt{5}}$

Answer 1

c is the correct answer

Answer 2

For the block to be in rest,

mg sinα = ma cosα

a = g tanα

here, α = 30⁰

a = g tan 30⁰

a = g/√3

This is the required answer.