Question

A smooth cylinder of mass m and radius R Is resting on two corner edges A and B as shown in the figure. the relation between normal reaction at the edges A and B is.

Answer 1

Answer:

NB=√3NA.

Explanation:

If we take the two component of the two sides along the weight of the ball. So, the two components will be for B we will have NB=NBcos30 and NBsin30 and for the A we will have NAsin60 and NAcos60 where NA is the normal reaction of A and the normal reaction on B is NB. So, we get from  the diagram and the components that the vertical components will get cancelled so the cos components get cancelled and only we will have the sin components.

Thus, NBsin30=NAsin60 or NB/2 = NA√3/2 which on solving we will get that the NB=√3NA.