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.
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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.
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