Question

All the blocks shown in the figure are at rest. The pulley is smooth and the strings along
of friction at all the contacts is 0.2. A frictional force of 10 N acts between A and B.The DIOCR Sasoul
to slide on block B.The normal reaction and frictional force exerted by the ground on the block D IS.I ​

Answer 1

Answer:

The frictional force exerted by the ground on the block B is zero.

Explanation:

From making the free body diagram of the boxes in the above figure we will get that frictional force on block A is μN1 =10 so, N1 = 10/0.2 = 50 N.

Since, in the vertical direction the net force on block B is zero so,

N2 = 50 + N1 + 10 = 110N

Hence, we know that the normal reaction exerted by ground  itself on the block B will be 110N.

So,in the horizontal direction the net force on the block B will be

f + 10 – 10 = 0.

Hence, we can finally conclude that the frictional force exerted by ground on block B is zero.