Question

Two blocks of mass m and M are lying on a smooth horizontal surface,making a contact with each
other.If the blocks are pushed by applying force F, find the force at the common interface.​

Answer 1

Answer:

Method 1 :

Since the two blocks always remain in contact with each other, they must move with the same acceleration. Using Newton's second law, we get

For block m

1

:F−N=m1a......(i)

For block m 2:N=m2a......(ii)

On adding the two equations, we get

F=(m 1+m )a⇒a= F/m 1 +m 2

Substituting the value of a in (ii), we get

N=m 2 a= m1 +m2a-Fm 2/m1+m2

Method 2 :

The situation may be considered as follows: Instead of drawing the free-body diagrams of each block, we can draw the free-body diagram of both blocks together as shown in Fig.6.33.

The net force acting on the system is F, and the total mass of the system is

m1+m2 . Thus, a= F/m1 +m 2

To find out the normal reaction N between the two blocks, we can imagine the following; Block m2 is moving with an acceleration a; therefore, the net force acting on it must be m2 a, which is nothing but the normal reaction applied by the block m1 . Thus,

N_m2a= m 2F/m1+m2