Question

Two blocks of masses $m_1$ and $m_2$ are placed side by side on a smooth horizontal surface as shown in $Fig.6.31$. A horizontal force F is applied on the block.
$a$. Find the acceleration of each block.
$b$. Find the normal reaction between the two blocks.

Answer 1

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 $m_1:F-N=m_{1}a$......(i)

For block $m_2:N=m_{2}a$......(ii)

On adding the two equations, we get

$F=(m_1+m_2)a\Rightarrow a=\frac{F}{m_1+m_2}$

Substituting the value of a in (ii), we get $N=m_{2}a=\frac{F{m}_2}{m_1+m_2}$

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

$m_1+m_2$. Thus, $a=\frac{F}{m_1+m_2}$

To find out the normal reaction N between the two blocks, we can imagine the following; Block $m_2$ is moving with an acceleration a; therefore, the net force acting on it must be $m_{2}a$, which is nothing but the normal reaction applied by the block $m_1$. Thus,

$N=m_{2}a=\frac{m_{2}F}{m_1+m_2}$