Question

A force $F$ is applied horizontally on mass $m_1$ as shown in figure. Find the contact force between $m_1$ and $m_2$.

• A. $\frac{m_{2}F}{m_1+m_2}$
• B. $\frac{m_{1}F}{m_1+m_2}$
• C. $m_{1}F$
• D. $m_{2}F$

Answer 1

The correct option is A $\frac{m_{2}F}{m_1+m_2}$

Considering both blocks as a system to find the common acceleration.

Common acceleration
$a=\frac{F}{(m_1+m_2)}\dots \dots \left(1\right)$
To find the contact force between $A$ and $B$ we draw F.B.D. of mass $m_2$.
$\sum F_x=m{a}_x$
$N=m_2.a$
$N=\frac{m_{2}F}{(m_1+m_2)}$