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Second Law of Motion

A force F is ...

Question

A force $F$ is applied horizontally on mass $m_{1}$ as shown in figure. Find the contact force between $m_{1}$ and $m_{2}$ .

\frac{m_{2} F}{m_{1} + m_{2}}

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\frac{m_{1} F}{m_{1} + m_{2}}

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m_{1} F

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m_{2} F

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Solution

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
$a = \frac{F}{(m_{1} + m_{2})} \dots \dots (1)$
To find the contact force between $A$ and $B$ we draw F.B.D. of mass $m_{2}$ .
$\sum F_{x} = m a_{x}$
$N = m_{2} . a$
$N = \frac{m_{2} F}{(m_{1} + m_{2})}$

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