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Inclined Planes

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Question

Three blocks are placed on a smooth inclined plane with force acting on $m_{1}$ parallel to the inclined plane. Find the contact force between $m_{2}$ and $m_{3}$ .

A

\frac{(m_{1} + m_{2} + m_{3}) F}{m_{3}}

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B

\frac{m_{3} F}{m_{1} + m_{2} + m_{3}}

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C

F - (m_{1} + m_{2}) g

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D

None of these

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Solution

The correct option is B $\frac{m_{3} F}{m_{1} + m_{2} + m_{3}}$
Analysing the motion of $(m_{1} + m_{2} + m_{3})$ as a system,
Acceleration of the system, $a = \frac{Net pushing force}{Total mass}$
$= \frac{F - (m_{1} + m_{2} + m_{3}) g sin θ}{m_{1} + m_{2} + m_{3}} - - (i)$

Let $N$ be the contact force between blocks $m_{2}$ and $m_{3}$ .
Drawing FBD of $m_{3}$ and analysing its motion,

Using the above figure, we can write
$N - m_{3} g sin θ = m_{3} a - - (i i)$

Using equation (i) in (ii), we get
$N = m_{3} g s i n θ + m_{3} [\frac{F - (m_{1} + m_{2} + m_{3}) g sin θ}{m_{1} + m_{2} + m_{3}}]$
$⟹ N = \frac{m_{3} F}{m_{1} + m_{2} + m_{3}}$

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