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Limiting Friction

Three blocks ...

Question

Three blocks of masses $m_{1}, m_{2}$ and $m_{3}$ are lying in contact with each other on a horizontal frictionless plane as shown in the figure. If a horizontal force $F$ is applied on $m_{1}$ then the force due to $m_{2}$ on $m_{1}$ will be:

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

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

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m_{1} F

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

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Solution

The correct option is A $\frac{F (m_{2} + m_{3})}{(m_{1} + m_{2} + m_{3})}$
Consider the three blocks as a single block of mass $(m_{1} + m_{2} + m_{3})$ and $a$ be the acceleration of the system.

So, $(m_{1} + m_{2} + m_{3}) a = F$

$\Rightarrow a = \frac{F}{(m_{1} + m_{2} + m_{3})}$

Now, net force on mass $m_{2}$ is $F_{2} = F_{21} - F_{23}$

Here, $F_{2} = m_{2} a$ ,

Also, $F_{32} = F_{23} = m_{3} a$

Thus, we get:

$m_{2} a = F_{21} - m_{3} a$

$\Rightarrow F_{21} = (m_{2} + m_{3}) a = \frac{(m_{2} + m_{3}) F}{(m_{1} + m_{2} + m_{3})}$

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