Two blocks $m_{1}$ and $m_{2}$ , are acted upon by the forces $F_{1}$ and $F_{2}$ as shown in the fig. If there is no relative sliding between the blocks and the ground is smooth, find the magnitude of static friction between the blocks.Make necessary assumptions.

$∣ ∣ \frac{m_{1} F_{1} - m_{2} F_{2}}{m_{1} + m_{2}} ∣ ∣$

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$∣ ∣ \frac{m_{2} F_{2} - m_{1} F_{1}}{m_{1} + m_{2}} ∣ ∣$

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$∣ ∣ \frac{m_{2} F_{1} - m_{1} F_{2}}{m_{1} + m_{2}} ∣ ∣$

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$∣ ∣ \frac{m_{1} F_{1} + m_{2} F_{2}}{m_{1} + m_{2}} ∣ ∣$

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Solution

The correct option is C
$∣ ∣ \frac{m_{2} F_{1} - m_{1} F_{2}}{m_{1} + m_{2}} ∣ ∣$

As there is no sliding between the blocks the friction will be static. Assuming the frictional forces on $m_{1}$ as f( $\leftarrow$ ) and $m_{2}$ as f( $\to$ ), we have drawn the Free body diagram without normal reactions and weights.

Referring to Free body diagram as shown in figure, we have the following equations of motion.

For m₁:

$F_{1}$ - f = $m_{1}$ $a_{1}$ . . . . . . (i)

For $m_{2}$ : $F_{2}$ + f = $m_{2}$ $a_{2}$ . . . . . . (ii)

Since the friction is static, $a_{1}$ = $a_{2}$

Substituting $a_{1} f r o m e q u a t i o n (i) a_{2} f r o m e q u a t i o n (i i) i n ‘ e q u a t i o n (i i i), w e h a v e \frac{F_{1} - f}{m_{1}} = \frac{F_{2} + f}{m_{2}}$

This gives f = $\frac{m_{2} F_{1} - m_{1} F_{2}}{m_{1} + m_{2}}$

case(ii)

Friction on $m_{1} i s a c t i n g i n r i g h t w a r d d i r e c t i o n .$

For $m_{1} : F_{1} + f = m_{1} a_{1} a n d f o r m_{2} : F_{2} - f = m_{2} a_{2}$

As static, $a_{1} = a_{2}$

$∴ \frac{F_{1} + f}{m_{1}} = \frac{F_{2} - f}{m_{2}}$

$\Rightarrow f = \frac{m_{1} F_{2} - m_{2} F_{1}}{m_{1} + m_{2}}$
$∴ M a g n i t u d e o f f = \frac{m_{2} F_{1} - m_{1} F_{2}}{m_{1} + m_{2}}$

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