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Concept of Force

Two masses m1...

Question

Two masses $m_{1}$ and $m_{2}$ which are connected with a light string are placed over a frictionless pulley. This set up is placed over a weighing machine as shown in figure. Three combination of masses $m_{1}$ and $m_{2}$ are used, in first case $m_{1} = 6 kg$ and $m_{2} = 2 kg$ , in second case $m_{1} = 5 kg$ and $m_{2} = 3 kg$ , in the third case $m_{1} = 4 kg$ and $m_{2} = 4 kg$ . Masses are released from rest. If $W_{1}, W_{2}$ and $W_{3}$ be the reading of weighing machine in three cases, then:

W_{1} > W_{2} > W_{3}

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W_{1} < W_{2} < W_{3}

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W_{1} = W_{2} = W_{3}

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W_{1} = W_{2} < W_{3}

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Solution

The correct option is B $W_{1} < W_{2} < W_{3}$
If we consider (masses+pulley) and support rod as a system, then weighing machine will experience force due to tension and weight of support.

$F.B.D :$

If $F$ be the force due to support. Thus reading of weighng machine is,

$N = 2 T + F$

Where, $N$ is the normal reaction measured and $T$ is tension in the string.

As we know that, tension in the string due pulley mass system, $T = \frac{2 m_{1} m_{2} g}{m_{1} + m_{2}}$ Since, the force due to support $F$ is same in all three cases.

we can say that, $N \propto T$

$\Rightarrow N \propto \frac{2 m_{1} m_{2} g}{m_{1} + m_{2}}$

Since, sum of masses is same in all cases i.e. $8 kg$

$\Rightarrow N_{W . M} \propto m_{1} m_{2}$

Product $m_{1}$ and $m_{2}$ is highest in 3rd case,

$∴ W_{1} < W_{2} < W_{3}$ is correct order of reading of weighing machine.

Why this question?

Tip:- For any arrangement of masses whether the frame is at rest or accelerating. The weighing machine reads the normal reaction exerted at its contact points.

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