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Question

Below Figure shows a man of mass 60 kg standing on a light weighing machine kept in a box of mass 30 kg. The box is hanging from a pulley fixed to the ceiling through a light rope, the other end of which is held by the man himself. If the man manages to keep the box at rest, what is the weight shown by the machine? What force should he exert on the rope to get his correct weight on the machine?


A

scale reading - 150 N; Tension - 1800 N

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B

scale reading - 1800 N; Tension - 150 N

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C

scale reading - 1800 N; Tension - 15 N

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D

scale reading - 15 kg; Tension - 1800 N

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Solution

The correct option is D

scale reading - 15 kg; Tension - 1800 N


In situation 1: The man and the box are at rest.

Let's draw the Free Body Diagram of the man and see what all forces are acting on him

Tension by rope

Normal by the weighting scale mg - pull of gravity

Since the body id at rest

T+N=mg=600 ----------------(I)

Tension by string only one end of the string is connected to the box so only one tension on the box also normal that the box applied on man in upward direction will be applied to it in downward direction in accordance with newton's 3rd law.

T=mg+N ----------------------(II)

Solving (I) & (II) N = 130N

Situation 2: If the weighing scale is showing the correct weight this means the normal reaction is equal to the actual weight of the man (weighing scales measure normal reaction on them). So if we assure he is at rest.

Free Body Diagram of man

T + N = mg

N = mg

T = 0

Then

Free Body Diagram of box

If tension is 0, then net external force on the box is in downward direction. So the box will fall down, and so will the man and hence there still be no contact between the man and the box so normal will also become 0.

So clearly this isn't the case and tension 0 for normal to be equal to mg so lets assume some tension 'T0' in the string now.

Clearly man has a net external force in upward direction as N = mg and T0 in left

Body will accelerate it has to (2nd law)

So let's assume and upward acceleration of a0

Now T0 + N - mg = ma0

As N = mg

T0 = ma0

T0 = 60 a0 --------------------(III)

Now the box has also to accelerate up with the exactly same acceleration in order to keep the contact between the man and box maintained or else the weighing scale won't measure the exact weight.

T0 - Mg - N = Ma0

T0 - 300 - 600 = 30a0

T0 - 900 = 30a0 -----------------(IV)

Solving (III) and (IV)

T0 = 1800N


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