Question

A weighing machine of mass 20 kg is fixed in a lift of mass 10 kg a man of mass 60 kg is standing on the weighing machine and is pulling the lift through light strings passing over light fixed pulleys.
a)-If lift is stationary then reading of weighing machine is
b)-If reading of weighing machine is equal to true weight of man the force exerted on the string by each hand of man is
c)-If the mass of a man is 20 kg the reading of weighing machine is ​

Answer 1

Answer:

34. D 15 kg

35. B  900 N

36. A  Zero (if machine can't display negative acceleration value)

         

Explanation:

Let T be the tensions in the strings.  

The downward forces acting on the system of the man and the lift are

i.) the weight of the man

ii.) the weight of the box.

The upward forces are 2T and 2T on each of the strings with T & T on each side of string.

Using condition of static equilibrium in the first case as the lift is stationary we can get the reading of the weighing machine.

Using Newtons' Second Law of Motion we can get the force exerted by the hands of man on the strings.

36. -5g or 5 kg (for static equilibrium condition)

But Zero as machine can't display negative acceleration values.

Detailed Solution is in the adjoining attachment. Kindly Refer.

Answer 2

Explanation:

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