Question

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 he manages to keep the box at rest (i) what is the weight shown by the machine ? (ii) what force should the man exert on the rope to get his correct weight on the machine ? (Please explain the FBDs you draw for each body)

Answer 1

    Let the tension force be T in the rope.  The downward forces acting on the system of the man and the box are the weight of the man and the weight of the box. The upward forces are T and T on each side of the rope.

   As the system is in a static equilibrium:
        T + T = (60 + 30) g 
=>       T = 45 kg f   or   45 g Newtons    = 450 N      if g = 10m/s^2

The forces acting on the man are: Tension T, reaction from the weighing machine, and weight 60 g.  So,
   Normal reaction of the weighing machine = 60 - T = 15 kg f.

(i)  The weighing machine shows its normal reaction force as the weight of the person standing on it.  So weight displayed = 15 kg.

(ii) Suppose the man pulls down the rope with a force F and the tension becomes T. .
The forces acting downwards on the rope at the two ends are:  F, weights of man and the box.  The upward forces are Tension T on each side of rope.

The system being in equilibrium:  2 T = F + 60 kg f + 30 kg f = F + 90 kg f
=>     T = 45 kg f + F/2
=>  Normal reaction force from the weighing machine on the man
              N = 60 + F - T = 15 kg f + F/2

If N is to become 60 kg f, then  F has to be 90 kg f or 900 Newtons.

Answer 2

2nd part :-
Since we have to calculate the force with which the man should hold the rope this means we have to calculate the tension 'T' on the rope
in earlier case we got T = ma
and a = 3g
putting the value of a we get
T= m× 3g
T= 60 ×3×10
T =1800 N
This is our answer for 2nd part ..

For 1st part see all the attachments