Question

In 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 ?

Answer 1

(i)Given, Mass of man = 60kg

Let R' = Apparent weight of man in this case

From free body diagram of man,

R' + T - 60 kg = 0

$\Rightarrow$ T = 60g - R' ....(i)

From free body diagram of box,

T - R' - 30g = 0 ....(ii)

From equation (i),

60g - R' - R' - 30g = 0

$\Rightarrow$ R' = 15g

Hence weight shown by the machine is 15 kg.

(ii) To get his correct weight, suppose the applied force is T and so, accelerates upward with a. in this case,

Given that ; correct weight = R = 60 g where g= acceleration due to gravity

From the free body diagram of the man

T' + R -60 g -60 a = 0

$\Rightarrow$ T' - 60 a = 0 [since R= 60 g]

$\Rightarrow$ T' = 60 a ....(i)

From the free body diagram of the box,

T' - R - 30g -30a = 0

$\Rightarrow$ T' - 60g - 30g -30a = 0

$\Rightarrow$ T' - 30a + 90g = 0

$\Rightarrow$ T'= 30a - 900 ....(ii)

From equations (i) and (ii), we get

T' = 2T' - 1800

$\Rightarrow$ T' = 1800 N

So, he should exert 1800 N force on the rope to get correct reading.