Question

a man of mass 50kg standing on a light weighing kept in a box of mass 30kg. 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 managres to keep the box at rest the weight shown ny the machine is

Answer 1

50 as the weight will remain same

Answer 2

let

weight of man on scale = W

force exerted by man on rope = F

tension on the rope = T

now,

the equation of motion of the stationary man will be written as

T - mg = -Wg

[here m = 60kg]

so,

T - 60g + Wg = 0  ..........................(1)

and

the equation of the stationary box will be written as

T - m'g = Wg

[here, m = 30kg]

so,

T - 30g - Wg = 0 ..........................(2)

now,

by subtracting (1) from (2), we get

[T - 60g + Wg] - [T - 30g - Wg] = 0

so,

2Wg = 30g

thus, we have

W = 15kg

now,

For the scale to show the correct weight [i.e. 60kg] let the man standing on it changes the force on the rope and in addition to that, constantly accelerates with an acceleration 'a'. Then , in this case the equation of motion will be given as

for the man

T - mg + mg = ma

or

T - 60g + 60g = 60a

so,

T = 60a ..........................(3)

and

for the box

T - m'g - mg = m'a

or

T - 30g - 60g = 30a

or

T - 90g = 30a

so,

T = 90g + 30a ..........................(4)

now,

from (3) and (4), we can conclude that

T = 2T - 180g

T = 180g

or finally

T = 1800 N