Question

Figure shows a man of mass 70 kg standing on a light weighing
machine kept in a box of mass 35 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 manager to keep the box at rest.
The weight shown by the machine​

Answer 1

Given:

Figure shows a man of mass 70 kg standing on a light weighing machine kept in a box of mass 35 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.

To find:

Reading shown by weighing machine.

Calculation:

Weighing Machines always show the normal reaction of the person standing on it, so let normal reaction be N:

Let mass of man be m , mass of box be M;

According to FBD of man:

$\rm{ \therefore \: mg = T + N}$

According to FBD of box:

$\rm{ \therefore \: Mg + N = T}$

Putting value of T in 2nd Equation:

$\rm{ = > \: Mg + N = T}$

$\rm{ = > \: Mg + N = mg -N }$

$\rm{ = > \: 2N = mg -Mg }$

$\rm{ = > \: 2N = (m -M)g }$

$\rm{ = > \: N = \dfrac{ (m -M)g}{2}}$

Putting the available values:

$\rm{ = > \: N = \dfrac{ (70 -35) \times 10}{2}}$

$\rm{ = > \: N = \dfrac{ 35 \times 10}{2}}$

$\rm{ = > \: N = \dfrac{ 350}{2}}$

$\rm{ = > \: N = 175 \: Newton}$

So, reading shown is:

$\boxed{ \bf{\: N = 175 \: Newton}}$