Question

A person of mass \[50kg\] stands on a weighing scale oil a lift if the lift is descending with a downward acceleration of $9m{s^{ - 2}}$ . What would be the reading of the weighing scale?

Answer 1

Hint: To answer this issue, we must first rewrite the supplied facts, and because we have the acceleration value and also know the gravitational (acceleration due to gravity), we must first get the apparent weight, and then we can determine the weighing scale reading using the apparent weight.

Formula used:
$N = mg - ma$
Here, $N$ = normal force acting upward along the body , $m$ = the mass of the person , $a$ = acceleration of the lift and $g$= acceleration due to gravity.


Complete step-by-step solution:
Here, we are given that;
Mass of the person is = \[50kg\]
Acceleration of the lift = \[a\] = $9m{s^{ - 2}}$
For the following question, if the lift descends with an acceleration the apparent weight on the weighing scale will decrease.

When a person steps on the weighing scale, the person experiences a normal reaction. The magnitude of a normal reaction is determined using a calibrated weighing scale reading.
It is possible to assume that the weighing scale reading corresponds to the person's visible weight.
Normal reaction acts upward along the upward direction of the individual, while gravitational force acts downward on the person.
Thus,
 $
  W' = N = \left( {mg - ma} \right) \\
  \,\,\,\,\,\,\,\,\,\,\,N = m\left( {g - a} \right) \\
 $
The apparent weight of the lift on the weighing scale would now be; due to reaction force.
$
  W' = 50\left( {10 - 9} \right) \\
  W' = 50N \\
 $
Thus, the reading of the weighing scale would be
$\dfrac{R}{g} = \dfrac{{50}}{{10}} = 5kg$
Therefore, the reading of the weighing scale would be $5kg$.

Note:During a free fall of a body under gravity in an elevator, the apparent weight of the body reaches zero. As the force of reaction between the individual and the plane with which he is in contact vanishes, the body might be said to become weightless.