Question

A person of mass 70kg stands on a weighing machine in a lift which is
$(A)$ Moving upwards with the uniform speed of 10m/s
$(B)$ Moving down with a uniform acceleration of 5m/s2
$(C)$ Falling freely under gravity. What would be the reading of the weighing machine in each case?

Answer 1

Hint: If you stand on a scale in a lift that offers a reaction R accelerating upward, you feel heavier because the lift's floor presses harder on your feet, and the scale will show a higher reading than when the lift is at rest. On the other hand, when the lift accelerates downward, you feel lighter. And when the lift falls freely then you will be in a state of Weightlessness.


Complete answer:
$(A)$ For condition A) we have to take
Mass of man \[m = 70kg\] and reaction force $R$
Acceleration \[a = 0\]
Using Newton’s second law of motion we can write the equation as
$R - mg = ma$ Where,
($ma$is the net force acting on the man inside the lift and as the lift is moving with a uniform speed, acceleration = 0)
Therefore we get $R = mg = 70 \times 10 = 700N$. Hence Reading on the weighing scale is $\dfrac{{700}}{{10}} = 70kg$
$(B)$For condition B) we have to take
Mass of man \[m = 70kg\] and reaction force $R$
Acceleration \[a = 5m{s^{ - 2}}\] downwards
Using Newton’s second law of motion we can write the equation as
$R + mg = ma$
$R = m(g - a) = 70(10 - 5) = 350N$ Hence Reading on the weighing scale is $\dfrac{{350}}{{10}} = 35kg$
$(C)$ For condition C) we have to take $a = g = 10N$ and reaction force $R$
Using Newton’s second law of motion we can write the equation as
$R + mg = ma$
$R = m(g - a) = 70(10 - 10) = 0N$ Hence Reading on the weighing scale is $\dfrac{0}{{10}} = 0kg$
The man will be in a state of Weightlessness.

Note: The true weight is caused by gravity and it is the force exerted by earth's gravity. The force applied by the scale is called apparent weight and it does not change with constant speed. When both the parties are falling freely it undergoes this same type of Weightlessness as observed when a satellite is orbiting the earth.