A man of mass 70 kg stands on a weighing scale in a lift which is moving
A. upwards with a uniform speed of 10 m s-1,
B. downwards with a uniform acceleration of 5 m s-2,
C. upwards with a uniform acceleration of 5 m s-2.
What would be the readings on the scale in each case?
D. What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?
Let the force experienced by the weighing scale needle be F,
Given,
Mass of the man = 70 kg
Acceleration due to gravity, g = 10 m/s2
A. If the lift if moving upwards with a uniform velocity,
The acceleration of the lift ‘a’ is zero,
Using the Newton’s second law of motion, The net force is thus gravitational force acting downward,
F – m× g=0
⇒ F= m×g =70 kg × 10 m/s2
⇒ F=700 N
The apparent weight m’ shown on the weighing machine is that of mass which is calculated as,
⇒ m’= 70 kg
which is same as the original weight of the man.
B. If the lift is moving downwards
Given,
Acceleration of the lift, ‘a’=5 m/s2
The motion is in the same direction as the acceleration due to gravity,
Using the Newton’s second law of motion, the net force thus is,
F + mg = ma
⇒ F = m× (a-g) = 70 kg× (5– 10) m/s2
⇒ F = 70 kg× (-5 m/s2) = -350 N
Thus, the apparent weight of the man is,
C. If the lift is moving upwards,
Given,
Acceleration of the lift, ‘a’= 5 m/s2
The motion is in the opposite direction as the acceleration due to gravity,
Using the Newton’s second law of motion, the net force thus is,
F – mg = ma
⇒ F = m (a+g) = 70 kg × (5+10) m/s2
⇒ F = 70 kg × 15 m/s2 =1050 N
The apparent weight of the am on the weighing machine thus is,
D. When the lift is moving downward freely under gravity,
The acceleration of the lift, ‘a’ = g= 10 m/s2
Using the Newton’s second law of motion, the net force thus is,
F + mg = mg
⇒ F = 0 N
Thus, the body is under the state of free fall with a reading of zero in the weighing machine.