A man of mass 70kg stands on a weighing scale in a lift which is moving (a) upwards with a uniform speed of 10ms^1,
A man of mass 70kg stands on a weighing scale in a lift which is moving (a) upwards with a uniform speed of 10ms-1,
(b) downwards with a uniform acceleration of 5ms-2,
(c) upwards with a uniform acceleration of 5ms-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?
1 Answers
(a) Mass of the man, m = 70 kg
Acceleration, a = 0
Using Newton’s second law of motion, we can write the equation of motion as,
R – mg = ma
where
ma is the net force acting on the man.
As the lift is moving at a uniform speed, acceleration a = 0
∴ R = mg
= 70 × 10
= 700 N
Therefore, reading on the weighing scale =700/g
=100/10
=70 kg
(b) Mass of the man, m = 70 kg
Acceleration, a = 5 m/s2 , downward
Using Newton’s second law of motion, we can write the equation of motion as:
R + mg = ma
R = m(g – a)
= 70 (10 – 5)
= 70 × 5
= 350 N
∴ Reading on the weighing scale = 350 g =350/10 = 35 kg
(c) Mass of the man, m = 70 kg
Acceleration, a = 5 m/s2 upward
Using Newton’s second law of motion, we can write the equation of motion as:
R – mg = ma
R = m(g + a)
= 70 (10 + 5)
= 70 × 15
= 1050 N
Therefore,
Reading on the weighing scale =1050/g
=1050/10
= 105 kg
(d) When the lift moves freely under gravity, acceleration a =g
Using Newton’s second law of motion, we can write the equation of motion as,
R + mg = ma
R = m(g – a)
= m(g – g)
= 0
∴ Reading on the weighing scale =0/g = 0 kg
The man will be in a state of weightlessness.