Question

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?

Answer 1

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

Reading on the weighing scale = 700/g = 700/10 = 70 kg

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

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

Reading on the weighing scale = 1050/g = 1050/10 = 105 kg

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.