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

(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

∴Reading on the weighing scale =

(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 =

(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

∴Reading on the weighing scale =

(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 =

The man will be in a state of weightlessness.