# A man of mass 70 kg stands on a weighing scale in a lift which is moving

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 $10 \mathrm{~m} \mathrm{~s}^{-1}$,

(b) downwards with a uniform acceleration of $5 \mathrm{~m} \mathrm{~s}^{-2}$,

(c) upwards with a uniform acceleration of $5 \mathrm{~m} \mathrm{~s}^{-2}$.

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?

solution:

(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

$\therefore R=m g$

= 70 × 10 = 700 N

$\therefore$ Reading on the weighing scale $=\frac{700}{\mathrm{~g}}=\frac{700}{10}=70 \mathrm{~kg}$

(b) Mass of the man, $m=70 \mathrm{~kg}$

Acceleration, $a=5 \mathrm{~m} / \mathrm{s}^{2}$ downward

Using Newton's second law of motion, we can write the equation of motion as:

$R+m g=m a$

$R=m(q-a)$

$=70(10-5)=70 \times 5$

$=350 \mathrm{~N}$

$\therefore$ Reading on the weighing scale $=\frac{350}{\mathrm{~g}}=\frac{350}{10}=35 \mathrm{~kg}$

(c) Mass of the man, $m=70 \mathrm{~kg}$

Acceleration, $a=5 \mathrm{~m} / \mathrm{s}^{2}$ 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 $=\frac{1050}{\mathrm{~g}}=\frac{1050}{10}=105 \mathrm{~kg}$

(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

$\therefore$ Reading on the weighing scale $=\frac{0}{g}=0 \mathrm{~kg}$

The man will be in a state of weightlessness.