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 $10m{s}^{-1}$,
b) downwards with a uniform acceleration of $5m{s}^{-1}$,
c) upwards with a uniform acceleration of $5m{s}^{-1}$.
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

(i) 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
$\therefore R=mg=70\times 10=700N$
$\therefore$ Reading on the weighing scale $=\frac{700}{g}=\frac{700}{10}=70kg$
(ii) Mass of the man, m = 70 kg
Acceleration, $a=5m/s^2$ downward
Using Newton's second law of motion, we can write the equation of motion as:
$mg-R=ma,\therefore R=m(g-a)=70(10-5)=70\times 5=350N$
$\therefore$ Reading on the weighing scale $=\frac{350}{g}=\frac{350}{10}=35kg$
(iii) Mass of the man, m = 70 kg
Acceleration, $a=5m/s^2$ upward
Using Newton's second law of motion, we can write the equation of motion as:
$R-mg=ma,R=m(g+a)=70(10+5)=70\times 15=1050N$
$\therefore Readingontheweighingscale=\frac{1050}{g}=\frac{1050}{10}=105kg$
(iv) 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:
$mg-R=maR=m(g-a)=m(g-g)=0$
Reading on the weighing scale $=\frac{0}{g}=0kg$
The man will be in a state of weightlessness.