Question

A man of mass 70 \(\text {kg}\) stands on a weighing scale in a lift which is moving 

1. upwards with a uniform speed of 10 \(\text m \,\text s^{-1}\) , 
2. downwards with a uniform acceleration of 5 \(\text m \,\text s^{-2}\) , 
3. upwards with a uniform acceleration of 5 \(\text m \,\text s^{-2}\) . What would be the readings on the scale in each case? 
4. What would be the reading if the lift mechanism failed and it hurtled down freely under gravity ?

Answer 1

(a) Mass of the man, \(\text m\) = 70 \(\text {kg}\) 
Acceleration, \(\text a\) = 0 
Using Newton’s second law of motion, we can write the equation of motion as:
\(\text R - \text {ma}\) = \(\text {ma}\)
Where, \(\text {ma}\) is the net force acting on the man.
As the lift is moving at a uniform speed, acceleration \(\text a\) = 0 
\(\therefore\) \(\text R\) = \(\text {mg}\) 
= 70 × 10 = 700 N
Reading on the weighing scale = \(\frac{700}{\text g}\) = \(\frac{700}{10}\) = 70 \(\text {kg}\)

---

(b) Mass of the man, m = 70 \(\text {kg}\)
Acceleration, \(\text a\) = 5 \(\text m/\text s^2\) downward
Using Newton’s second law of motion, we can write the equation of motion as:
\(\text R+\text{mg} = \text {ma}\)
\(\text R\) = \(\text m\)(\(\text g\) – \(\text a\)) 
= 70 (10 – 5) 
= 70 × 5 = 350 N 
Reading on the weighing scale = \(\frac{350}{\text g}\) = \(\frac{350}{10}\) = 35 \(\text {kg}\)

---

(c) Mass of the man, \(\text m\) = 70 \(\text {kg}\) 
Acceleration, \(\text a\) = 5 \(\text m/\text s^2\) upward 
Using Newton’s second law of motion, we can write the equation of motion as:
\(\text R\) – \(\text {mg}\) = \(\text {ma}\) 
\(\text R\) = \(\text m\)(\(\text g+\text a\)) 
= 70 (10 + 5) 
= 70 × 15 
= 1050 N
Reading on the weighing scale = \(\frac{1050}{\text g}\) = \(\frac{1050}{10}\) = 105 \(\text {kg}\)

---

(d) When the lift moves freely under gravity, acceleration \(\text a\) = \(\text g\)
Using Newton’s second law of motion, we can write the equation of motion as:
\(\text R+\text{mg} = \text {ma}\)
\(\text R\) = \(\text m\)(\(\text g\) – \(\text a\))  
= \(\text m(\text g-\text g)\) 
= 0
Reading on the weighing scale =\(\frac{0}{\text g}\)= 0 \(\text {kg}\)
The man will be in a state of weightlessness.