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

Principle

Scale reads normal force = apparent weight:

$$N - mg = ma \quad \Rightarrow \quad N = m(g \pm a)$$ $$\text{Scale reading} = N \quad (g = 9.8 \, \text{m/s}^2)$$

Results Summary

Case	Motion	Apparent Weight (N)	Scale Reading
1	Up, \(v = 10\) m/s (constant)	\(70 \times 9.8 = 686\)	686 N
2	Down, \(a = 5\) m/s²	\(70 \times (9.8 - 5) = 343\)	343 N
3	Up, \(a = 5\) m/s²	\(70 \times (9.8 + 5) = 1030\)	1030 N
4	Free fall (\(a = g\))	\(70 \times (9.8 - 9.8) = 0\)	0 N

Case 1: Constant Upward Speed

\(a = 0\) → \(N = mg = 70 \times 9.8 = 686\) N (normal weight)

Case 2: Downward Acceleration

\(a = 5\) m/s² down → \(N = m(g - a) = 70 \times 4.8 = 343\) N (half weight!)

Case 3: Upward Acceleration

\(a = 5\) m/s² up → \(N = m(g + a) = 70 \times 14.8 = 1030\) N (~1.5× weight)

Case 4: Free Fall Failure

\(a = g\) down → \(N = m(g - g) = 0\) N (weightlessness)

Final Scale Readings

686 N, 343 N, 1030 N, 0 N

Physical Insight

• Scale measures contact force, not true mass
• Acceleration changes effective gravity
• Free fall = elevator in orbit = zero-g