A spring weighing machine inside a stationary lift reads 50 kg when a man stands on it. What would happen to the scale reading if the lift is moving upward with (i) constant velocity, and (ii) constant acceleration?

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Solution

(i) In the case of constant velocity a = 0
therefore, N = mg so $W_{a p p} = W_{a c t}$
(ii) In this case the upward acceleration
N - mg = ma
So, N = mg + ma = m(g + a)
So, $W_{a p p} > W_{a c t}$
Hence scale reading will be = $\frac{50 (g + a)}{g} = 50 (1 + \frac{a}{g})$

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A spring weighing machine inside a stationary lift reads 50 kg when a man stands on it. What would happen to the scale reading if the lift is moving upward with (i) constant velocity, and (ii) constant acceleration?

(i) In the case of constant velocity a = 0 therefore, N = mg so Wapp=Wact(ii) In this case the upward acceleration N - mg = ma So, N = mg + ma = m(g + a) So, Wapp>WactHence scale reading will be = 50(g+a)g=50(1+ag)

(i) In the case of constant velocity a = 0
therefore, N = mg so $W_{a p p} = W_{a c t}$
(ii) In this case the upward acceleration
N - mg = ma
So, N = mg + ma = m(g + a)
So, $W_{a p p} > W_{a c t}$
Hence scale reading will be = $\frac{50 (g + a)}{g} = 50 (1 + \frac{a}{g})$