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A spring weig...

Question

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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