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Laws of Motion

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Question : 13 of 40
Marks: +1, -0
A man of mass 70 kg stands on a weighing scale in a lift which is moving
(a) upwards with a uniform speed of 10 m s−1^{-1},
(b) downwards with a uniform acceleration of 5 m s−2^{-2},
(c) upwards with a uniform acceleration of 5 m s−2^{-2}.
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?
Solution:  
Here, m = 70 kg, g = 10 m s−2^{-2}
The weighing machine in each case measures the reaction R i.e. the apparent weight.
(a) When the lift moves upwards with a uniform speed, its acceleration is zero.
∴R=mg=70×10=700 N\therefore R = mg = 70 \times 10 = 700\,\mathrm{N}
(b) When the lift moves downwards with a = 5 m s−2^{-2}
R=m(g−a)R = m(g - a)
=70(10−5)=350 N= 70(10 - 5) = 350\,\mathrm{N}
(c) When the lift moves upwards with a = 5 m s−2^{-2}
R=m(g+a)R = m(g + a)
=70(10+5)=1050 N= 70(10 + 5) = 1050\,\mathrm{N}
(d) If the lift was to come down freely under gravity, downwardacceleration
a=ga = g
∴R=m(g−a)=m(g−g)=zero\therefore R = m(g - a) = m(g - g) = \text{zero}
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