Question

A person is standing on a weighing machine placed on the floor of an elevator. The elevator starts going up with some acceleration, moves with uniform velocity for a while and finally decelerates to stop. The maximum and the minimum weights recorded are 72 kg and 60 kg. Assuming that the magnitudes of the acceleration and the deceleration are the same, find (a) the true weight of the person and (b) the magnitude of the acceleration. Take g = 9.9 $m/s^2$

Answer 1

When the elevator is accelerated upwards, maximum weight will be recorded

$R-(W-m_a)=0$

$\Rightarrow R=W+m_a$

= m (g+a) maximum weight

When decelerating upwards, maximum weight will be recorded.

$\Rightarrow R+ma-W=0$

$\Rightarrow$ R = W - ma

$\therefore$ m (g+a) = $72\times 9.9$ ....(1)

and $m(g-a)=60\times 9.9$ .....(2)

Now, mg + ma = $72\times 9.9$

and mg -ma = $60\times 9.9$

$\Rightarrow$ 2 mg = 1306.8

$\Rightarrow m=\frac{1306.8}{2\times 9.9}=66$kg

So, the true weight of the man is 66 kg.

(b) Again, to find the acceleration

mg + mg = $72\times 9.9$

$\Rightarrow a=\frac{72\times 9.9-66\times 9.9}{66}$

$=\frac{9.9\times 6}{66}=\frac{9.9}{11}$

=$0.9m/s^2$