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 $72kg$ and $60kg$. 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 $g=9.9m/s^2$

Answer 1

When the elevator is accelerating upwards, maximum weight will be recorded.

$R-(w+ma)=0$

$\Rightarrow R=W+ma=m(g+a)$ max.wt

When decelarating upwards, minimum weight will be recorded.

$R+ma-W=0$

$\Rightarrow R=W-ma=m(g-a)$

So, $m(g+a)=72\times 9.9$$m(g-a)=60\times 9.9$

Now $mg+ma=72\times 9.9$

$\Rightarrow mg-ma=60\times 9.9$

Therefore, $\Rightarrow 2mg=1306.8$

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

So, the true weight of the man is $66kh$

Again to find the acceleration

$mg+ma=72\times 9.9$

$\Rightarrow a=\frac{72\times 9.9-66\times 9.9}{66}=\frac{9.9}{11}=0.9m/s^2$