# Assignment question for 1st year physics (Review) - Am I answering it logically?

A woman stands on a scale in a moving elevator. Her mass is 60.0 kg, and the combined mass of the elevator and scale is an additional 815 kg. Starting from rest, the elevator accelerates upward. During the acceleration, the hoisting cable applies a force of 9410 N. What does the scale read during the acceleration?

Mass of woman: 60 kg

Mass of elevator and scale: 815 kg

Starting from rest means Vo = 0 m/s

Force applied is 9410 Newtons in the positive y direction (making up positive y and down negative y)

g = -9.8 m/s^2

Sigma Fy = m*a

Solution:

Two forces acting on the elevator:

Force of gravity against the force of the tension in the cable.

Therefore:

Sigma Fy = m*a
Force of cable on the elevator + Force of gravity = (mass of the woman + mass of the elevator + mass of the scale)*Acceleration of the elevator

We have everything except acceleration of the elevator therefore we can solve for that.

9410 N + (mass of elevator + mass of woman + mass of scale)*(-9.8 m/s^2) = (mass of elevator + mass of woman + mass of scale) * a

9410 N + (815 kg + 60 kg)* (-9.8 m/s^2) = (815 kg + 60 kg)*a
9410 N + (-8575 N) = 875*a
835 N = 875 kg*a
835 N / 875 kg = a
a = 0.95 m/s^2 upwards

Before I solve for the scale's readings, I will assume that it will read a larger mass since it went from rest to an acceleration. The scale will read:
Force felt by the scale = (m)*(acceleration of elevator)
835 N = (m) * ( 0.95)
835/0.95 = m
m = 878.9 kg
Now subtract 815 from it.
878.9 - 815 = 63.9 kg

Is this correct?

- Mohammed.