What Does the Scale Show When an Elevator Accelerates Upward?

[Mohammed17]

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.

 

[TSny]

g = -9.8 m/s^2Sigma 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

This looks OK. However, most textbooks use the symbol ##g## to represent the magnitude of the acceleration due to gravity. Thus, most books take the value of ##g## at the surface of the Earth to be 9.8 m/s² and not -9.8 m/s². In this case, if positive direction for ##y## is upward, then the equation ##\sum F_y = M_{\rm sys} \, a_y## would be set up as ##F_{\rm applied} - M_{\rm sys} \, g = M_{\rm sys} \,a##. Here, ##g## is a positive number and the minus sign is indicating that the direction of the force due to gravity is downward.

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.

This part is incorrect. It is not correct to say
Force felt by the scale = (m)*(acceleration of elevator)

The hint for this part is to apply ##\sum F_y = m a_y## to just the person. ##\sum F_y## will include the force of gravity on the person and the normal force that the scale exerts on the person. The normal force that the scale exerts on the person can then be determined.

What does Newton's 3rd law say about the relation between the force that the scale exerts on the person and the force that the person exerts on the scale?

What is the relation between the force that the person exerts on the scale and the reading of the scale?

The problem statement does not tell us if the scale is calibrated in Newtons or kg.