Apparent weight of a man standing in an elevator

[Tavon]

Homework Statement

An 80.0 kg man is standing on a scale calibrated in Newtons inside an elevator. Determine the apparent weight of the man as shown by the scale as the elevator is
a. stationary. [ans.785 N]
b. accelerating downwards at a rate of 1.10 m/s^2. [ans. 697 N]
c. accelerating upwards at a rate of 1.10 m/s^2. [ans. 873 N]
d. moving upwards at a constant velocity. [ans. 785 N]

Homework Equations

Fnet= ma

3. The Attempt at a Solution [/B]
For a. I used the equation Fnet=ma and multiplied 80 by 9.81 and got 785 N, but when I tried the same method for b. (80 kg * 1.10 m/s^2) I got 88 N. This is wrong and I can't figure out how to solve it, I now that the apparent weight will decrease since the elevator is going downwards. I feel like the method for solving c. and d. is going to be the same as the method for solving b..

 

[PeroK]

For part a) the acceleration is 0, as the elevator and man are stationary. So, why didn't you calculate the apparent weight as 0?

 

[Tavon]

That's what I thought at first but the answer in my Physics 20 workbook said 785 N and I tried the method mentioned earlier to get the answer but I didn't understand how it worked. I thought it might be 785 N since gravity is 9.81 and it is still acting on the elevator even though it isn't moving.

 

[PeroK]

That's what I thought at first but the answer in my Physics 20 workbook said 785 N and I tried the method mentioned earlier to get the answer but I didn't understand how it worked. I thought it might be 785 N since gravity is 9.81 and it is still acting on the elevator even though it isn't moving.

Does gravity stop acting when the elevator moves?

 

[Tavon]

I don't think so. It doesn't say so in the question so I guess it still applies force on the elevator.

 

[PeroK]

I don't think so. It doesn't say so in the question so I guess it still applies force on the elevator.

You may be missing the key concept about forces. If you are standing on the ground, how many forces are acting on you?

 

[Tavon]

2? Normal Force and Fg?

 

[PeroK]

2? Normal Force and Fg?

Yes. What about in an elevator accelerating downwards?

 

[Tavon]

Force applied as well or Tension since the cables lower it?

 

[PeroK]

Force applied as well or Tension since the cables lower it?

I meant the man in the elevator. What are the forces on him?

 

[Tavon]

Fg and Normal Force but if the elevator is accelerating down the man's apparent weight will decrease.

 

[PeroK]

Fg and Normal Force but if the elevator is accelerating down the man's apparent weight will decrease.

Yes. So what can you say about the normal force?

 

[Tavon]

It stays the same.

 

[Tavon]

Wait it increases...?

 

[PeroK]

Wait it increases...?

Think about the forces on the man. Gravity pushes him down and the normal force pushes him up.

If he is accelerating downward, then ...?

 

[Tavon]

Then normal force has to try harder to push him up, but its magnitude is still lower than the magnitude of gravity.

 

[PeroK]

Then normal force has to try harder to push him up, but its magnitude is still lower than the magnitude of gravity.

I would say the normal force must be less than gravity. Acceleration is caused by an imbalance in forces.

Anyway, the other important question is what is the reading on the scales in relation to these forces?

 

[Tavon]

As gravity increases apparent weight decreases and as normal force increases apparent weight increases?

 

[PeroK]

As gravity increases apparent weight decreases and as normal force increases apparent weight increases?

I'm going offline now, so maybe someone else can pick this up.

No. Gravity is constant. It doesn't change.

The reading on the scales is the normal force. You need to have a think about this perhaps.

 

[Tavon]

Ok. Thankyou.

I solved a), b), and c) using the method in the attachment, but I am not sure how to solve d).

 

[jbriggs444]

I solved a), b), and c) using the method in the attachment, but I am not sure how to solve d).

If the man is moving upward with a constant velocity, what is his acceleration?

 

[Tavon]

0?

 

[jbriggs444]

0?

Right. So if Fnet=ma, what does that tell you about Fnet?

 

[Tavon]

Fnet= ma
Fnet= 80.0 kg x 0
Fnet = 0

 

[jbriggs444]

Can you see your way from there to the answer to d) ?