Physics Q: Force Exerted by 50-kg Student in Stationary/Accelerating Elevator

[celect]

I'm doing self study on physics course I'm reviewing self test and
have question on force.

A 50-kilogram student stands in an elevator. How much force does she
exert on the elevator floor if:


The elevator is stationary?


For the above would it be true to say that if the elevator is stationary
that there is zero force exerted. Is this an example of the upward and downward forces balancing?



The elevator accelerates upward at 1 meter per second squared (m/s2)?
50-kilogram.

I used (F=ma)to solve
(50kg)(9.8m/s^2)=490N
Am I using the correct formula?

 

[jamesrc]

Originally posted by celect

For the above would it be true to say that if the elevator is stationary
that there is zero force exerted. Is this an example of the upward and downward forces balancing?

No, the force exerted by the girl onto the floor is her weight (mg). The force exerted by the floor on the girl is usually called the normal force, N. Since the system is stationary, its acceleration is 0. Using F_net = ma (Netwon's second law), we see that the vector sum of the forces must be 0 (the forces balance out). You would use that if you wanted to find out the value of the normal force (N-mg = 0).

Originally posted by celect
The elevator accelerates upward at 1 meter per second squared (m/s2)?
50-kilogram.

I used (F=ma)to solve
(50kg)(9.8m/s^2)=490N
Am I using the correct formula?

This is not correct either. You are asked for the force that the girl exerts on the elevator, which is unaffected by the motion of the elevator. The answer for this part is still mg. You would need the acceleration to find the normal force, which is no longer equal to mg. A free body diagram should help you to see that ma = N - mg, where a is the upward acceleration.

(edited for stupidity in last sentence; should be error-free now)

 

[celect]

In my text it does not talk about normal force it list Applied force, net force, and support force.

the formula (mg) would be (50kg*g) how do you determine g=1

 

[jamesrc]

Originally posted by celect
In my text it does not talk about normal force it list Applied force, net force, and support force.

What I called normal force is most likely what they refer to as support force then. You could also call it a reaction force.

g does not change (unless you care about altitude changes which you don't here), so it is 9.81 m/s² (or however many digits you round it off to). That's why the answer for the first part and the second part are the same. 1 m/s/s would be used as a in the expression (m)a = N - mg, where you would be solving for N (the normal force/support force/reaction force). We can even give it another name: apparent weight. The apparent weight is what a scale would read if the person was standing on it in the elevator.

(I just realized I wrote something stupid in my first post; going to fix it now.)

 

[Doc Al]

I think jamesrc makes a few statements that might be confusing. (For example: "..the force that the girl exerts on the elevator, which is unaffected by the motion of the elevator") Here's my version.

Originally posted by celect
A 50-kilogram student stands in an elevator. How much force does she
exert on the elevator floor if:


The elevator is stationary?

If the elevator is stationary, its acceleration is zero. Now consider the forces on the girl. There are only two: the force that the elevator pushes her up with (call that N) and the force that the Earth pulls her down with (which equals her weight, mg). These forces must balance:
N-mg=0, so N=mg=(50)(9.8)=490 N, which means the elevator pushes up on her with a force of 490 Newtons.

Now, by Newton's third law, she pushes down with an equal and opposite force on the elevator. So she pushes down on the elevator with 490 N.

The elevator accelerates upward at 1 meter per second squared (m/s2)?
50-kilogram.

This time the acceleration (of the girl and the elevator) is not zero. So the forces on the girl add up to:
N-mg=ma, so N=mg+ma=(50)(10.8)= 540 N. The elevator pushes her up harder now, so she pushes back down harder! She pushes down on the elevator with a force of 540 Newtons. (Newton's third law again.)

 

[jamesrc]

Originally posted by Doc Al
I think jamesrc makes a few statements that might be confusing. (For example: "..the force that the girl exerts on the elevator, which is unaffected by the motion of the elevator") Here's my version.

Yeah, you're right. Sorry about that. I don't know why I interpreted the question so awkwardly. Newton's 3rd law is the way to go. The question effectively asks what the apparent weight of the girl is. Whatever you do, don't ask the girl how much she weighs; nothing good can come of that.

 

[Doc Al]

Originally posted by jamesrc
Whatever you do, don't ask the girl how much she weighs; nothing good can come of that.

A wise policy.

 

[nautica]

A self-study physics course? Good luck.

Nautica