Riding the Elevator: Forces on Passenger & Floor

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AI Thread Summary
The discussion revolves around calculating the forces acting on a passenger in an elevator based on its acceleration profile. The maximum upward force on the passenger occurs during the elevator's positive acceleration, while the minimum force is experienced when the elevator is at rest, factoring in gravitational force. Participants clarify that the forces should be considered in relation to gravity, with upward acceleration adding to the force and downward acceleration subtracting from it. There is also a realization that the forces on the passenger and the floor are equal in magnitude due to Newton's third law. Overall, understanding the relationship between acceleration and gravitational force is key to solving the problem.
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Homework Statement


A 48 kg passenger rides in an elevator that starts from rest on the ground floor of a building at t = 0 and rises to the top floor during a 10 s interval. The acceleration of the elevator as a function of the time is shown in Fig. 3-32, where positive values of the acceleration mean that it is directed upward. Give the magnitude and direction of the following forces.

Fig. 3-32 (in words)
-max positive acceleration = 2 m/s^2 at t = 2 s
-max negative acceleration = -3 m/s^2 at t = 8.5 s
-0 acceleration from 4 s to 7 s


(a) the maximum force on the passenger from the floor

(b) the minimum force on the passenger from the floor

(c) the maximum force on the floor from the passenger



Homework Equations


F = ma

The Attempt at a Solution


For part a), I thought it would just be F =(48kg)(2), but that was not correct. I know it's in an upward direction, but I can't seem to find the magnitude.

Part b) I figured that the minimum force should be when the acceleration = 0, so I tried F=(48)(0) = 0. But it just doesn't seem logical to me...

Part c) I think the maximum force should be when a = -3, but I plugged it into the formula and it was not correct.

Any help would be appreciated, thanks!
 
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pmelt said:

Homework Statement


A 48 kg passenger rides in an elevator that starts from rest on the ground floor of a building at t = 0 and rises to the top floor during a 10 s interval. The acceleration of the elevator as a function of the time is shown in Fig. 3-32, where positive values of the acceleration mean that it is directed upward. Give the magnitude and direction of the following forces.

Fig. 3-32 (in words)
-max positive acceleration = 2 m/s^2 at t = 2 s
-max negative acceleration = -3 m/s^2 at t = 8.5 s
-0 acceleration from 4 s to 7 s(a) the maximum force on the passenger from the floor

(b) the minimum force on the passenger from the floor

(c) the maximum force on the floor from the passenger

Homework Equations


F = ma

The Attempt at a Solution


For part a), I thought it would just be F =(48kg)(2), but that was not correct. I know it's in an upward direction, but I can't seem to find the magnitude.

Part b) I figured that the minimum force should be when the acceleration = 0, so I tried F=(48)(0) = 0. But it just doesn't seem logical to me...

Part c) I think the maximum force should be when a = -3, but I plugged it into the formula and it was not correct.

Any help would be appreciated, thanks!

Welcome to PF.

What do you figure is the effect of gravity on the passenger?
 
Thanks :)

Since the elevator is accelerating upward for a) and b), would you subtract the acceleration from gravity?
 
pmelt said:
Thanks :)

Since the elevator is accelerating upward for a) and b), would you subtract the acceleration from gravity?

No. You would add the forces. But ... one of the forces is negative because 1 of the accelerations is negative. Accelerating up would add the force. Decelerating (slowing at the top) would subtract.

(Draw from your own experience in an elevator.)
 
Okay I got it now.. so then would a) and c) have the same magnitude because of Newton's 3rd law?

Thank you very much by the way, much appreciated.
 
pmelt said:
Okay I got it now.. so then would a) and c) have the same magnitude because of Newton's 3rd law?

Thank you very much by the way, much appreciated.

Looks like it don't you think?
 
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