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A man in an elevator at the forces affecting him

  • Thread starter aeromat
  • Start date
  • #1
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Homework Statement


A 75kg man stands in an elevator. What will be the force the elevator exerts on him when:
a) the elevator is at rest
b) the elevator is moving upward with a uniform acceleration of 2.0m/s^2
c) the elevator is moving downward with a uniform acceleration of 2.0m/s^2.


Homework Equations


Fg = mg
FNet = m(a)


The Attempt at a Solution


I got the following answers that matched the ones on my worksheet:
a) = 735.75 N [up]
The other two I got from just randomly pluging and adding/subtracting the acceleration of gravity with the uniform acceleration of the elevator.
However, I don't understand what I did and WHY this is the answer. I am the type that doesn't just take the answer as correct from getting it right, but until it makes sense to me..

b) F = (75)(9.81 + 2.0)
c) F = (75)(9.81 - 2.0)
 

Answers and Replies

  • #2
114
0
Just use Newton's Laws to know that a force corresponds to an acceleration.
As the elevator moves, a new force is exerted on the man. The direction of this force depends on the direction of motion of the elevator.
Does it make sense?
 
  • #3
114
0
No sorry it doesn't make sense. If you were to draw an FBD for b), and c), how would it look like? ;|
 
  • #4
If the elevator is going down, then the force on the man would be smaller, and vice versa for going up in an elevator.
 
  • #5
jhae2.718
Gold Member
1,161
20
Consider it from the man's (mass m) frame of reference. When the elevator (mass M) is at rest, in the man's frame there is a normal force (N) counteracting his weight (force of gravity, mg) applied to the bottom of the elevator. Using Newton's Second Law and calling up positive,
Fnet=manet
where Fnet=N-mg. Since the elevator is at rest, anet=0.
This implies: 0=N-mg leading to: N=mg.

However, when the elevator starts to move, the man's frame is no longer at rest, but is moving with some acceleration. The man only sees the forces of his weight and the normal force of the elevator, but the reference frame is accelerating. Thus, while the man sees no net acceleration with respect to the elevator, the frame has its own acceleration, which we'll call aF.

Our equation of motion becomes:
Fnet=manet+maF
which simplifies to:
Fnet=maF
then:
N-mg=maF
N=m(aF+g)

Now, if we expand our system to look at the elevator, our system becomes locally inertial again with two forces acting on the elevator, a tension in the cable that pulls/lowers the elevator, and the force of gravity on the elevator plus the man. The man's weight is added to the elevator's, and the now-internal forces between the elevator and man are ignored.
 
Last edited:

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