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Elevators and Force

  1. Apr 6, 2009 #1
    1. The problem statement, all variables and given/known data
    A 60 kg person is standing in an elevator. Calculate each one of the forces acting on the person while the elevator is: a) at rest, b) moving up with a constant velocity of 3m/s,
    c) accelerating down at 4m/s^2, d) accelerating up at 4m/s^2.

    2. Relevant equations
    Force = mass times acceleration. im not exactly sure what "each one" is supposed to imply, i just know that for the first one, the net force is zero. i also know that the normal force is equal to the weight of an object, W= mg, when the object is in equilibrium.

    3. The attempt at a solution
    im not sure what formulas/figures to use in order to show each one of the forces for each situtation. should i write F= 0 for the forces in the first example? such as: F_normal = 0, for the force of the elevator down, F_g = 0 for the force of gravity, and then F_net = 0 for the net force?
  2. jcsd
  3. Apr 6, 2009 #2
    Just do a sum of forces, fill in the knowns, and solve for the unknowns. Keep the equation F = ma in mind.

    I'll give you a second hint. If the elevator is moving at a constant velocity (for part b), what is the acceleration?
  4. Apr 6, 2009 #3


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    I think it's best if we start working through each one of these questions in turn and discuss any questions as we go.

    So for part (a), name all the forces acting on person and specify their value.
  5. Apr 6, 2009 #4
    the sum of forces would something like: Fg + Fn?
    the accleration for constant velocity would be zero
  6. Apr 6, 2009 #5
    okay-for part a: the forces acting are the force of gravity down, and the force of the elevator pushing up on the person. both would be zero i think, because it is at rest (net force is zero)
  7. Apr 6, 2009 #6
    Ok, be careful there... what do you mean by "both?"
  8. Apr 6, 2009 #7
    oops i mean the force of gravity is 9.8 m/s^2
  9. Apr 6, 2009 #8
    the force of gravity pushing down on the person is 9.8m/s^2, and the force of the elevator pushing up on the person is zero. The net force is zero i think.
  10. Apr 6, 2009 #9
    Incorrect. Force = ma. You claimed that Force = a in your first statement. Fix that.

    You're right that the net force is zero, but write that mathematically. If you do that, you'll find that the normal force CANNOT equal zero.
  11. Apr 6, 2009 #10
    im still confused about this. im not sure how to write it even though i have notes from class. i know that Fg = mg, and -Fg + Ft = -ma but the t stands for tension which doesnt apply to this problem. what the heck is the normal force really and what does it have to do with the F= ma???
  12. Apr 6, 2009 #11
    this makes no sense, how can you add two things which dont equal zero, and get a net force of zero??
  13. Apr 6, 2009 #12
    Ok, forget forces. What's 3 + (-3)?
  14. Apr 6, 2009 #13
    zero. would it have something to do with: -1Newton and + 1Newton?
  15. Apr 6, 2009 #14
    You're on the right track. Keep in mind that force is a vector, not a scalar. What that means is it has direction. If you have a force going straight up and a force going straight down with the same magnitude, the vector sum will be zero.

    Now what's pushing upwards? Well, you asked "what's a normal force." I find it hard to believe it wasn't taught in class, but I'll give you an example.

    Lets say you have a book sitting on the table with a mass of 1 kg. The force of gravity is pulling down on that book with a force of 9.8 newtons. What's keeping that book from falling?
  16. Apr 6, 2009 #15
    the table is. Is that the normal force? would it be 1 Newt.? 9.8m/s^2 = 1 N right?
    then Fg= mg and Fg + Fn = 1 N - 1N = 0 = net force?
    mg = weight which is 60 kg times 0
    force = ma 60 kg times -9.8 m/s^2
  17. Apr 6, 2009 #16
    oops 9.8 m/s^2 times 1 kg = 9.8 newtons = Fg
  18. Apr 6, 2009 #17
    so would it be (588 newtons -588 newtons) = 0 net force because mg - ma =
    weight - force im still confused the stupid ma and mg
  19. Apr 6, 2009 #18
    Fnet = ma
    Wa + (-Wa) = ma = 0 (apparent weight) for at rest
    and, Wa = W = 588 newtons
    velocity = 0
    accel. = 0
  20. Apr 6, 2009 #19
    Keep in mind that g is just an acceleration. Nothing special about it.

    Anyway, it looks like you've got the basic idea down for the first problem. You'll probably need to write it out stating specifically what each force is, but it looks like you've got it.

    Keep in mind there exist more forces than just gravity, tension, and normal force. There's friction, you can push or pull something... It helps to draw a picture with all the force vectors acting on the object. This is called a "Free Body Diagram." http://en.wikipedia.org/wiki/Free_body_diagram
    Last edited: Apr 6, 2009
  21. Apr 6, 2009 #20
    i think i understand how to find the forces now. the Fnet = Fg + Fa and Fg(force of gravtity)) = mg = 60kg times 9.8 m/s^2 = 58.8 kg m/s^2
    F = ma = 60kg times 0 m/s^2 = 0 N (force of elevator)
    so, Fnet = 58.8 kg m/s^2 + 0 = 58.8 kg m/s^2 Net force

    for a constant velocity, it is the same.
    for an acceleration down of 4 m/s^2, F =ma= 60 kg times 4 m/s^2 = 240 N
    Fnet = mg - ma = 58.8 kg m/s^2 - 240N = 34.8 kg m/s^2

    for an acceleration up of 4m/s^2, F = ma = 60 kg times 4 m/s^2
    and Fnet = mg + ma = 58.8 kg m/s^2 + 240 N = 82.8 kg m/s^2
    i hope i did that stuff right
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