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Homework Help: Elevator Weight

  1. Jul 26, 2007 #1
    1. The problem statement, all variables and given/known data

    A scale is placed in a normal elevator on earth. A person with a mass of 60kg steps onto the scale. What force would be exerted on the person by the scale during the following occurances?
    -Moves up at constant speed
    -Slows at 2.0m/s/s while moving upwards
    -Speeds up at 2.0m/s/s while moving downward
    -Moves downward at a constant speed
    -Slows to a stop while moving downward with constant acceleration

    2. Relevant equations

    3. The attempt at a solution
    I'm somewhat stuck. I would like to answer the problem myself, but guidance towards the correct methods would be greatly appreciated. I am confused on how to decide which forces and vectors should be used. I also am confused on how to calculate the normal forces. Any help is appreciated.
  2. jcsd
  3. Jul 26, 2007 #2
    First, you should find the weight of the person. Then simply use the formula you posted to go through each scenario. In terms of forces, you only have the person and the scale, as far as I can tell you just have the weight and normal force, where Normal force = - Weight

    Here's a few hints:

    Remember that constant speed means [tex]a = 0[/tex].
    When you're in an elevator and going UP, do you feel heavier/lighter?
    how about when you're in an elevator and going DOWN?

    Hope this helped.
  4. Jul 26, 2007 #3
    Thanks for the Normal weight formula, can't believe I overlooked that. Up in an elevator means that you feel heavier and down means you feel lighter. So would one go about calculating the actual force that the scale reads?
  5. Jul 26, 2007 #4
    Yes, you're right. So now look at your scenarios. In each scenario, something changes with your acceleration, nothing else. Your mass is constant, so the only variable here is the acceleration. In each case, just think of the direction of the acceleration and which way you're moving.

    Two important things:

    First, the equation is [tex]F = m a_t[/tex], where [tex]a_t[/tex] is the total acceleration.

    Second, you have to realize what the signs and directions mean. What I mean is, gravity is [tex]9.8 m/s^2[/tex] DOWN, in your second case, your slowing down so you have [tex]- 2.0 m/s^2[/tex] UP. Notice how the sign changes. The sign means ACCELERATING or DECELERATING.

    So you would have:

    [tex]F = ma[/tex]

    [tex]F = (60)[9.8 - (- 2.0 ) ][/tex]

    [tex]F = 708 N[/tex]

    Just to check if this makes sense, look at your weight while the elevator isn't moving. Like you said before, when you move up, you should be HEAVIER. So you want the new weight to be greater than the original weight, which it is. So you can check yourself that way.
  6. Jul 26, 2007 #5
    Wow, thanks. That makes MUCH more sense. When the elevator is moving at a constant speed, is the Force the same as if they were not moving at all since the acceleration is zero?

  7. Jul 26, 2007 #6
    Yup, exactly. You got it now.
  8. Jul 26, 2007 #7
    Ok. While I do understand the two equations that I/you have done above, I am still somewhat confused on the others. Would the "Moves up at constant speed" and "Moves downward at a constant speed" have the same values? That's the way that it would seem if acceleration is the only variable. For the "Speeds up at 2.0m/s/s while moving downward", what would the sign be? I don't understand how you determind the sign of the change in acceleration.
  9. Jul 26, 2007 #8
    You're absolutely right.

    Here's the thing. Acceleration is how fast the speed changes. Speeding up is like accelerating positively at [tex]2.0 m/s^2[/tex], while slowing down is like accelerating negatively [tex]2.0 m/s^2[/tex]. In other words, speeding up means + sign, slowing down means - sign.

    If the problem was that it speeds up at [tex]2.0 m/s^2[/tex] while moving up, the equation would be:

    [tex]F = (60)[9.8 - ( 2.0 ) ][/tex]

    [tex]F = 468N [/tex]

    which makes sense.

    I think you might be confused about the equation. The equation is really:

    [tex]F = m(9.8 - a)[/tex]

    where a is your given acceleration. Since that's the case, you just need to figure out if it's accelerating or decelerating.
  10. Jul 26, 2007 #9
    Thanks, that cleared everything up as well. For the "It slows to a stop while moving downward with a constant accleration, do you think it's implying that you are reading your weight while it is stopped? If not I don't see how it can be answered since an acceleration value isn't given.
  11. Jul 26, 2007 #10
    I'm a bit confused about that one. If the acceleration is constant, it means there IS a force, simply because there's acceleration. So if there's a force that you need to calculate, then you must be given an acceleration or some other information. Sorry but it seems to me that either your not given enough information and then this question has no answer, or I'm missing something.
  12. Jul 26, 2007 #11
    Thanks. Yeah that question is worded strangely and may be a mistype in the text. Thanks for your help!
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