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Homework Help: Basic Mechanics

  1. Apr 22, 2005 #1
    Hi everyone, I am reviewing physics...and here comes a question that has confused me for a long time.

    The question is about the elevator (lift) and a weighting scale...what is the force diagram on a person standing in the elevator on a weighting scale...1. when the elevator is going upwards? and 2. when the elevator is going downwards? How should i calculate the acceleration and the weight appear on the weighting scale if relative values are given?

    Can anyone show me the force diagram and tell me how to do the calculatoins...thanks
  2. jcsd
  3. Apr 22, 2005 #2


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    The diagrams are the same as long as the elevator is going at constant speed, and they are the same as for a person in the elevator at rest. Did you mean to ask about upward acceleration and downward acceleration? You can have either one when the elevator is going up and when it is going down.
  4. Apr 22, 2005 #3
    I know that the force diagram is the same when the elevator is at rest and when its at constant velocity. What i dont understand is that when the elevator is accelerating...when it speeds up AND when it slows down. What is the change in the normal force? and how will the weighting scale change?
  5. Apr 22, 2005 #4


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    Well, the elevator has to provide its own acceleration to the person inside it, right?
  6. Apr 22, 2005 #5
    So...the accelereation from the earth is 10N/kg, when the elevator provides its own acceleration...can i add the two accelerations together?
    What will be the force diagram?
    Still don't get it!
  7. Apr 22, 2005 #6
    If the elevator is accelerating upwards, then the person will experience a normal force greater than what he would normally. Therefore the scale will read a higher reading.

    If the elevator is descending, the person wil lexperience a normal force less than he would if he were standing on a non-accelerating platform. The scale will measure him as lighter.
  8. Apr 22, 2005 #7


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    g' = g + a

    Where g' is the observed acceleration of gravity, g is 9.8 m/s^2 and a is the acceleration of the elevator (if we define up to be the positive direction).

    So if you are going down in an elevator accelerating at 2.0 m/s^2, g' is 9.8 + (-2.0) = 7.8 m/s^2. And if you are going up at the same a, g' is 9.8 + 2 = 11.8 m/s^2.

    And then when you step on a scale and measure your weight F = mg, it becomes F = mg'
  9. Apr 22, 2005 #8
    I see. Thanks
  10. Apr 23, 2005 #9
    Here is more on basic Newtonian Mechanics...with lots of exercises


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