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Using newtons 2nd law in an elevator

  1. May 24, 2005 #1
    if a guy is in an elevator (55 kg for the person), and then he jumps straight uo- the scale will read 622 N. what is the acceleration of the guy at this instant?

    how do you deal with this b/c i don't know how to do it unless he is still standing on the scale. i know there should be an acceleration. could someone help me start this question?

    thanks, Lisa
  2. jcsd
  3. May 24, 2005 #2


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    I don't see any connection to the elevator, Maybe this is part of a bigger problem. The scale reads the force it is applying to the guy. Gravity is also acting on him. His acceleration is the result of the net force, which is force from scale less his weight.
  4. May 24, 2005 #3
    You perhaps want to know how the laws can be applied to noninertial frames. Well, I recommend (especially if you are new) that you do this problem using the conventional method. What are the forces on a body in an elevator? And what is the acceleration of the body as seen by an observer in an inertial frame?
  5. May 25, 2005 #4
    Referring to your Question,Liza.
    By Newon's 2nd Law's of Motion, using the formula F= ma, you can find the acceleration.
  6. May 25, 2005 #5
    This is an interesting question because if he is in the air when the scale reads 622 N, then the man's mass has nothing to do with that force. Are you sure that's the exact question? You might need to know other things.
  7. May 26, 2005 #6
    thats what the question asked. but i'm thinking maybe it really means the instant right before he jumps b/c you really need to use the weight in the -x direction or else the question is riddiculous. Make sense??
  8. May 26, 2005 #7
    pardon me: weight in the -y direction
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