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Elevator 2 part Question - Box Dragging

  1. Oct 24, 2007 #1
    If an elevator accelerates upwards at say about 1.96m/s/s, and say I need to find what the minimum force is required to move the box while the elevator is in motion. Would I find the force of the elevator (using F=ma) moving upwards and add it to the boxes weight and then the total weight of the force of elevator and weight of box would be my Fn. If everything is correct so far... How do I figure out how much force is needed to pull the box?
     
  2. jcsd
  3. Oct 24, 2007 #2

    nrqed

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    There are ways to get the right answer following the lines you describe but the real way to do it is to start from a Free body diagram.
    The force required to get it to move (assuming the force is horizontal ) is [itex] \mu_s N [/itex] where N is the normal force. To find the normal force on the box, use [tex] \sum F_y = m a_y [/tex] This is where the acceleration of the elevator will come into play. the box obviously has the same vertical acceleration as the elevator. from this, find the normal force on the box.
     
  4. Oct 24, 2007 #3
    I was wondering would the acceleration of the elevator affect the weight of the box?, and isnt Fn=W because of newtons law?
     
  5. Oct 24, 2007 #4

    nrqed

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    The weight of the box is simply mg and it does nto depend on anything (as long as you don't go in outer space!). what is affected by the acceleration of the elevator is the normal force on the box. *Sometimes* people call the normal force the "apparent weight" of an object which is a source of confusion.
     
  6. Oct 24, 2007 #5
    thnx alot for the help.
    I have one more short question.
    Would the weight of the elevator come to be use in this question or should i just leave the weight of elevator alone and just use its mass?
     
  7. Oct 24, 2007 #6

    nrqed

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    You are welcome.

    If you just want the force required to move the box you don't need at all the mass or weight of the elevator itself. It plays no role. You just need the vertical acceleration in order to find the normal force on the box. What ou do need is the mass of the *box*.
     
  8. Oct 25, 2007 #7
    umm yea i just had my test and i asked my teacher how to find Fn of the box. I told him i used Fn=ma he said it was wrong. can u tell me what formulas used to find fn
     
  9. Oct 25, 2007 #8

    nrqed

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    By fn you mean the normal force?

    I said in my post number 2 to draw a free body diagram and to apply [tex] \sum F_y = m a_y [/tex]. You never commented on this so I assumed you knew how to do this. What are the forces along y acting on the box?
     
  10. Oct 26, 2007 #9
    O LOL my teacher never taught me what y was... but i did draw the free body diagram and by Fn i do mean normal Force, but i didnt use the sigma the sum of forces
     
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