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. Suppose you are pushing a 60 N uniform crate at rest with a horizont

  1. Apr 16, 2013 #1
    1. The problem statement, all variables and given/known data

    . Suppose you are pushing a 60 N uniform crate at rest with a horizontal force of 40 N
    exerted at the top as shown. The crate is a cube with length of L. If the coefficient of
    static friction between the crate and floor is 0.75, which statement below is correct?

    (Hint: If the crate is about to pivot at point O, the
    normal force will act at point O, and not through the
    center of mass.)

    a. The crate will slide along the surface with constant speed.
    b. The crate will slide along the surface speeding with constant acceleration.
    c. The crane will slide along the surface speeding to a maximum speed and
    then continue to with that speed.
    d. The crate does not budge to your force, it neither slides nor pivots.
    e. The crate will pivot at point O.

    pic : http://i49.tinypic.com/16p2x1.png


    2. Relevant equations



    3. The attempt at a solution

    I did Fs = us * Fn = 0.6 x 60 = 45
    So since applied force is greater than frictional force, the crate should pivot?? then how is this using torque to solve the quesiton?
     
    Last edited: Apr 16, 2013
  2. jcsd
  3. Apr 16, 2013 #2

    CWatters

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    The friction means it won't slide (so that rules out a to c) but you also need to check the torque about point O to work out if it's d or e.

    EDIT:

    Where did you get the 0.6 from in ..

     
  4. Apr 16, 2013 #3

    sorry its a typo.
    0.75 x 40 = 30N

    but I dont understand how is it torque..
    cuz in the textbook, it says something about cancel out the torque produced by friction because it acts on the pivot or something..?


    or is this question just simply because the applied force is greater than the friction so the crate will pivot?
     
  5. Apr 16, 2013 #4
    Fapp * L and mg L/2
    but they are both tilting towards clockwise position..

    imagining a force push the crate, then the crate will be moved the same direction as the direction of the force...
     
  6. Apr 16, 2013 #5

    CWatters

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    Lets look at the Torque about point O and define clockwise as +ve.

    Torque = force * distance

    You have the force pushing on the crate at the top which produces a torque = 40 * L

    You have the weight acting through the center of gravity producing an anticlockwise torque = - 60 * L/2

    You have friction acting through point O so that produces a torque = 30 * 0 = 0. It's zero because the displacement from point O is zero.

    Total Torque is

    40L + (-30L) + 0 = 10L

    That's a positive torque so the crate rotates clockwise.
     
  7. Apr 16, 2013 #6
    why is the weight anticlockwise? I just couldn't picture these rotational torque... How do you picture it?
     
  8. Apr 16, 2013 #7

    CWatters

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    Will post picture in a moment.
     
  9. Apr 16, 2013 #8

    CWatters

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    Like this..
     

    Attached Files:

  10. Apr 16, 2013 #9
    oh right, i get why is it CW and CCW now thks
     
  11. Apr 16, 2013 #10

    CWatters

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    Have drawn it slightly better. This shows the crate at the point of tipping. The normal force has moved to point O.
     

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