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Crate sliding on a flatbed truck

  1. Sep 11, 2005 #1
    A crate of oranges weighing 180N rests on a flatbet truck 2.0m from the back of the truck. The coefficients of friction between the crate and the bed are us=0.30 and uk=0.20. The truck drives on a straight, level highway at a constant 8.0 m/s. (a) What is the force of friction acting on the crate?
    (b) If the truck speeds up with an acceleration of 1.0 m/s^2, what is the force of friction on the crate?
    (C) What is the maximum acceleration the truck can have without the crate starting to slide?

    Please help asap!!!
     
  2. jcsd
  3. Sep 11, 2005 #2
    F=MA

    [itex] \sum[/itex] F = [itex] \vec F_{fric} + \vec F_{car} [/tex]
     
  4. Sep 11, 2005 #3
    I still don't understand You can't use that equation for all three answers can you?
     
  5. Sep 11, 2005 #4

    Tom Mattson

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    Staff Emeritus
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    Actually, he shouldn't be telling you much of anything, since you haven't shown how you started the problem and where you got stuck. :tongue:

    Rather than tell you how to do it, I am going to ask you some questions to get you going in the right direction.

    Have you drawn a free body diagram of the crate?
    What is the crate's acceleration?
    What must the net force be equal to?

    Again: Have you drawn a free body diagram?
    Given the acceleration, what must the net force be equal to?
    What is the maximum static frictional force?
    Is that frictional force exceeded here?

    If you can get (b) then you should be able to get this.

    Give it a shot and let us know what you come up with.
     
  6. Sep 11, 2005 #5
    Alright I used your questions and I got answers for all the parts but now the question is did I use your suggestions correctly. So could you tell me if I did this stuff right. On (a) I got the acceleration to be zero therefore the net force is zero. (b) F static friction equals 54N. (c) I get 54N = 180/9.8 *a which equals 2.94 m/s^2 for the max acceleration Let me know how I did please!
     
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