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Finding Work Done by Friction

  1. Oct 11, 2007 #1
    The last part of this problem has been bothering me. I don't understand how the work done by friction can be found if you are not given the coefficient of friction. Can anyone help me with this? There is a picture attached.

    1. The problem statement, all variables and given/known data
    Sally applies a horizontal force of 567N with a rope to drag a wooden crate across a floor with a constant speed. The rope tied to the crate is pulled at an angle of 48 degrees relative to the floor. The acceleration of Gravity is 9.8 m/s^2. How much force is exerted by the rope on the crate in Newtons? What work is done by Sally if the crate is moved 34.1m? Answer in units of Joules. What work is done by the floor through force of friction between the floor and the crate? Answer in units of Joules.


    2. Relevant equations
    567cos(theta) = F
    W = F*d


    3. The attempt at a solution
    So far I've found the answer to the first and second question.
    Force of rope = 567/cos(48) = 847.3682038 N
    Work done by Sally = 567*34.1 = 19334.7 J

    I don't think that this last part can be done without the coefficient of friction. Am I wrong?
     

    Attached Files:

  2. jcsd
  3. Oct 11, 2007 #2

    Doc Al

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    Staff: Mentor

    You don't need the coefficient of friction. Hint: The crate moves at constant speed.
     
  4. Oct 11, 2007 #3
    That's something that I thought of too when I was working on the problem, but that makes me think that the answer is 0. But, I know that's not right. The only equation I know that gives the work done by friction is Wfk = -fk*d.
     
  5. Oct 11, 2007 #4

    Doc Al

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    Staff: Mentor

    Nothing wrong with that equation. What's fk?
     
  6. Oct 11, 2007 #5
    The Force of kinetic friction. But, I can't get that value without the coefficient.
     
  7. Oct 11, 2007 #6

    Doc Al

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    Staff: Mentor

    Again, you don't need the coefficient since you're not going to calculate fk using (mu)*N. Instead you're going to figure out what fk must equal by analyzing the horizontal forces acting on the crate and taking full advantage of the fact that the crate is not accelerating.
     
  8. Oct 11, 2007 #7
    The horizontal force on the crate is 567 N. The crate is not accelerating. The acceleration equals 0. Every sample problem in my book gives me the coefficient of friction. I really have no idea what to do. Maybe I do 567N/(9.8m/s^2), but that doesn't give me Joules.
     
  9. Oct 11, 2007 #8
    In relation to the horizontal force being applied to the crate, what direction is the Force of kinetic Friction?
     
  10. Oct 11, 2007 #9
    The opposite direction. That leads me to believe that the force of friction is -567 N, but that doesn't make sense to me because if the forces are opposite and equal, then the crate shouldn't be moving.
     
  11. Oct 11, 2007 #10
    Not moving or not accelerating? If acceleration=o, then there are two possible cases: There are NO forces acting on the object or is no NET force on the object.


    Casey
     
  12. Oct 11, 2007 #11
    So, the crate can be moving, but have 0 force acting on it? And if this is true, then is -567N correct?
     
  13. Oct 11, 2007 #12
    Is 567 the horizontal component of the Force exerted by Sally? Hint: No.
    And [tex]F_k[/tex] is not WORK it is FORCE. If you have a force, how do you find the work done by it?

    Casey
     
  14. Oct 11, 2007 #13
    But that's what it says in the problem. "Sally applies a horizontal force of 567 N..."
    Ah, so I would do -567N*34.1m? The work done by friction would be -19334.7J then?
     
  15. Oct 11, 2007 #14
    Is that what it says in the problem?! It looks to me like it says there is a 48 degree angle relative to the floor...
    She is dragging the box horizontally; however, not ALL of that Force is directed horizontally...you need to decompose the vector.

    Casey
     
  16. Oct 11, 2007 #15
    Yes, it does say that she pulls a rope 48 degrees above the floor, but I'm not given that force, I'm given the horizontal component, which shows 567N. I attached a picture, but maybe it isn't working. I found the force along the rope to be about 847N, but that is not horizontal.
     
  17. Oct 11, 2007 #16

    I am very sorry. I just went downstairs and looked at my PC instead of my Mac and I can see the image. -567N*displacement is the correct answer.

    Apologies,
    Casey
     
  18. Oct 11, 2007 #17
    Alright. Thanks a lot for the help!
     
  19. Oct 25, 2007 #18
    I'm New

    Hi I'm new to physics forums and I have the general idea of how it works. But if I have a homework problem do I just post a new thread?
     
  20. Oct 26, 2007 #19

    dynamicsolo

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    Homework Helper

    Yes. It is also generally preferred that if you have a second problem, you should start another thread, rather than continue on an existing one. The idea is to (try to) end up with one solved problem in each thread...
     
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