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Simple Free Body Force Diagram + Friction Question

  1. Feb 20, 2010 #1
    I just want to check if I am doing the right thing with this force diagram.

    Forgetting the force of friction for the moment, if I know that the body is accelerating at 6.34ms^-2 with a mass of 6000kg; and the opposing force of mgsin3 is 3077.4N, then would the I have to add that to the acceleration force? Therefore Force acceleration would be 38040 + 3077.4 = 41117.4N...


    My actual question is about friction. Do I have sufficient values to calculate the Force of friction or the friction co-efficient?

    F of Normal = mu x N
     

    Attached Files:

  2. jcsd
  3. Feb 20, 2010 #2

    PhanthomJay

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    Your terminology for 'acceleration force is a bit unorthodox. It is most always best to identify all real forces acting on a body, determine the NET of those forces in the x and y direction, and apply newton's laws in that direction (F_net =ma, which, for the x direction, is F_x_net = 38040 ). So what are all the forces acting in the x direction (including friction)? Do you have enough info to solve for the friction force?
     
  4. Feb 20, 2010 #3

    I thought F_x_net would be 41117.4N.

    Um, I don't think so.
     
  5. Feb 20, 2010 #4

    PhanthomJay

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    No, and you are looking at 2 cases, with and without friction, so let's not confuse the two. When you look at the problem as if it were written without friction, the net force is 38040, which is made up of the applied pulling force acting up the incline, and the weight component acting down the incline. Thus
    F_net =ma = 38040
    F_applied - mgsin3 = 38040
    F_applied - 3077 = 38040
    F_applied = 41117 N
    which i think is what you were trying to say before; but you were missing the concept of an applied force (by a person, or machine, etc.) required up the plane inorder to overcome gravity and accelearte the block.

    Now for the real problem with friction, use the same approach and respond to your own question> do you have enough info to solve for the friction force?
     
  6. Feb 20, 2010 #5
    F_net = 38040 = ma
    F_applied - mgsin3 - F_friction = 38040
    F_applied - F_friction = 41117N
    F_applied - F_normal*mu = 41117N

    I have two unknown variables now.
     
  7. Feb 20, 2010 #6

    PhanthomJay

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    yes, good. So can you solve for the friction force or not??
     
  8. Feb 20, 2010 #7
    I would say no.
     
  9. Feb 20, 2010 #8

    PhanthomJay

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    And so would I:cry:
     
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