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Forces and components

  1. Feb 18, 2008 #1
    If a block of mass m is pulled in the direction of (x-component) Tcos(theta) and (y-component) Tsin(theta) across a rough surface with a constant velocity. What is the magnitude of the frictional force?

    A) μkmg B) μkT cos θ C) μk(T – mg) D) μkT sin θ E) μk(mg – T sin θ)

    Since fk is in the opposite direction as the motion and since there is no acceleration F=0=Tcos(theta) - fk
    So
    fk= Tcos(theta)

    So is the answer B?
     
  2. jcsd
  3. Feb 18, 2008 #2

    Doc Al

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    Staff: Mentor

    clarification

    Is the x-y plane parallel to the surface? (I would assume yes.) Or is the y-direction vertical?
     
  4. Feb 18, 2008 #3
    Edit: (supposing x is horizontal+ to right and y is vertical+ to top)

    So, you know that friccional force is proporcional to Normal force exerced to the block;

    you know that if dv/dt=0 => Sum of all Forces=0, so,
    begin with making a diagram of the problem, putting every vector force on it with magnitudes that correspond to total F=0...
    than you need to understand why isn't B)
     
  5. Feb 18, 2008 #4
    y is vertical and x is horizontal
    If the fnet = 0 then
    0=Tcos(theta) - fk
    which makes fk= Tcos(theta)

    If it is not B then I do not understand how to solve this problem.
     
  6. Feb 18, 2008 #5

    Doc Al

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    This is true, but it's not one of the choices. But there is another correct choice.
     
  7. Feb 18, 2008 #6
    I am at a loss
    I know that fk usually equals ukmg on a horizontal force. Any help would be appreciated.
     
  8. Feb 18, 2008 #7

    Doc Al

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    fk = ukN. But does N always equal mg?
     
  9. Feb 18, 2008 #8
    If N= Tsin(theta)
    then Fk= uTsin(theta)

    Answer D
     
  10. Feb 18, 2008 #9

    Doc Al

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    Figure out the normal force by considering the vertical forces acting on the block.
     
  11. Feb 18, 2008 #10
    ok
    the normal force is the y-component and the force in the opposite direction is the force of gravity on the mass
    so 0=mg-Tsin(theta)
    so mg=Tsin(theta) so N=Tsin(theta)
    so fk = uk (Tsin(theta))

    what am I missing???
     
  12. Feb 18, 2008 #11

    Doc Al

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    There are three vertical forces acting on the block: The vertical component of T, the weight, and the normal force. They must add to zero. Solve for N.
     
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