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Torque problem -- friction force on nail between 2 boards

  1. Jul 8, 2014 #1
    Could some please help me with this problem?

    Boards X and Y are both massless and 4 m in length. A 4 N force is applied to board Y as shown. Board X is held stationary. The two boards are nailed together at 1 m from the left end of board Y. If the boards do not move, what is the static frictional force between the nail and board X?

    (A) 4 N
    (B) 8 N
    (C) 12 N
    (D) 16 N
     

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  2. jcsd
  3. Jul 8, 2014 #2

    CWatters

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    If this is homework it should be in the homework section and you should show your own attempt first.

    Hint. Levers.
     
  4. Jul 8, 2014 #3
    No, not homework. This is test prep/self-study.
     
  5. Jul 8, 2014 #4
    I haven't attempted any math for this problem yet. However, I do know that torque= F x d. My confusion is that the beams are massless. Thanks in advance
     
  6. Jul 8, 2014 #5

    A.T.

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    You don't have to consider the weight of the boards.
     
  7. Jul 8, 2014 #6

    jtbell

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    It doesn't matter. If you're trying to find the solution for a specific exercise like this, we consider it "homework-like" and it belongs in the homework forums.

    You've already received some help, and you've used the appropriate forum before, so I'll simply move this there.
     
  8. Jul 8, 2014 #7

    CWatters

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    So ignore the mass. What other forces or torques must exist in addition to the applied 4N ? Are the beams moving? Accelerating? What does that imply?
     
  9. Jul 8, 2014 #8
    fulcrum of rotation is at the end of board y
    4newtons * 4metres = 16N*m

    T=F*S
    where s = 1
    16=F*1
    F=16Newtons
     
  10. Jul 8, 2014 #9

    CWatters

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    Ry122. Please see...

    https://www.physicsforums.com/showthread.php?t=414380

     
  11. Jul 8, 2014 #10
    Okay. I am slowly understanding this problem but I still need some clarification...

    My understanding of torque as a static problem requires the sum of all torques must equal zero. So, initially, I set the problem up as:

    T (clockwise) = T (counterclockwise)

    My initial question was: why can't the nail be a point of rotation? Now I realize that the nail exerts a force, therefore, it cannot be a point of rotation. If the point is chosen to be at the end of y, then:

    T (clockwise) = T (counterclockwise)

    F (1 m) = (4 N) (4 m)
    F = 16 Nm

    Is this an accurate way to express the equation?
     
  12. Jul 8, 2014 #11

    CWatters

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    That all look fine to me.

    You could pick the nail as the point of rotation about which to sum the torques.. but you would then have to take into account other torques such as the force that board Y applies on board X at the left hand end of Y (and that's not known).

    In general.. if there is an unknown force in a problem try putting the point of rotation there so that it's taken out of the equation (eg T=F*d but d=0)
     
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