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Torque and ladder problem

  1. Mar 27, 2015 #1
    1. The problem statement, all variables and given/known data
    wcEDaVN.png

    2. Relevant equations
    F = μ FN

    τ = r F

    3. The attempt at a solution
    how do I approach this problem? Mass isn't given. And If i set the top of the ladder to be the pivot point, the pivot point will move vertically downwards if the ladder slides - is that even allowed?
     
  2. jcsd
  3. Mar 27, 2015 #2

    ehild

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    If the ladder does not slip, the pivot point is also in rest. Collect all the forces and all the torques. What is the condition that the ladder does not start to move?
     
  4. Mar 27, 2015 #3
    dEtntjD.png

    does this look correct?
     
  5. Mar 27, 2015 #4

    haruspex

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    No.
    First, what is "Fgy"? The horizontal component of the gravitational force?!
    Secondly, what force stops the ladder from slipping (until it does)?
     
  6. Mar 27, 2015 #5
    Force of the wall should equal force of friction.
     
  7. Mar 27, 2015 #6

    ehild

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    Is the x axis horizontal, and the y axis vertical? I think, you mean Fg the force of gravity. It acts at the centre of mass of the ladder, not at the ends. What is its direction? What are the x and y components?

    The normal force at the wall is not the same as the normal force from the ground.

    And you completely ignored the force of friction at the ground.
     
  8. Mar 27, 2015 #7
    mg = FN at the bottom of the ladder.

    so Force of wall = force of friction since ladder doesn't move.

    Fwall = Ffriction = μ FN = μ mg

    torques : Torque of wall pushing ladder should equal torque of ladder at the center of mass.
    let's pick the bottom of the ladder as the pivot point since the ladder doesn't move.
    Fwall⋅ d ⋅sin 55.2 = mg ⋅ d/2 ⋅ cos 55.2

    plugging in μ mg for Fwall

    μ mg ⋅ d ⋅sin 55.2 = mg ⋅ d/2 ⋅ cos 55.2

    canceling out mg and d from both sides, μ should come out to be 0.347
     
  9. Mar 27, 2015 #8

    ehild

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    It looks correct.
     
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