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Magnitude of force to keep stick in equilibrium.

  1. Jun 13, 2013 #1
    1. The problem statement, all variables and given/known data[/b
    A straight and homogenous stick with mass m is pressed against a wall with the force F. The stick is horizontal perpendicular against the wall. Given that the friction between the wall and the stick is μ, determine the horizontal component of F in order for the stick to not fall down.

    http://i.imgur.com/QFJq7wT.png

    3. The attempt at a solution

    Forces involved:

    We have:
    The gravitational force mg in the negative y-direction.
    The normal force from the wall, N.
    The friction force in the positive y-direction which is f=μN
    The force F which acts in the positive x-direction
    ??From the figure, the force F should have a component in the y-direction but if I follow the description I don't necessarily think it has...

    Anyway;

    ƩF_x; F-N=0 <---> F=N
    ƩF_y; mg-μN=0 <---> N=F=mg/μ

    Hence F=mg/μ. Correct answer is however F=mg/(2μ)
     
  2. jcsd
  3. Jun 13, 2013 #2

    gneill

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    Staff: Mentor

    Assume that the force does have some vertical component. Can you tell from the setup what value that component would have? Hint: Through what point does the weight of the stick act? How much of that weight has to be supported at each end of the stick?
     
  4. Jun 15, 2013 #3
    That's precisely what I have done, leading to F_x: F-N=0 , F=N. Through what point does the weight of the stick act? Hmm, the midpoint I guess. Do you mean that the "torque-equation" should be used? Because I believe all "lines of action" for forces(F, N, f) go through this point.
     
  5. Jun 15, 2013 #4

    gneill

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    Staff: Mentor

    The stick is horizontal and of uniform composition and density so you should be able to draw an informed conclusion based upon symmetry (without even resorting to calculation!) about how much of the stick's weight needs to be supported at its ends... Its full weight is mg and acts vertically through the midpoint of the stick. How much of the full weight needs to be supported at either end? Your initial calculation seemed to imply that the stick is pressing down by its full weight mg at both ends.
     
  6. Jun 15, 2013 #5
    I am very sorry gneill about my errors. Of course you are right, the vertical component of force F holds up half the stick, so the friction force only needs to hold up the other half of the stick's weight, not the whole weight. Of course....again, my apology.

    My mistake was that I assumed the force to be horizontal with no vertical component. In that case, my calculations is correct I think.
     
  7. Jun 15, 2013 #6

    gneill

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    Staff: Mentor

    Hey, no need to apologize! The important thing is getting to the right answer and hopefully gaining experience along the way!
    If F had no vertical component, how could the stick remain horizontal?
     
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