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Bar of uniform mass hinged to the ceiling.

  1. Dec 28, 2012 #1
    Hello,
    1. The problem statement, all variables and given/known data
    A uniform bar of mass m is hinged to the ceiling and rests on a smooth floor at an angle α with respect to the vertical axis (please see attachment). A thin sheet of paper, with coefficient of friction μ, is inserted between the floor and the bar.
    What is the minimum force necessary to pull the paper to the right?
    What is the minimum force necessary to pull the paper to the left?

    2. Relevant equations



    3. The attempt at a solution
    Obviously I must figure out which forces are in action before analysing the torques.
    The forces acting on the bar are:
    N1, in the positive y direction (i.e. up), at the point where the bar is in contact with the floor; fs, acting on the bar at that contact point with the floor, in the negative x direction (i.e. to the left); W, from the center of mass of the rod, in the negative y direction. Is that all? Are there any forces acting on the bar at the point where it is hinged to the ceiling?
     

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  3. Dec 28, 2012 #2

    TSny

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    If you pull on the paper to the right, in which direction does the force of friction act on the bar?
    In general there will be both a vertical and horizontal component of force on the bar from the hinge.
     
  4. Dec 28, 2012 #3
    Is the question not tantamount to a bar hinged to the ceiling and lying at rest on a floor with COF mu?
    When referring to forces on the hinge in your answer, are these forces strictly a result of Newton's third law?
     
  5. Dec 28, 2012 #4

    TSny

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    In this problem, there's the added feature that the bar is resting on a sheet of paper that is being pulled to the right (or left) with just sufficient force to start the paper slipping.
    The forces, of course, obey the 3rd law. I guess it's semantics as to whether or not you would say that the forces are a result of the 3rd law. There must be a force holding up the left end of the rod. That force is the force that the hinge exerts on the left end of the rod. The rod exerts a 3rd law "reaction force" on the hinge.
     
  6. Dec 28, 2012 #5
    And what exactly is the difference between pulling the rod/bar to the right in the alternative set up I delineated above, and pulling the sheet of paper in this one, as presented in the question?
     
  7. Dec 28, 2012 #6

    TSny

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    I didn't know you were thinking of pulling the bar to the right in the alternative set up. But I still don't believe the two scenarios are equivalent. Suppose I apply a horizontal force to the lower right end of the rod. If the question is what force would I have to apply to get the right end of the rod to move it, then I don't think it's equivalent to the original question. The only way the right end of the rod can move if I pull to the right is for the end of the rod to lift off of the floor. That's different than what happens with the paper. If I push to the left on the right end of the rod, then I wouldn't be able to get the rod to move no matter how hard I push (unless the hinge breaks :smile:).
     
    Last edited: Dec 28, 2012
  8. Dec 28, 2012 #7
    Alright, so let's pull the paper to the right :-).
    First, are the "vertical and horizontal components" referred to those of a normal force exerted on the bar by the hinge? I am not sure I completely understand the origin of this force. Would you please care to explain?
     
  9. Dec 28, 2012 #8

    TSny

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    Yes, you can think of the force acting on the rod at the hinge as a normal force from the surface of the pin that goes through the rod. See attached figure. The rod rests on the surface of the blue pin. The pin supports the rod with a normal force ##\vec{R}## acting at point p where the rod makes contact with the pin. The 3rd law implies that the rod exerts an equal but opposite force on the blue pin.
     

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  10. Dec 28, 2012 #9
    Alright, that's clearer. Our focus is nevertheless restricted, vis-a-vis this particular question of course, to the forces exerted on the rod, is it not?
    Now that we're synchronised, and before I complete the force (and torque) equations, would you please explain why the static friction would not be to the left?
     
  11. Dec 28, 2012 #10

    TSny

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    If you stand on a rug on the floor and then I slowly pull the rug horizontally to the right over the floor, you will ride the rug to the right. What force caused you to move to the right?
     
  12. Dec 28, 2012 #11
    I see where you're going, but isn't the direction of the friction generally opposed to that of the movement?
     
  13. Dec 28, 2012 #12

    TSny

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    As far as the paper is concerned, the friction opposes the motion of the paper. As you pull the paper to the right, the friction force on the paper acts to the left. But friction on the rod acts to the right.

    Similarly, if you have a block sliding to the right along a rough floor, the friction force on the block will be to the left, but the friction force on the floor will be to the right (the same way the block is sliding.)
     
  14. Dec 28, 2012 #13
    Makes sense :-).
    All right then, moving on to the equations I believe we would have:

    N1, in the positive y direction (i.e. up), at the point where the bar is in contact with the paper; fs, acting on the bar at that contact point with the paper, in the positive x direction (i.e. to the right); W, from the center of mass of the rod, in the negative y direction; N2cos(alpha), in the negative y direction and N2sin(alpha) in the positive x direction, both designating the forces exerted by the hinge on the rod.

    Now for the torques (choosing the point where the bar is in contact with the paper as my reference):
    (L/2)*mg*sin(alpha) in the positive z direction

    Am I missing something?
     
  15. Dec 30, 2012 #14

    TSny

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    We don't know the angle that N2 makes to the vertical. It's probably not alpha. I would recommend replacing N2 by it's horizontal and vertical components: N2x and N2y
    You'll also need to include the torques due to N2x and N2y.

    [EDIT: You might find it easier to set up torques about the point where the rod is hinged to the ceiling. Then you won't need to worry about N2x and N2y.]
     
    Last edited: Dec 30, 2012
  16. Dec 31, 2012 #15
    Hi,
    Would you then review the following equation for the torques in the positive z direction?:
    -1/2*l*mg*sin(alpha) + l*f_s*cos(alpha) + l*N_1*cos(alpha)

    Once the paper begins moving to the right, f_s becomes mu*N_1, doesn't it?
     
  17. Dec 31, 2012 #16

    TSny

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    Check that you have the correct trig function for N1. Otherwise, looks good!
    Yes, just at the point where the paper is ready to slip, fs = μsN1. That will be the minimum force required to start the paper slipping.
     
  18. Dec 31, 2012 #17
    And to the left, may we simply replace f_s with -f_s in all the above equations?
     
  19. Dec 31, 2012 #18

    TSny

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    Yes.
     
  20. Dec 31, 2012 #19
    And do we equate force and torque equations to zero?
     
  21. Dec 31, 2012 #20

    TSny

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    Yes, we have equilibrium all the way until the paper slips.
     
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