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Statics - Forces on posts

  1. Dec 13, 2006 #1
    1. The problem statement, all variables and given/known data
    One end of a post weighing 400N and with height h rests on a rough horizontal surface with (mu) = .30. The upper end ish help by a rope fastened to a surface and making an angle of 36.9° with the post. A horizontal force F is exerted on the post. (a) If the force F is applied at the midpoint of the post, what is the largest value it can have without causing the post to slip. (b) How large can the force be without causing the post to slip if its point of application is 6/10 of the way from the ground to the top of the post? (c) Show that if the point of application fo the force is too high, the post cannot be made to slip, no matter how great the force. Find the critical high for the point of application

    2. Relevant equations

    F = ma

    F_Friction = (mu)N

    3. The attempt at a solution

    (a) Net force = F cosθ -(mu) F

    Net force = F (cos(36.9°) - .30)

    Net force = F (~.50)

    Where to go from here?
     
  2. jcsd
  3. Dec 13, 2006 #2
    Add on:

    Moment of inertia is: 1/12 (ML²)

    for (b) Moment of inertia is 1/12 (ML²) + M (L/10)² = (1/12+1/100)(ML²)

    And torque = I(alpha)
     
  4. Dec 13, 2006 #3

    OlderDan

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    What is the orientation of the post? It could be anything based on the problem description you have given.
     
  5. Dec 13, 2006 #4
  6. Dec 13, 2006 #5
    Oh, and after sleeping on it, I got this far:

    Normal force = mg + T cos(θ)

    Force = Friction + T sin(θ)

    And I know I need a torque equation, but how does that work out?
     
    Last edited: Dec 13, 2006
  7. Dec 13, 2006 #6

    OlderDan

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    As long as nothing is moving, you can calculate torque about any point and it must be zero. The point of connection of the rope looks good to me because that eliminates T from the torque equation.
     
  8. Dec 13, 2006 #7
    But at the top, would the torque just be:

    F(height) = 0 ??

    Or am I missing another force?
     
  9. Dec 13, 2006 #8

    OlderDan

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    Every force acting is a potential torque producer, but torque is the force times the perpendicular distance between the line of the force and the point about which torque is being calculated. For any point along the center of the vertical pole, the weight and the normal force produce no torque because the distance is zero. At the top, the tension produces no torque because the distance is zero, but F and friction both produce torques. What are they?
     
  10. Dec 13, 2006 #9
    Would it just be:

    FR - Friction(R) = 0

    Thus, the force = friction

    If this is true, does it hold true for every spot on the stick, or is my initial thought incorrect?
     
  11. Dec 13, 2006 #10

    OlderDan

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    What is R? The distances from the top of the post to where F and friction are acting are not the same.
     
  12. Dec 13, 2006 #11
    F(H/2) - Friction(H) = 0

    F = 2(Friction)

    Would this be correct then?
     
  13. Dec 13, 2006 #12

    OlderDan

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    That should do it. Combine this with your other equations and you should be on your way.
     
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