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Constrained Plane of Motion Problem

  1. Mar 26, 2017 #1
    1. The problem statement, all variables and given/known data
    A uniform slender rod of length L = 1180 mm and mass m = 5.4 kg is suspended from a hinge at C. A horizontal force P of magnitude 75 N is applied at end B.

    Determine the distance 13252705286110469&Q_13252705286110469_rnd=1%3D7%3B1%284%29%3D71%3B1%281%29%3D1180%3B1%282%29%3D5.gif for which the horizontal component of the reaction at C is zero.

    I have attached the image below.

    2. Relevant equations
    M = Iα

    3. The attempt at a solution
    FBD:
    W = mg = (5.4)(9.81) = 52.974 downwards, at G
    Cy at C
    75 N left, at B (already on the diagram)
    Angular Acceleration at B?
    Angular Velocity at B?

    I'm kind of stuck at the setup and don't really know how to proceed beyond this.
     

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  3. Mar 26, 2017 #2

    haruspex

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    The question is only concerned with the instantaneous circumstance, so there is as yet no velocity at B.
    The will be accelerations. If the angular acceleration is α, what will be the linear acceleration of the mass centre?
    What equations can you write relating forces to accelerations and torques to angular accelerations?
     
  4. Mar 26, 2017 #3
  5. Mar 26, 2017 #4
    -Since there is no angular velocity the normal acceleration will be equal to 0.
    -Likewise, the tangential acceleration will be equal to the product of r and α. However, I am unsure of the direction it will move in.
    -Also since that is the total net force, ma = mrα = summation of all forces.
    -Also, the torque will equal Iα + mad (the perpendicular d).

    So I'm assuming I solve for I, which I think would be equal to (1/12)mL^2 = (1/12)(5.4 kg)(1.180 m)^2 = 0.531 kg*m^2. And that would act at the bar's center.
    For the net force, I'm assuming I use this to solve for alpha (in terms of r). Also, I assume the tangental acceleration will be completely to the left, in which case:
    mrα = P --> α = P/m/r = (75)/(5.4)/r = 13.889/r.

    However, I'm not sure what else I can do from here. Was my assumption about the direction of the acceleration wrong? Was my inertia calculation wrong?
     
  6. Mar 26, 2017 #5

    haruspex

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    I guess you mean the radial acceleration.
    That will be the linear acceleration of the rod's mass centre, yes.
    Correct.

    Not sure what you meant by your discussion of torque. For angular accelerations and torques you need to pick a reference axis. It can be any stationary point or the mass centre. If we take the pivot as reference, what is the torque from P about that? What is the moment of inertia of the rod about that? What equation does that give you?
     
  7. Mar 26, 2017 #6
    Okay so if I understand correctly, we need to sum the torques at C, to solve for r. So now two formulas:
    mrα = P --> α = P/(mr)
    τ = Iα + mar = (IP)/(mr) + P = (r+L/2)P
    Is this setup correctly? Also, I'm unsure about how to calculate the moment of inertia. Is it simply (1/12)mr^2?
     
  8. Mar 26, 2017 #7

    haruspex

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    P=mrα, not mra.
    Yes.
     
  9. Mar 26, 2017 #8
    Okay so in that case:
    Iα + mar = (r+L/2)P
    (1/12)(mr^2)α + mar = rP + PL/2
    a = rα
    (1/12)(mr^2)α + mαr^2 = rP + PL/2
    α = P/(mr)
    (1/12)(mr^2)P/(mr) + mPr^2/(mr) = rP + PL/2
    (1/12)rP + rP = rP + PL/2
    (1/12)r = L/2
    r = 6L

    Obviously that's not possible, so what did I mess up?
     
  10. Mar 26, 2017 #9

    haruspex

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    Sorry, I did not check that properly. You do not mean r there. What should it be?
     
  11. Mar 26, 2017 #10
    Got it, thanks!
     
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