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Statics: two rods connected by frictionless collar

  1. Jun 6, 2016 #1
    1. The problem statement, all variables and given/known data
    For problems 102 and 103, find the value of Mc required to maintain equilibrium, and find the forces at C. Ma is given as 20 Nm for both problems.

    2. Relevant equations
    Start with AB. The geometry is a 3/4/5 triangle, so AB is 100mm or 0.1 M.
    So 20NM=0.1B
    B=20/0.1 or 200N
    So the force at B has to be 200N at the 53.1-degree angle dictated by the geometry.

    3. The attempt at a solution

    I found, online, a solution to problem 103; the parameters are different but I understand the logic given for problem 103.

    I see the subtle difference between 102 and 103: in 103, the collar can only move along AB, and in 102, the collar can only move along CB.
    For problem 102, the book's answer is Cx = 217N.
    Intuitively, this makes no sense.
    How can Cx be more than the 200N force at B? The force at B, which is 200N, would be the hypotenuse of a triangle that would intersect Cx and Cy. So Cx would have to be smaller than 200N.

    Is the 200N force at B not correct for problem 102?

    Thank you in advance for any insight! :^)


    Attached Files:

  2. jcsd
  3. Jun 6, 2016 #2


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    No, it is not 200 N.
    When you draw the FBD of AB, the resultant force at B is not perpendicular to AB. It is perpendicular to what?
  4. Jun 9, 2016 #3
    Oh, I see.
    In problem 103, we took B (before it was broken down into Bx and By) perpendicular to AB because that's the only way the collar can move.
    So in problem 102, we take B perpendicular to BC, because again the collar can only slide up and down along that axis.

    OK, so we determine from the geometry that angle ACB is 21.8 degrees.
    The lever arm AB is 100mm or 0.1 meter, and it lives at a 36.9-degree angle below horizontal; it forms a 53.1-degree angle above the horizontal.
    Our interest is in the force which is 21.8 degrees above the horizontal.
    The difference in these 2 angles is 31.3 degrees.
    So we now have a triangle with a side of 200N, and we want to find the corresponding hypotenuse at 31.3 degrees. The hypotenuse turns out to be 233, which exactly matches the book's answer.

    Now that I think about it, this makes perfect sense. From CB's point of view, you have a 200N force at 31.3 degrees off perpendicular; what corresponding force would you need to render the same impact at a perpendicular? It would have to be a greater force, and it would have to be a function of that 31.3-degree angle.

    Thank you for giving me that hint.
    It was the perfect nudge I needed to get pointed in the right direction (and that direction was at 21.8 degrees!).

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