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Statics Forces/Moments Problem

  1. Nov 21, 2009 #1
    1. The problem statement, all variables and given/known data

    In the attachment

    2. Relevant equations

    Sum of the forces in the X, Y, and Z
    Sum of the Moments in the X, Y, and Z.

    3. The attempt at a solution

    I don't believe I need help solving the problem, I just need help setting it up.

    I can find the vector of CD, which came out to 400i-100j-500k (I know I need to get the unit vector), and the tension in W is -(250*9.8)k. But after that I get lost. I tried to make a free body diagram, but I am not sure because the problem wants the force in B, so I think I have to break it down in components and what not, but I think it's the rod that throws me off.

    How would I go about getting an equation for A and B, or would they just simply be Ax, Ay, Az, Bx, By, Bz and I simply take the sum of forces and moments? Like I said, I get lost after finding the vector of CD and tension in W.
     

    Attached Files:

  2. jcsd
  3. Nov 22, 2009 #2

    tiny-tim

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    Hi akhmed966! :smile:
    A free body diagram probably doesn't help when there are so many forces.

    Yes, break it into Ax, Ay, Az, Bx, By, Bz and take the sum of forces and moments …

    I don't think there's any short way of doing it (though you may be able to make it slightly shorter by eg taking moments about the most convenient point, and taking components perpendicular to one of the cables)
     
  4. Nov 22, 2009 #3
    Ok, thanks.

    I got three equations so far.

    Sum of X = Tcd*.617+Ax+Bx = 0
    Sum of Y = -Tcd*.154+Ay+By = 0
    Sum of Z = -Tcd*.77+Az+Bz = 2450

    7 unknowns and 3 equations. Not looking too good so far.

    So, I was thinking I should take the moment about point A since that would reduce 3 unknowns, but I don't exactly remember how to do 3D moments with bars, I'm pretty sure I have to do the cross product, I just don't remember how to get the sum. How would I go about taking the moment at point A?
     
  5. Nov 22, 2009 #4

    tiny-tim

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    Hi akhmed966! :smile:
    Good point, but you've forgotten the extra information "rests against a smooth vertical wall at end B" … so the reaction at B is … ? :wink:
    Yes, definitely take moments about somewhere, to get three extra equations (btw, taking moments about two points won't give you any extra information you can't get from the three linear equations) … personally, I'd go for D rather than A, since that eliminates two forces, and keeps things more symmetrical.

    And the easiest way to do cross product is probably to use i x j = k etc. :smile:

    btw, you might like to consider using components in the three directions along the bar, and the string CD, and perpendicular to both (instead of the usual x y and z) … I have a feeling that might make the equations simpler.
     
  6. Nov 22, 2009 #5
    B? That's what he expects the answer to be in, but I have never handled a problem in that manner so I don't know how to modify the equation. Unless that is stating that it only has a force normal (N) opposite to that of the wall, so all I would need would be Bx (N = Bx) and not By and Bz?

    If I take the moment about point D, wouldn't I not have anything usable data? If B is one variable, it would be like Az = 0, wouldn't it?

    Yeah, that's the only method I actually use in 3d problems (or remember how), but it's like

    |i j k|
    |1 2 3| distance
    |Tcd Tcd 3| tensions

    And it would give me the moments in i, j, and k, and then I just sum up all the i, j, k moments and set them equal to zero, right? I remember doing that 2 months ago, when we started these problems.

    I think I can solve this problem, but I just need to know if my assumption that Bx is all I need for point B, and if taking the moment about point D is valid or invalid.

    Thanks again.
     
  7. Nov 22, 2009 #6

    tiny-tim

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    That's right! There's no friction, so there can't be any force parallel to the wall, and so N = Bx. :wink:
    I don't understand. :confused: You'll have the reaction forces at both A and B.
    Yes, that's right (but I was actually not think ing about that matrix method, but the rather simpler method of writing r x A = (rxi + ryj + rzk) x (Axi + Ayj + Azk))
    Yes, just Bx. And you can always take moments about any point. :smile:
     
  8. Nov 22, 2009 #7
    Oh, thank you so much.

    I took the moment about D, and set the two remaining forces equal to zero. Sum of Moments about point D = r x Fb + r x Fa = 0, would that be right? Then I just gather the i, j, and k terms and set them each equal to 0, and have a total of 6 equation (3 from the sum of the x, y, and z).

    Then I simply solve, would that be correct?
     
  9. Nov 22, 2009 #8

    tiny-tim

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    Yes, except you can't use the same r for both forces, can you? :wink:
    Yes. :smile:
     
  10. Nov 22, 2009 #9
    No, thanks a ton.
     
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