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3d Objects in Equilibrium

  1. Oct 20, 2015 #1
    1. The problem statement, all variables and given/known data
    The 1.1m bar is supported by a ball and socket support at A and 2 smooth walls. The tension in the vertical cable CD is 1kN.
    IMG_1335.JPG
    Draw the Free Body Diagram of the Bar.
    Determine the Reactions at A and B.


    2. Relevant equations
    Equillibrium equations.
    Perhaps the vector projections...??


    3. The attempt at a solution
    I drew the FBD:
    IMG_1336.JPG
    and you can see the beginnings of my equillibrium equations. I am fairly sure these are all correct except for the fact By should not be there.
    The tension CD can only pull normal to the location its given - ie, along the y axis if its directly under the object. The smooth surface gives a normal force to the object, I experimented by resting pencils together that this would exert an x, z force so i broke them into components. The ball and socket is basically a hinge, so it exerts axial forces in the x, y, z.

    My confusion is this:
    The tension will obviously produce a moment about A pulling the rod downward along the y axis.
    The only thing preventing this descent is the combined reactions of the wall's x and y forces.
    Together they must produce a positive force along the y direction.
    This force must be equal to the relationship:
    0.4mTCD = 1.1mBy
    0.4kN-m= 1.1mBy
    By = 0.4/1.1 kN-m

    Now, By does not exist because the smooth surfaces can only exert forces normal to their plane.
    However, could it be the cross product of the two forces...? If so, then how do we find it? By projecting By along the two surfaces?
     
  2. jcsd
  3. Oct 20, 2015 #2

    SammyS

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    The wall being "smooth" exerts zero force in the y-direction.
     
  4. Oct 20, 2015 #3
    I know, but moments are the result of cross products... If there's no relationship between the horizontal and vertical forces then there's no reason for then walls to even exist. The ball and socket can't exert a moment. Something has to. This is statics lol.
     
  5. Oct 20, 2015 #4

    SammyS

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    The walls do exert force, but the y-component of the force is zero.
     
  6. Oct 20, 2015 #5
    I know that, my intention is to find these two forces and I am looking for things I can find to get me there. Currently, the socket resists all of the y-direction pull of the cable. However, the cable produces a moment and That must be resisted by the walls - it's the only thing that makes sense.
     
  7. Oct 20, 2015 #6

    SammyS

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    What is the moment produced by the cable? Have you calculated that?
     
  8. Oct 20, 2015 #7
    Is it not just a vertical line of action against the rod? I said it was, which would just be 0.6*1kN
     
  9. Oct 20, 2015 #8

    SammyS

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    It's a vector. What are its components or what is its direction?
     
  10. Oct 21, 2015 #9
    Okay, I think I see where this is going...
    It is a vector <0, 0.27, 0> by my calculations. The rod and the cable create two similar right triangles. Pythagoreans theorem gives us a height of 0.5 for the larger one, and the ratio between the hypotenuse is 0.64 which gives us a height of 0.32 for the smaller one. It is vertical, so it has no slant.

    Note, the rod has a vector <0.7, -0.5, 0.6>

    This allows us to make the tension a normal vector: <0.1, -o.07, 0.08>

    I'm going to stop there and make sure I did it right because I don't think that that's the answer I should be getting...
     
  11. Oct 21, 2015 #10

    SammyS

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    The rod, as a vector from B to A, is <0.7, -0.6, 0.6> .

    The tension is purely in the negative y direction.
     
  12. Oct 21, 2015 #11
    Can you show me how you got the y component of the rod? I found -0.5..
     
  13. Oct 21, 2015 #12
    I noted that the tension (cable) is purely in the -y direction, I found the normal vector because that would give the components of the force on the rod. The rod is angled, and the tension is normal to the plane, not the rod which is what we must account for.
     
  14. Oct 21, 2015 #13

    SammyS

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    A vector normal to the plane determined by the rod and the cable is perpendicular to each of these. Since the cable is purely along the y direction, any vector perpendicular to the cable has a y-component of zero.

    Therefore, your normal vector is incorrect.
     
  15. Oct 21, 2015 #14

    SammyS

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    (.7)2 + (y)2 + (.6)2 = (1.1)2
     
  16. Oct 21, 2015 #15
    I'm sorry, maybe I should show you a diagram of what I interpreted from that. Are you saying the vector I am looking for is normal to the plane while also being perpendicular to the cable and the rod? This does not make sense to me.

    I understand how a normal vector from the cable wouldn't have a y-component, but I originally intended to find one perpendicular to the rod to represent the resultant force for the moment equation
    M=Fd, but I believe you are saying to find one perpendicular to the cable in order to see the X,Z components instead?
     
  17. Oct 21, 2015 #16

    SammyS

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    The moment produced about A by the cable tension is the cross product of the vector from A to D (you can instead use A to C) and the tension vector. That cross product is a vector perpendicular to each of those vectors, thus normal to the plane determined by the rod and the tension vector (or any vertical line intersecting the rod).

    In order for the reaction at B to produce a moment opposite to this, the reaction force must be in the same plane as the rod and the tension force.
     
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