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Finding the reaction force - moments and equilibrium

  1. Aug 8, 2013 #1
    The problem statement

    A uniform rod of length 2m and mass 5kg is connected to a vertical wall by a smooth hinge at A and a wire CB as shown. If a 10 Kg mass is attached to D, find:

    a) the tension in the wire
    b) the magnitude of the reaction at the hinge A.

    attached is my copy of the diagram from the textbook, (with forces added by me) and a force diagram


    3. The attempt at a solution

    The tension in the wire will be the anticlockwise moment about A. This is balanced by the clockwise moment - which comes from the weight of the beam and the mass attached to the end.

    Clockwise moments [tex] 49 cos 26.77^\circ + 98 * 2 cos 26.77^\circ =218.74N[/tex]

    Anticlockwise moment: [tex] 0.5 T[/tex]

    [tex]0.5T = 218.74N[/tex]

    [tex] T = 437.5N[/tex]



    It is the next part where I am unsure. I need to get the magnitude of the reaction force at A. This is going to be a result of what I've called force D, which runs along AD.

    I have illustrated the forces diagram in the attachment and the thing that is confusing me is that the weight force is apparently the largest on in the triangle. But from the forces in the original diagram, tension is a larger force that the combined weights.

    As far as I can see there are no other forces acting on the system.

    Can anyone give me some pointers as to where my thinking on this is going wrong/what my approach should be for finding the force along AD, and thus the reaction at A?

    I would be very grateful for any information!
     

    Attached Files:

  2. jcsd
  3. Aug 8, 2013 #2
    Why do you think that the reaction force at the hinge will be parallel to the rod? Observe that the force of tension and the weights together tend to rotate the rod, and that can only be opposed if the force has a component perpendicular to the rod.
     
  4. Aug 8, 2013 #3
    Thanks Voko.

    I don't think the reaction force is paralell to the rod but I hadn't included it in that diagram. For some reason, I didn't think it needed to be included in the diagram of forces.

    The force I drew paralell to the rod was the force D which is the force of the wall against the base of the rod.

    Surely the tension and the weights are tending to rotate the rod in opposite directions?

    Would this be a more appropriate diagram?
     

    Attached Files:

  5. Aug 8, 2013 #4
    The tension and the weight are applied at two different points. So it is invalid to say they tend to rotate the rod in opposite direction. Take a pencil, and hold one end with your left hand, and the other with your right hand. Now move your right hand up, and your left hand down. Do you see that as you hands just start moving, they do not oppose each other at all, and they work together to rotate the pencil?

    Back to the problem. You do not need to introduce two reaction forces. One is enough, just drop the assumption that it is directed along the rod.
     
  6. Aug 8, 2013 #5
    OK, thank you for clearing that up Voko.

    I have removed the force paralell to the rod and modified the force diagram, but can't understand why the value for the reaction force is still not what it should be according to the answer given in the book.

    I make R to be 221N; the book makes the answer to be 314N.
     

    Attached Files:

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