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Regid body equalibrium

  1. Nov 14, 2006 #1
    A 1090-N uniform beam is attached to a vertical wall at one end and is supported by a cable at the other end. A 1960-N crate hangs from the far end of the beam. Using the data shown in the figure, find (a) the magnitude of the tension in the wire and the magnitudes of the (b) horizontal and (c) vertical components of the force that the wall exerts on the left end of the beam.

    [​IMG]


    I was able to get part A but I'm having trouble on b and c, I just don't know how to draw this free body diagram?

    Thanks for all ur help
     
  2. jcsd
  3. Nov 14, 2006 #2
    Okay -- to work you through the free body diagram of the beam...

    First can you tell us all the forces that will have to be on the diagram?
    Where, and in what direction, are all of these forces?

    Maybe attach a picture of this so we can check it out.
     
  4. Nov 14, 2006 #3
    well at tutoring today the guy didn't explain it too sell. But I know there is the weight of the ladder and the weight of the box going down oh and the tension force of course of the top string. which I solved to equal 2202 N

    I can't attach a picture of my hand drawing cause I don't have a scanner :( I really just need a equation for this problem for I'm scared that it'll pop up on the exam tomorrow :(

    Thanks a lot
     
    Last edited: Nov 14, 2006
  5. Nov 14, 2006 #4
    The problem also told you there were two forces at place it joins the wall... those need to go in there. Since you don't know the magnitudes of them (that's what you are looking for) you need to just use them as variables...

    Then -- to find two unknowns, you'll need to use two equations.
    Use -- net force is zero, net torque is zero.

    Does that help?
     
  6. Nov 14, 2006 #5

    PhanthomJay

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    Since you already have determined the tension in part A, and since the reaction at the top of the wall must be equal and opposite to (in line with) that tension force, it just becomes a matter of solving for the horizontal and vertical forces at the lower part of the wall by Newton 1 : sum of x component of forces = 0, and sum of y component of forces = 0.
     
  7. Nov 14, 2006 #6
    Yeah! that's even easier -- you get two equations just by the net force equals zero. No need for torque. Must be getting late here for me. :yuck:
     
  8. Nov 14, 2006 #7
    you guys//lady are awsome :)

    Thanks a lot
     
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