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Force acting on struts

  1. Oct 2, 2009 #1
    the problem is actually for a MATLAB programming exercise. the math is really simple but I want to know if my alternate solution works compared to the "hints" the prof gave us. (sorry if we aren't supposed to put images directly into threads but I'm in a hurry and I didn't see anything relevant in the rules)

    1. The problem statement, all variables and given/known data
    "As shown in the following figure, we know the angle between the two struts, denoted by ∂, in the range 0-90 degrees. Determine the angle ø so that a 500-lb horizontal force has a component of 600-lb directed from A towards C. Also, you need to determine how much the component of force acting along member BA is."

    sm8r54.gif


    2. The attempt at a solution
    I figure that the horizontal force must form a right triangle with the force along AC, so the angle between them (phi) must be cos (x) = 500 / 600 = 33.56 degrees.
    2s1roew.gif

    the solution given to us (below) was to use the law of sines, which makes sense to me, but what is wrong about my solution?
    21o7ef.gif

    the difference seems to be that when using the right-angle method, a steeper angle = greater force. The closer the angle is to being horizontal, the closer the force is to 500- makes sense to me. My real question is, what effect would a 500lb-horizontal force have on the strut if it hit it at 90 degrees? 0lb (0/500), or 500(500/1)?
     

    Attached Files:

  2. jcsd
  3. Oct 2, 2009 #2

    Delphi51

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    Homework Helper

    The key to the problem is to realize that the three forces acting on point A must add up to zero, because that point is not accelerating. In the second solution, these forces are arranged in head-to-tail fashion beginning and ending at A so the total is zero. It is unfortunate that the arrowhead on the A to C vector is backwards so it is hard to see this.

    In your first solution, I do not see anything to indicate that the 3 forces are adding to zero. The AB vector is not even shown. It does not seem to have anything to do with the problem at all.
     
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