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Homework Help: Suspended beam

  1. Mar 20, 2015 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations
    I do not know exactly.

    3. The attempt at a solution
    I honestly have no idea how to approach this.
  2. jcsd
  3. Mar 21, 2015 #2
    Do you know what the forces acting upon the beam are?
  4. Mar 21, 2015 #3


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    State a law relating forces to accelerations.
  5. Mar 24, 2015 #4
    Gravity and the Tension in the wire
  6. Mar 24, 2015 #5
    It can't be just those. Gravity points downwards, and the tension force points up and to the left. Since this beam is at rest, you've gotta have something canceling that leftward force out.

    Here's a hint. The hinge produces two reaction forces: a horizontal and vertical component. Take those into account, draw a free body diagram, and write the expressions for the net forces and moments.
  7. Mar 24, 2015 #6
    Torque = Fd
    Tsinθ(2.41) - mg(2.9) = 0
    Find the angle through trig?
  8. Mar 24, 2015 #7
    Look at the hint again and factor that into the equations. You will need to find the net force in the x-direction and the net force in the y-direction. You don't need the angle, as you are given the sides of the triangle formed by the cable.
  9. Mar 24, 2015 #8
    I don't understand why I don't need the angle. Can I get a little more help than just this?
  10. Mar 24, 2015 #9


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    You can get the angle from trig, but you have assumed incorrectly that the beam's weight acts at the free of the beam.
  11. Mar 24, 2015 #10
    And therefore you should probably correct me right?
  12. Mar 24, 2015 #11
    My apologies. Yes, you need the angle in order to determine the x- and y- components of the tension force.

    You have a tension force, the weight, and the two forces from the hinge I mentioned acting on the beam. Now write equations for the x and y net forces.
  13. Mar 24, 2015 #12
    Is it
    Fx = Tcosθ
    Fy = Tsinθ
  14. Mar 24, 2015 #13
    Yes, now write the equations for the net force in the x direction, the net force in the y direction, and the net torque.
  15. Mar 24, 2015 #14
    Tsinθ(2.41) - mg(1.45) = 0
    I don't know what to do with Fx and Fy, I don't think I'm properly understanding.
  16. Mar 24, 2015 #15
    That's almost the net torque. You're forgetting the things I mentioned in the hint from earlier.
  17. Mar 24, 2015 #16
    Wait I think I got it here.
    Tsinθ(2.41) - mg(1.45) = 0
    Fx = Tcosθ
    Fy = Tsinθ - mg
  18. Mar 24, 2015 #17
    I'm sorry, I misread something up there. To clear things up:

    Your first formula, the formula for the torque, is correct.

    Your net x and y force equations are incorrect. The hinge produces a vertical and horizontal reaction force that you need to factor into your equation. Notice that you need to find these for part 2.
  19. Mar 24, 2015 #18
    But I just used those and got the right answer?
  20. Mar 24, 2015 #19
    You can get your torque from what you have now. You're supposed to factor in the reaction forces from the hinge to solve this, but I can see what steps would have led to the correct answer.

    If you left your equations as they are and just plugged your numbers in, you would get the reaction forces. The "correct" way to solve is to put the reaction forces into the equations, set them all equal to 0, then solve. I imagine it worked for you because of the simplicity of this system.
  21. Mar 24, 2015 #20
    It worked on a similar problem that had a weight handing on the end too. We don't do anything more complicated than that so it should work. I mean the parts of the question are split so I have to find the tension first.
  22. Mar 25, 2015 #21


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    You were right the first time. You almost never need to find the angle, as such. Only cos and sin of the angle are needed, and these can be found directly from the linear dimensions.
    Since the reaction at the hinge and the horizontal component of the tension all act through the hinge, they produce no moment about the hinge. But they must be taken into account in ##\Sigma F_x## and ##\Sigma F_y##.
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