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Turning effects of forces

  1. Jul 3, 2009 #1
    1. The problem statement, all variables and given/known data
    A uniform, horizontal 300N beam, 5.00m long, is attached to a wall by a hinge. Its far end is supported by a cable that makes an angle of 53.0° with the horizontal. A 600N person stands 1.50m from the wall. Find the tension in the cable (force).


    2. Relevant equations
    What do I have to do? I calculated the moment/torque of 600N*1.50m and than add it to 300N*3.5m. This gives me 1950Nm. But I don't know how to get the right result (413N).
     
  2. jcsd
  3. Jul 3, 2009 #2

    Doc Al

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    Staff: Mentor

    Since the beam is in equilibrium, set the net torque equal to zero.
    That's the torque due to the person.
    That's supposed to be the torque due to the weight of the beam--but why 3.5m?
    Call the cable tension T. What torque does it produce?
     
  4. Jul 3, 2009 #3
    ok, so this gives me the equation: 600N*1.50m+300N*5.0m=2400Nm but what do I have to do with the angle of 53.0°? And I can't just say 2400Nm=0, can I? I mean because you said that I have to set the net torque equal to zero, so I suppose that 2400Nm isn't the net torque...
     
  5. Jul 3, 2009 #4

    Doc Al

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    Here you just added two of the three torques, but ignored the torque due to the cable tension. Also, where did you get the 5.0m? (Hint: Where does the weight of the beam act?)
    You need that angle to compute the torque due to the cable tension.
    That wouldn't make sense, would it? :wink:
    Right. You forgot about the cable tension.
     
  6. Jul 3, 2009 #5
    Be consistent about the point about which you're calculating your torques. What's the torque of the weight of the beam, relative to its center of mass, which is the point I'm assuming you're calculating with respect to?
     
  7. Jul 3, 2009 #6

    Doc Al

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    I presume that Miri is attempting to calculate torques with respect to the hinge. (But, as you say, all that matters is to be consistent.)
     
  8. Jul 3, 2009 #7
    Is it 300N*2.5m because the force acts in the middle of the beam downwards? So we get 600N*1.50m+300N*2.5m+T=0 so for T we get -1650. Isn't there a mistake? :D And how do I have to use the angle??
     
  9. Jul 3, 2009 #8

    Doc Al

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    Good.
    Assuming you are using T to stand for the cable tension, you need the torque not just T. Note that the torque from the cable tension goes in the opposite direction of the torques from the weight of person and beam, which means it has an opposite sign.
    You'll need it to calculate the torque. Review such calculations here: http://hyperphysics.phy-astr.gsu.edu/hbase/torq2.html#tc"
     
    Last edited by a moderator: Apr 24, 2017
  10. Jul 4, 2009 #9
    So for T I take 300N*5.00m*sin(53.0°)=1197,95Nm. And as you said, it is +1650Nm when you solve it for T. But now I have two results and the last one is wrong, because I didn't use the formula like in the first one. Can you tell me what to do?
     
  11. Jul 4, 2009 #10

    Doc Al

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    No, the tension force is T, not 300N. You are trying to solve for T.
     
  12. Jul 4, 2009 #11
    So what do I have to do instead of solving for T?
     
  13. Jul 4, 2009 #12

    Doc Al

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    :confused:

    What you do mean "instead"? Of course you must solve for T. The only question is how.

    Add up the torques for all three forces. Two of them will be clockwise (say), while the other will be counterclockwise. The torque due to the tension will be in terms of the unknown tension, T.

    Set the net torque = 0. Solve for T.
     
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