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Torque-related question

  1. Nov 10, 2004 #1
    I have the hardest homework ever for my college (algebra-based) physics class (took 4 hours for 8 problems!!) and I really need some help understanding how these problems work. The one's I can't get are all related so if I figure out one I can get more. It's all related to torque, inertia, etc. One that I'm having the most problems with is:
    A shop sign weighing 214 N is supported by a uniform 144N beam of length L=1.90 m. There is a guy wire connected D=1.20m from the backboard. Find the tension in the guy wire. Assume theta=39.5degrees. (the right angle of the right triangle made by the guy wire and the beam is against the wall of the building in the drawing). So what I have so far is that I know I need to find all the forces for Fx and for Fy. The tension (T) points upwards on the guy wire, in the negative x direction. The weight of the beam and the weight of the sign are also forces pointing downwards in the negative y direction. For Fx, I know for sure that Fx= -Tcos(theta) and there is also a force in the horizontal direction at the hinge of the beam and the wall. For Fy, I know that Fy= -Weight(beam) - Weight(sign)-Tsin (theta) and there is a vertical force in the positive y direction at the hinge. I get stuck at where the length comes into play when solving the equations and what do do because the guy wire isn't connected at the end of the beam, only 1.20 m down it. Can anyone give me some hints?!?! Thank you!
     
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  3. Nov 10, 2004 #2

    jamesrc

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    First thing: The signs on the reaction forces at the hinge (Fx and Fy) depend on how you draw them. If I go by what you've written for what you've worked on, that means you drew Fx pointing in the negative x direction and you drew Fy pointing down. I personally wouldn't have drawn it that way, but that's fine; all that is important is that you remain consistent. Look at your equation for Fy. This should come from a force balance in the y direction:

    [tex] \Sigma_y F = -F_y -W_{\rm sign} - W_{\rm beam} + T\sin\theta = 0 [/tex]

    Do you see where you made a mistake? The y-component of the tension in the guy wire points up, bearing some of the weight.

    To finish the problem you need to do a torque balance. You can pick any point to do this; I'll recommend picking the origin (where the hinge is). This way, you don't even need to find Fx and Fy. I'll write out the equation for the torque balance and you should think about the definition of torque when you're trying to see where it comes from:

    [tex] \Sigma_o \tau = -\frac L 2 W_{\rm beam} - L W_{\rm sign} + D T\sin\theta = 0 [/tex]

    The first term is the torque about the origin due to the weight of the beam (acts like a point load in the center of the beam) and is directed in a clockwise sense.

    The second term is also directed clockwise and is due to the weight of the sign. Note that I am assuming the sign is attached to the end of the beam since I didn't see you state explicitly where it is. (If it is connected in some other fashion, replace D with the appropriate lever arm).

    The final term acts in a counter-clockwise direction and is due to the tension in the guy wire. More specifically, it is due to the vertical component of the tension in the guy wire. D is the lever arm that is multiplied by this force to get the torque.

    Remember to work all of this out for yourself just in case I made any mistakes. Good luck.
     
  4. Nov 10, 2004 #3
    I am horrible at these problems, I don't understand how to do them very well but your explanation did help a great deal. The only thing is, what gave the impression I drew Fx and Fy in the negative x direction and the negative y direction? I did on my free-body diagram, so obviously I wrote it wrong? I had the tension pointing up, like you said, but it points to the left in general because of the way the problem drew the diagram. Did I still do it completely wrong? Thanks again!
     
  5. Nov 10, 2004 #4

    jamesrc

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    Well, let me back up a minute: are Fx and Fy your labels for the reaction forces at the hinge?

    --Rereading your post: are Fx and Fy meant to be the x and y components of the tension?
     
    Last edited: Nov 10, 2004
  6. Nov 10, 2004 #5
    Ooh, no. I didn't know how to do sigma Fx and sigma Fy and that's what I meant. Wow, really confused. On my paper, I do use Fx (pointing to the right) and Fy(pointing up) for the hinges too. I should've clarified that, I'm sorry.
     
  7. Nov 10, 2004 #6

    jamesrc

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    OK, don't worry about it.

    You should notice from the torque equation that we don't even need to find the reaction force at the hinge if all we have to do is find the tension in the wire.

    If we want to find the reaction forces at the hinge, let's call them Rx and Ry. Draw them into your diagram and let me know what directions you choose to give them; then write out the force balance equations for the x and y directions.
     
  8. Nov 11, 2004 #7

    jamesrc

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    P.S. I didn't mean to make that last post sound like an order. I've got to get some sleep now, but if you keep posting your work, either I or someone else here will be happy to help you out. Have fun.
     
  9. Nov 11, 2004 #8
    I understand that we don't need to find reaction forces from the equation. For some reason, it doesn't give me the right answer though. Maybe I calculated it wrong.

    Anyway...for Rx I pointed it to the right (same direction as the beam coming out of the wall to the right) and Ry I pointed upward in the positive y direction. Sigma Fx= Rx-Tcos theta and Sigma Fy=Ry+Tsin theta-Wbeam-Wsign . But if I pick the point of origin from the hinge, I don't need them like you said, so the sigma torque= DTsin theta-(L/2)Wbeam-LWsign=0 right?
     
  10. Nov 11, 2004 #9
    P.S. It didn't sound like an order, don't worry, I appreciate all your help. I wish my instructor was as good at explaining things. Staying up late has gotten to be a habit and I forget it's 1:00 am already. Sorry to keep you up and thanks again!
     
  11. Nov 11, 2004 #10

    jamesrc

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    I'm sorry to hear that your instructor has a hard time explaining things, but luckily there are places like this to help people learn in spite of bad instruction. The work you've shown above is all correct, so maybe you understand this stuff better than you think. I hope that will help your sleeping patterns improve.
     
  12. Nov 11, 2004 #11
    The work seems right, I thought about it in my sleep even (physics overload!), but I couldn't get the right answer. We have computerized homework that we type in online and we get 10 tries, and my answer hasn't been right yet. The one I got was 6976.7 and that's not right. Any suggestions??
     
  13. Nov 11, 2004 #12
    Never mind, I'm dumb, I got it :)
     
  14. Nov 11, 2004 #13
    Related to this question, if I wanted to find out what the vertical force on the hinge is, how would I do that? It's not just Tension sin theta is it?
     
  15. Nov 11, 2004 #14

    jamesrc

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    No, it's Ry, which you get from:

    [tex] \Sigma_y F = R_y + T\sin\theta - W_{\rm sign} - W_{\rm beam} = 0 [/tex]
     
  16. Apr 15, 2005 #15
    Humm i am trying to find the horizontal force exerted on the hinge but it is not working out for me :confused: Can someone please help?


    Sorry please disreguard the above I just figured out what I did wrong :rofl:
     
    Last edited: Apr 15, 2005
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