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AP Physics B Torque

  1. Oct 9, 2007 #1
    A uniform steel beam of length 5.50 m has a weight of 4.50 103 N. One end of the beam is bolted to a vertical wall. The beam is held in a horizontal position by a cable attached between the other end of the beam and a point on the wall. The cable makes an angle of 25.0° above the horizontal. A load whose weight is 12.0 103 N is hung from the beam at a point that is 4.60 m from the wall.

    (a) Find the magnitude of the tension in the supporting cable.
    N
    (b) Find the magnitude of the force exerted on the end of the beam by the bolt that attaches the beam to the wall.
    N

    my attempt:
    ETorque = 0
    Fl(D) + Wb (.5 D) - Ty(D) = 0
    12.0E3N(4.60 m) + 4.50E3N(2.75) - Ty(6.069m) = 0
    Ty = 11134.457 N

    Find Tx : tan25 = 11134.457 / x
    Tx = 23877.913 N

    T = sqrt(23877.913^2 + 11134.457^2)
    T = 26346.36 N

    Efx = 0
    23877.913 N - Fx = 0
    Fx = 23877.913 N

    EFy = 0
    Fy + Ty-Fl _fb = 0
    11134.457 N - 12E3 N - 4.5 E3 N + Fy = 0
    Fy = 5365.543 N

    F = sqrt(23877.913^2 + 5365.543^2)
    F = 24473.327 N

    When i tried the T and F in the online submission, there were wrong both times, and now im on my last try. Can someone please help me??
     
  2. jcsd
  3. Oct 9, 2007 #2

    Doc Al

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    Staff: Mentor

    Where did the distance 6.069 m come from?
     
  4. Oct 9, 2007 #3
    that is the length of the cable..i did 5.5 / cos25
     
  5. Oct 9, 2007 #4

    Doc Al

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    When calculating torques use the same pivot point for all forces. You want the length of the beam, not the length of the cable.
     
  6. Oct 9, 2007 #5
    they gave the lenght of the beam..5.5 m
     
  7. Oct 9, 2007 #6

    Doc Al

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    So use it! :smile:
     
  8. Oct 9, 2007 #7
    ETorque = 0
    Fl(D) + Wb (.5 D) - Ty(D) = 0
    12.0E3N(4.60 m) + 4.50E3N(2.75) - Ty(6.069m) = 0
    i used it here.....................^ the T is for the cable thats added to the beam
     
  9. Oct 9, 2007 #8

    Doc Al

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    That's true. You need to use distance from the pivot point for every force acting on the beam: its weight (in the middle), the load (at 4.60 m), the the tension in the cable (which is applied to the end of the beam).
     
  10. Oct 9, 2007 #9
    ya but then didnt i do the problem right...it says im not getting the correct answer...so im lost and i have only one try left
     
  11. Oct 9, 2007 #10

    Doc Al

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    No, you didn't do the problem correctly--you used the wrong distance when you calculated the torque due to the tension. So of course you'll get the wrong answer until you use the correct distance and do it over.

    Studying this example might help you: Support of a Boom
     
  12. Oct 9, 2007 #11
    Doc Al...I cant do it again..cause i only have one more trial left cool ill try the site
     
  13. Oct 9, 2007 #12
    ok i got the correct answers using the website you gave me
     
  14. Oct 9, 2007 #13

    Doc Al

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    I don't understand what you mean by "can't do it again". Just solve the problem on paper and post your solution here. Someone will check it.

    (If you don't do it again, how will you ever get the right answer?)
     
  15. Oct 9, 2007 #14
    can you explain to me what the website did that i didnt..cause i need to know how to do this for a test

    Correct Ans:
    29072.0130875 N for tension

    Sum of forces =
    26682.98896
     
  16. Oct 9, 2007 #15

    Doc Al

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    But I've already told you what you did wrong--twice!
     
  17. Oct 9, 2007 #16
    oh hahaha i c the mistake lol i wasnt reading properly thanks for EVERYTHING!!!!
     
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