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Calculating Tension in a horizontal cable

  1. Dec 19, 2008 #1
    1. The problem statement, all variables and given/known data
    The beam is 4.0 m long and has a mass of 14.0 kg. If the beam makes an angle of 41.0 degrees with the wall, what is the tension in the cable?

    The attachment is the picture to go along with visualizing the problem.

    2. Relevant equations

    sum forces = 0
    sum torqes = 0

    3. The attempt at a solution

    ok so since the object is not moving, both the net force and net torque = 0. I drew a FBD with the arrows indicated in the attachment.

    would I have to figure out the horizontal force from the beam and subtract it from the Tension of the cable which would be something like

    cos49 = adjacent / (3.02 / cos41) = 2.63

    That would be the length of the cable.

    how can i calculate the tensions? I'm kinda lost at this point. thanks for your help
     

    Attached Files:

    Last edited: Dec 19, 2008
  2. jcsd
  3. Dec 19, 2008 #2

    Doc Al

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    Not quite sure what you're doing there.

    Hint: If you choose your pivot point wisely, you can solve for the tension in the cable in one step by setting net torque = 0.
     
  4. Dec 19, 2008 #3
    i was trying to figure it out using the net forces = 0.

    if i chose the pivot point being the point where the beam touches the wall, would i have to figure it out in the x and y directions? or could i just take the d to be the length of the beam?
     
  5. Dec 19, 2008 #4

    Doc Al

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    That's not enough. Whenever you have an extended body (like a beam), be ready to use torques.

    Good choice.
    I don't understand what you're saying here. (What's d?)

    About that pivot point, what torques act on the beam?
     
  6. Dec 19, 2008 #5
    the only torque about that point would be the torque of the beam right?

    so the torque of the beam would be offset by the tension in the cable?

    sooo if i fiind the torque of the beam, that will equal the tension in the cable

    or am i making up my own laws of physics here?
     
  7. Dec 19, 2008 #6

    Doc Al

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    By "torque of the beam" I think you mean the torque due to the weight of the beam. That's one torque. The tension creates another torque.

    The torque due to the weight will be balanced by the torque due to the tension.

    Torque and force (such as tension) are different things with different units, so a torque cannot equal a force. (Of course they are closely related.)

    Just a little. :wink:
     
  8. Dec 19, 2008 #7
    ok i think i know what you are saying.

    the torque of the cable will be equal but opposite the torque of the beam, causing a net torque of 0. using the torque of the cable i should be able to figure out the tension of the cable. since tension is just a force, and force is just ma, in this case a = 9.8, Fbeam * rbeamsin(theta) = -Fcable * rcable


    since sin(90) = 1, that doesn't need to be taken into accord for the cable.

    Is that equation right so far?
    i would just solve for Fcable and that should be the tension.

    EDIT:
    would it be this equation instead?

    Fbeam*rbeamsin41 = Fcable*rcable*sin49?

    i realized the angle is probably 49 and not 90. i was looking at the incorrect place
     
    Last edited: Dec 19, 2008
  9. Dec 19, 2008 #8

    Doc Al

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    Good!
    Forget that last part about a = 9.8. Nothing's accelerating here.

    Why do you think the angle is 90 degrees?

    Put that sinθ back for the tension force.
    Yes, once you fix the equation you can solve for the tension.
     
  10. Dec 19, 2008 #9

    Doc Al

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    Much better! (What are rbeam and rcable equal to?)
     
  11. Dec 19, 2008 #10
    Rbeam = 2m
    Rcable = 3.02/sin41 = 4.6 / 2 = 2.3

    so it would be, with everything plugged in

    Fbeam*Rbeam*sin(41) = -Fcable*Rcable*sin(49)

    and therefore

    137.2N * 2m * .656 deg. = -Fcable * 2.3m * sin(49)

    solving for Fcable

    Fcable = -(137.2N * 2m * .656 deg.) / (2.3m * .755 deg.)

    Fcable = -103.7 N

    or would it be positive?
     
  12. Dec 19, 2008 #11

    Doc Al

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    Good. The weight acts in the center of the beam.
    No. The tension acts at the end of the beam, so Rcable should equal the full length of the beam.

    Get rid of the minus sign. What you're really doing is setting clockwise torques (from the weight) equal to counter-clockwise torques (from tension).
     
  13. Dec 19, 2008 #12

    ok so it would be

    137.2N * 2m * .656 deg = Fcable * 4.6 * .755

    Fcable = (137.2N*2m*.656) / 4.6 * .755
    Fcable = 51.83N ?!?!?
     
  14. Dec 19, 2008 #13

    Doc Al

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    How did you get 4.6? The beam is only 4 m long.
     
  15. Dec 19, 2008 #14
    4.6 was the length of the cable not the beam.

    since i know the angle between the wall and the beam, and i knew the length of the beam (the hypotenuse), i used sin41 = Lcable / 4....

    which i just calculated to be 2.62 haha wow strange mistake there....

    ok so 137.2N * 2m * .656 = Fcable * 2.62 * .755
    Fcable = (137.2N*2m*.656) / (2.62 * .755) = 90.999 so 91 N?
     
  16. Dec 19, 2008 #15

    Doc Al

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    The length of the cable is not what you need.

    Since you are finding torques on the beam, all distances are along the beam. R is the distance from pivot to the point of application of the force on the beam. Since the weight acts at the middle of the beam, Rweight = 2m (half the length of the beam). The tension acts at the end of the beam, so what must Rtension equal?

    Note: It might be less confusing if you used Fweight instead of Fbeam, since both forces act on the beam. Similarly, use Rweight instead of Rbeam.
     
  17. Dec 19, 2008 #16
    ok you lost me :(

    so Rweight is equal for both sides of the equation as well as Fweight?

    or is Rtension = Rbeam?

    i'm confusing myself i think
     
  18. Dec 19, 2008 #17

    Doc Al

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    No, Rweight is equivalent to Rbeam.

    All I'm suggesting here is to use the word "weight" (which describes the force) instead of the word "beam" when describing the force due to the weight of the beam.
     
  19. Dec 19, 2008 #18
    ok, so using your suggestion, the formula is:

    Fweight*Rweight* sin41 = Ftension*Rweight * sin49?

    are the Rweight supposed to cancel?

    if so,
    Fweight * sin41 = Ftension*sin49
    137.2 * sin41 = Ftension*sin49
    Ftension = 119.3 N?
     
  20. Dec 19, 2008 #19

    Doc Al

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    No, the formula should be:

    Fweight*Rweight* sin41 = Ftension*Rtension * sin49

    (I didn't realize that I changed "cable" to "tension" as well, so that might have confused you.)

    The key thing is: What is Rtension (which you called Rcable)? Rtension and Rweight are distances that you should know.
     
  21. Dec 19, 2008 #20
    as a pre-ps, thanks for putting up with me as long as you have haha

    Rweight = 2 because the weight acts on the middle of the beam.

    as for Rtension, like you said, acts on the end of the beam. so wouldn't it be the distance from the wall to the end of the beam, which happens to be Ltension?

    just to make sure, the unknown is still Ftension right? and Rtension = distance from pivot point to the point of application of force. Wouldn't that distance be the length of the cable? or is the point of application of force NOT where the cable and the beam meet?
     
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