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Statics Problem (tension, and a distance)

  1. Sep 10, 2009 #1
    2cpol0p.jpg

    Okay, i posted a picture of the problem above, sorry if blurry, do not have scanner. i will retype problem,

    The mass of block A is 250 kg. Block A is supported by a small wheel that is free to roll on the continuous cable between supports B and C the length of the cable is 42 m. Determine the distance x and the tension T in the cable when the system is in equilibrium.

    I tried doing a sum of forces in x and y components and equal to zero but i am not getting it.

    I am wondering, is the tension throughout the whole cable going to be 250(9.81)? or is it different for each side of the weight?

    Any help on where to go is helpful thank you.
     
  2. jcsd
  3. Sep 10, 2009 #2

    tiny-tim

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    Hi yopy! :smile:
    (As the question suggests, the tension should be the same all the way along a rope or cable)

    Sorry, but if you get 250(9.81), you're doing it completely wrong :redface:

    Show us your full calculations, and then we can see what went wrong, and we'll know how to help. :smile:
     
  4. Sep 10, 2009 #3
    250(9.81) was not my final answer but the calculated value i got for T(tension) in the wire. from there i do not know where to go.

    i tried doing sum of forces in x and y coordinates but my 2 unknowns were fetaA and fetaB which were the angles from the x axis to the cord. when i worked it out i got that the angles were the same but i think its wrong
     
  5. Sep 10, 2009 #4
    i worked through this above statement and found that both angles to the left and right of the weight, from the x axis to the line, is 30 degrees. im pretty sure i calculated it right but it seemed to be alot of work for the problem. but from 30 degrees i cant think of anywhere to go from there to get X so i think i did it the wrong way. Also, this is in a chapter that doesnt teach moments. moments are taught after this chapter so i dont think your supposed to use moments.
     
  6. Sep 10, 2009 #5

    nvn

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    yopy: You are correct that both angles are the same. Now it is just a matter of geometry to figure out the angle and lengths, and the value of x. Your 30 deg angle appears to be incorrect. Try again. Last, compute the cable tension T using static equilibrium. No moments are required.
     
  7. Sep 11, 2009 #6

    tiny-tim

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    Sorry, but that is wrong.

    Please show us the equations you got for the horizontal and vertical components of force.
     
  8. Sep 11, 2009 #7
    okay i did FBD, i have 3 forces, one acting on the string to the left, one acting on the string to the right and the weight in Y direction negative. i called the angles to the left and right FetaA and B

    sum Fx = 0 = -Tcos(fetaA)+Tcos(fetaB)
    sum Fy = 0 = -w + TsinFetaA +tcosfetaB

    from Fx i came to conclusion that both angles are the same?

    then i tryed to find T but i cant find it.... since angles are same i figured that x will come out to be the same value as long as i use the same 2 angles for fetaA and fetaB. i worked through alot of geometry of angles and law of sines and found X. but i cant get T.
     
  9. Sep 11, 2009 #8

    tiny-tim

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    Hi yopy! :smile:
    (i assume you meant to type sum Fy = 0 = -w + TsinFetaA + TsinfetaB ?)

    (oh, and have a theta: θ and try using the X2 tag just above the Reply box :wink:)


    Well, T is just w/2sinθ, so if you've found x, you should have also found sinθ …

    what did you get for x and θ? :smile:
     
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