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Tension in two ropes with mass

  1. Dec 10, 2014 #1
    1. The problem statement, all variables and given/known data
    A 1200 kg steel beam is supported by two ropes. Rope one is 20 degrees off of the vertical, and rope two is 30 degrees off of the vertical. What is the tension on rope two? (Answer should look like '2300 N'.)

    2. Relevant equations
    F=ma
    The weight of the mass is defined as the force of gravity on the object. therefore W=(1,200 kg)(-9.8 m/s squared). W= -11,760


    3. The attempt at a solution?
    W=(1,200 kg)(-9.8 m/s squared). W= -11,760

    Now I am not sure what do to.
     
    Last edited: Dec 10, 2014
  2. jcsd
  3. Dec 10, 2014 #2

    ehild

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    The beam is in equilibrium. What does it mean on the sum of forces acting on it? Draw the FBD.
     
  4. Dec 10, 2014 #3
    That the forces( or tensions of the ropes) are equal to each other?
     
  5. Dec 10, 2014 #4
    No!

    The W of the object is equal to the sum of the tensions of the ropes.

    W= T1 + T2

    I know W.

    But how am I to find the T1 or T2? I don't understand the vector/ breaking into two equations.
     
  6. Dec 10, 2014 #5
    Tension in Rope 1
    cos 20= (W/T), therefore T1=W/cos 20

    Tension in Rope 2
    cos 30=(W/T), therefore T1=W/cos 30
     
  7. Dec 10, 2014 #6

    CWatters

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    No!

    The beam isn't accelerating in any direction so the sum of the forces acting an any direction is zero. eg The horizontal components of all forces sum to zero. The vertical components of all forces sum to zero. Write these two simultaneous equations.
     
  8. Dec 10, 2014 #7
    T1 + T2 = 11760

    T1x + T2x = 0

    T1y + T2y = 0

    right?
     
  9. Dec 10, 2014 #8

    BvU

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    In which direction is W working ?
     
  10. Dec 10, 2014 #9
    vertically
     
  11. Dec 10, 2014 #10

    BvU

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    Means T1y + T2y + W = 0
     
  12. Dec 10, 2014 #11

    CWatters

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    I would write the two equations as:

    T1y + T2y - W = 0
    and
    T1x - T2x = 0

    However I don't get the answer T2 = 2300N.
     
  13. Dec 10, 2014 #12

    BvU

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    Poster had W = -11760

    2300 N can't be right: with rope 1 at 20 degrees, there's a minimum T2 of over 4000 N
     
  14. Dec 10, 2014 #13

    CWatters

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    Pity there isn't a drawing. I initially imagined a beam supported from the ends like this \___/ but I think that would imply a non uniform beam (otherwise the angles would be the same). So I assume the beam must be hanging like this __\/__ Can the rsrunner please confirm?
     
  15. Dec 10, 2014 #14

    ehild

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    Would you hang a heavy beam at its centre???? Supposing the beam is horizontal the setup migh look like in the figure. The problem did not state that the beam was uniform.

    beamropes.JPG
     
    Last edited: Dec 10, 2014
  16. Dec 10, 2014 #15

    ehild

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    does not mean that 2300 N is the answer. It means that you have to give the result in that form, perhaps, with two signifiacant digits.
     
    Last edited: Dec 10, 2014
  17. Dec 11, 2014 #16

    BvU

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    OP doesn't say ropes are attached at the ends
     
  18. Dec 11, 2014 #17

    ehild

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    True... So the beam can be uniform, but it does not matter.
     
  19. Dec 11, 2014 #18

    CWatters

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    Wouldn't a very non uniform beam allow T2 to be 2300N?
     
  20. Dec 11, 2014 #19

    BvU

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    ##11760 \; \sin(20^\circ) = 4022##
     
  21. Dec 11, 2014 #20

    haruspex

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    I don't see how. You can have the beam at an angle, non-uniform density, ropes attached anywhere - you still have two linear force equations containing only two unknowns, the tensions. The torque equation then puts constraints on points of attachment, etc.
     
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