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Help with physics homework on forces?

  1. Sep 28, 2014 #1
    1. The problem statement, all variables and given/known data
    A ball is held by a vertical tether C. The vertical tether C is held be tethers A and B (which are slanted). Tether A has an angle of 30 degrees. Tether B has an angle of 60 degrees. The ball has 150 Newtons. The system is in an equilibrium. What are the Newtons in theaters A, B, and C?

    2. Relevant equations
    Fnet = 0

    3. The attempt at a solution
    We know the vertical component force is equal to the two vertical components of the two slanted tethers.
    The two tethers' vertical components are 75 N because it is half of 150 N of the ball.
    sin 30 = (75 N)/ x // x is the tether A, we solve for x to get the newtons for A
    x sin 30 = 75 N
    x = (75 N)/(sin 30)
    x = 150 N

    sin 60 = (75 N)/ y // y is the tether B, we solve for x to get the newtons for B
    y sin 60 = 75 N
    y = (75 N) / (sin 60)
    y = 87 N

    The tether A has 150 Newtons and tether B has 87 Newtons and tether C has 150 Newtons.

    But my book says that tether A has 130 Newtons, tether B has 75 Newtons and tether C has 150 Newtons!!!
    I set the problem correctly though. I have an illustration of the same problem. Note they are in an equilibrium.
     

    Attached Files:

  2. jcsd
  3. Sep 28, 2014 #2

    mfb

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    With the same logic, if a mouse plus an elephant have a comined mass of 3000kg, then the mouse has 1500kg?

    The setup is not symmetric, there is no reason to assume their vertical components are the same.
    If you calculate the horizontal components of the tethers in your answer, you'll see they are not equal, so the system would be out of equilibrium.
     
  4. Sep 28, 2014 #3
    @mfb the textbook said they were in an equilibrium. but it didn't say what length the two tethers were so....

    Do you know how to at least set up the problem correctly?
     
  5. Sep 28, 2014 #4
    Equilibrium means it's not moving (or maybe it is moving with constant velocity, not too sure), but regardless the acceleration in the X and Y components equals 0. However, that does not mean it needs to be spread equally between both tethers.

    The scenario you're thinking of is when the angles of both tethers is 45 degrees. Then they both pull an equal amount in the X and Y direction.

    In order to set up the problem you need to think about what equilibrium means, and how that carries over to an equation.
     
  6. Sep 28, 2014 #5

    RMZ

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    mfb, can you please give me some analogy to help understand why the vertical components of A and B don't equal to 150? I have seen similar situations (such as pushing two objects of different masses that are touching each other) and I can explain those using Newtons second law, but this situation seems very strange.
     
  7. Sep 28, 2014 #6

    SteamKing

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    mfb is not saying that. You are not understanding what he is telling you.

    You are the one saying that because there are two tethers holding up the ball, that the vertical component of each tether must be equal to the other, and that is simply not the case here.

    Look, 150 = 75 + 75, but 150 = 90 + 60, or 150 = 100 + 50, ... etc.

    What you have to do is satisfy two conditions simultaneously here for the ball to remain in equilibrium:

    The vertical components of tethers A and B must sum to 150 N, and
    The horizontal components of tethers A and B must sum to 0 N.

    Use the trigonometry to resolve the unknown tensions in tethers A and B in terms of their horizontal and vertical components.
     
  8. Sep 29, 2014 #7

    RMZ

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    but if you look at the books answer, and use trig to find the vertical components of A and B's tension, they don't equal to 150
     
  9. Sep 29, 2014 #8

    RMZ

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    and i'm not the original poster here, i'm asking a completely different question
     
  10. Sep 29, 2014 #9

    SteamKing

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    Book answers have been known to turn out wrong.
     
  11. Sep 29, 2014 #10

    mfb

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    The length is not relevant. I added the important hint in my previous post: the horizontal forces have to cancel. This gives you two equations (horizontal and vertical) for two unknowns.

    Their sum is 150N. The book's answer is correct.
    By the way, you can edit your posts if you like to add something.
     
  12. Oct 1, 2014 #11

    RMZ

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    MFB, I'm doing:
    Using the book's answer (which he listed)
    TA,vertical+TB,vertical=
    130*sin(30degrees)+75*sin(60 degrees)=129.95N

    ....But if I use 60 degrees for A and 30 degrees for B, it comes out to 150. I wonder if that is part of petermojer's problem? Or maybe its just a typo in his post.

    and thanks
     
  13. Oct 3, 2014 #12

    mfb

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    If you swap the angles, you calculate the vertical forces. This is purely an accident here as the two given angles happen to add to 90°, in general swapping them would be meaningless.

    The horizontal forces go in opposite directions, you have to either subtract them or make one angle negative (as sin(-x)=-sin(x), this gives the same result).
     
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