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Vectors Help

  1. Feb 22, 2014 #1
    1. The problem statement, all variables and given/known data

    A mass of 100 kg is suspended by two ropes that make an angle of 60 degrees to the horizontal. If a horizontal pull of 200N, in a plane perpendicular to the plane of the other forces is applied find the tensions in the ropes after equilibrium has occurred.

    2. Relevant equations
    Component Method


    3. The attempt at a solution
    T1cos60=T2cos60
    T2=(T1cos60)/(cos60)

    Fx=-T1cos60+T2cos60
    Fy=T1sin60+T2sin60-Fg

    Sub in T2 into equation
    Fy=T1sin60+T2sin60-Fg
    = (T1sin60)+((T1cos60)/(cos60))(sin60)
    = T1sin60+T1(0.866)
    =T1(sin60+0.866)=(100)/(9.8)
    T1=T2=565.8

    Can someone please tell me if I did this right?
     
  2. jcsd
  3. Feb 22, 2014 #2

    vela

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    It doesn't look it. You solved the problem without the 200 N force.
     
  4. Feb 22, 2014 #3
    What would the 200 N force be?
     
  5. Feb 22, 2014 #4

    vela

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    The one in the problem statement:
     
  6. Feb 22, 2014 #5
    Yes but how would I put that in the equation? When using the component method and drawing the diagram where would I place the 200N?
     
  7. Feb 22, 2014 #6

    vela

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    "In the plane perpendicular to the plane of the other forces." If the tensions and weights are initially lying in the xy-plane, the horizontal 200-N force would point in the z direction.
     
  8. Feb 22, 2014 #7
    Would the 200N be for both ropes? And if the 200N was for the z direction, how would you find the tension of the ropes? Wouldn't the 200N be the tension then?
     
  9. Feb 22, 2014 #8
    This is a bit of a complicated problem geometrically. Before you start worrying about the forces involved, you need to consider in detail the 3D geometry (after the 200 N force has been applied). When the 200 N force pushes horizontally on the mass, the plane that had originally contained the two ropes and the mass rotates out of the vertical by an angle of, say, θ. The two ropes now have components in all three spatial directions. The simplest thing to do is to resolve the unit vector pointing in the direction along each of the ropes into components in the x, y, and z directions (in terms of i, j, k, and the angle θ). You can then draw a free body diagram of the mass, and do force balances in the x, y, and z directions in terms of θ. This will allow you to solve for the tensions in the ropes and the angle θ. Figuring out the geometry and resolving the rope unit vectors into components is the complicated part of this problem.

    Chet
     
  10. Feb 22, 2014 #9

    vela

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    You can also reduce it to a two-dimensional problem in the plane of the ropes. That makes the geometry a lot simpler.
     
  11. Feb 22, 2014 #10
    Yes. Excellent suggestion. I like that much better.

    Chet
     
  12. Feb 22, 2014 #11
    I'm still not understanding the concept of the 200N. Is it not possible to solve it without taking account of the horizontal pull. By using the horizontal pull, how would you use the component method? Can someone please try explaining this to me, I'm so confused. :0
     
  13. Feb 22, 2014 #12
    Step 1 is drawing a figure looking edge-on on the plane containing the ropes, after the 200 N force is applied. Can you please show us what your figure looks like.

    Chet
     
  14. Feb 22, 2014 #13

    vela

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    Of course not. The problem is asking you to find the tension in the ropes when the system is in equilibrium when the 200-N force is applied. You can't possibly get the right answer if you omit that force.

    It seems to me you might not even understand the basic configuration of the ropes, mass, and force. If you don't understand that, you can't possibly hope to even begin analyzing the situation. It would help if, as Chet suggested, you show us a sketch (or two) of what you think is going on.
     
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