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Finding Tension with angles

  1. Nov 18, 2011 #1
    1. The problem statement, all variables and given/known data
    A ball of weight 5 newtons is suspended by two strings as shown above. Determine the magnitude of each of the forces, T1 and T2. (Note: sin 37° = 0.6; cos37° = 0.8)

    t7hpvk.jpg


    2. Relevant equations
    T = mg


    3. The attempt at a solution
    I couldn't go anywhere because I need a formula that uses sin or cos, and I can't remember any right now. Can anyone supply me with the equations and a start? And I'll attempt at the solution with that. Please.
     
    Last edited: Nov 18, 2011
  2. jcsd
  3. Nov 18, 2011 #2
    Would this work? T x cos θ1 + T x cos θ2 = mg, where θ1 and θ2 are the angles the two sides.
     
  4. Nov 18, 2011 #3

    berkeman

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    Staff: Mentor

    I'm having trouble accessing your image. Can you try uploading it to the PF?
     
  5. Nov 18, 2011 #4
    Oops, there I edited the first post and it should be there now.

    Also, after I posted that second post, I realized that nowhere in there it had any variables for the length of the string, which likely play a part in here. So I think I'm not supposed to use that equation.
     
  6. Nov 18, 2011 #5

    berkeman

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    On these problems, always start by drawing a Free Body Diagram (FBD) of the object (the mass in this case). Show the forces on the object, including the force down due to gravity. Then write the two equations to sum all of the forces in the x and y directions to zero (since the object is not moving). then solve away!
     
  7. Nov 18, 2011 #6
    I have one drawn right in front of me right now, but I just don't know what formula to use, or how to find ALL of the forces in both axis.
     
  8. Nov 18, 2011 #7

    berkeman

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    How many forces act in the horizontal direction? What are they? Label forces in the +x direction positive, and forces in the -x direction negative.

    How many forces act in the vertical direction? What are they? Forces up are positive, and forces down are negative.

    Since the object is not accelerating, you know that the sum of the forces in the x direction needs to equal zero. Same thing for the y direction. Those are the two equations you write and solve for the tension forces.
     
  9. Nov 18, 2011 #8
    In the x direction T1 - T2 = 0?

    And in the y direction. T1 + T2 = mg?

    So i'll have to use Trig to find the numbers for each axis, but the only number I have is length, how would that help me?
     
  10. Nov 18, 2011 #9

    berkeman

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    That is much closer. Good.

    But you need to use the "components" of the tensions in the x and y directions, and that is where the sin and cos functions come in. Those are used to resolve vectors (like the tension forces applied at angles) into their x and y components.

    See the "Intro to Vector Mathematics" link at the bottom of this FBD tutorial, for example:

    http://physics.about.com/od/toolsofthetrade/qt/freebodydiagram.htm

    .
     
  11. Nov 18, 2011 #10
    I am still back to the beginning. I know 'what' to do, I just don't know 'how' to do it. What numbers do I use, what equation?
     
  12. Nov 19, 2011 #11
    Your first step is to break up the tensions into vertical and horizontal parts, then since the ball isn't moving the vertical parts add to balance the force of gravity and the two horizontal parts must cancel with each other :)
     
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