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Another tension force question

  1. Feb 11, 2013 #1
    1. The problem statement, all variables and given/known data
    See the attachment for the diagram and problem:
    Ch4C WA2.jpg

    2. Relevant equations
    So the magnitude of the tension in the wire is 13.38 N. And each angle is 14 degrees below the x-axis, one on the positive side and one on the negative side.
    Would I use my Tcos(theta) and Tsin(theta) equations for this? If so, which angle do I use? Because it would be counter clockwise from the x-axis, so NOT 14*, and they wouldn't be the same angle either.

    3. The attempt at a solution
    I don't really know how to start; I think I have everything I need, I just don't know which step to take first.
  2. jcsd
  3. Feb 11, 2013 #2
    Well ƩF=ma, so what are the forces acting on this?
  4. Feb 11, 2013 #3
    I don't have a mass or acceleration though...is there a normal force exerted on it?
  5. Feb 11, 2013 #4
    Ok, how about this, what are the forces in the x and y direction? There is no acceleration, therefore a=0 right? You have to find the force in each direction set them equal to 0 and you'll have the tension. ƩTx=what? ƩTy=what?
  6. Feb 11, 2013 #5
    I think I got that part...
    would you find the x and y components using Tcos(theta) and Tsin(theta) for each one? Do you have to do it for each angle? For the angle, would you do counter clockwise from the x-axis?
  7. Feb 11, 2013 #6
    Well since the tensions are the same, it'd just be 2Tsinx for the y direction right? What happens to the x direction?
  8. Feb 11, 2013 #7
    What angle would you use though???

    And I'm not really sure for the x direction...is it 0? Or is it 2Tcos(theta)?
  9. Feb 11, 2013 #8
    Ok this about the x direction. One Tx is going in the negative and one is going in the positive. If you sum them up, they become Tcosθ-Tcosθ and that equals 0 (because force is a vector, so it has a magnitude and direction, one positive and the other negative). So the forces in the x direction are 0. The y direction is different though.
    Think about the direction of the y components.
  10. Feb 11, 2013 #9
    The sum of the y components would be the 2Tsin(theta). I really just don't know what angle to use for theta.
  11. Feb 11, 2013 #10
    First off, what is the direction of the y component.
    Second, the degree is in the picture? 14 degrees
  12. Feb 12, 2013 #11
    Is the degree in the picture the correct one to use in the equation? Since you have to put in the x and y coordinates; it might be 14 degrees BELOW the x-axis on either side, but the angle is measured counter-clockwise from the positive x-axis. So that's why I'm questioning it. I have the magnitude to plug into the 2Tsin(theta) equation, I just need an angle.
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