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Angle on Tension?

  1. Oct 19, 2011 #1
    1. The problem statement, all variables and given/known data

    Two dogs pull horizontally on ropes attached to a post; the angle between the ropes is 60 degrees. If dog A exerts a force of 270N and dog B exerts a force of 300 N, find the magnitude of the resultant force and the angle it makes with dog A's rope.

    2. Relevant equations



    3. The attempt at a solution

    We know that the resultant force between 2 forces is

    [tex]f_{res} = \sqrt{F_{a}^2 + F_{b}^2 + 2 F_{a} F_{b} cos \theta}[/tex]
    [itex] = 493.86N [/itex]

    This part I got write, the part I don't understand is how to find the angle the resultant force makes with the dog A's rope.

    I grabbed the problem on cramster http://www.cramster.com/solution/solution/1339563 and the person uses this formula

    [tex]tan \theta = \frac{F_{b} sin 60}{F_{a} + F_{b} cos 60}[/tex]
    1. The problem statement, all variables and given/known data

    But after 30min+ grappling with how he got to that formula I still don't know how!

    Please help, I'm going crazy over this lol.
     
    Last edited by a moderator: Apr 26, 2017
  2. jcsd
  3. Oct 19, 2011 #2
    I prefer to use components rather some formula.

    But since you used the cosine formula for the resultant you can use the sine formula to get the angle:

    300/sintheta = resultant/sin120
     
  4. Oct 19, 2011 #3
    You can't because they only give you the angle between the rope, not the direction of pull.

    How did you get sin 120?
     
  5. Oct 19, 2011 #4
    If you look at the pararellogram whose sides are the tensions, one can use the sine formula in the traingle made up by the resultant and the two tensions. The angle opposite the resultant is 120deg.
     
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