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2 people pulling on ropes or cables with a third person trying to equal out

  1. Oct 7, 2012 #1
    1. The problem statement, all variables and given/known data

    We have a person pulling on a string or rope in the negative y negative x quadrant (call it left) this person is called Wesson, they have a force of 300N at 30°

    Same situation but pulling the other way, positive y and positive x quadrant (call it right), this person is called Canola, they have a force of 200N at 30°

    The third person pulling is the boss called Olive, they have a force of 250N at 60°

    What I'm seeking is this: What is the magnitude and direction of the force the boss needs to apply to make the system at equilibrium?

    2. Relevant equations

    First make a FBD then..

    I have the sum of Fx = -tw + tc + to = 0

    Pithagoram's theory

    arc tan........Tan-1 (Fy/Fx)


    3. The attempt at a solution

    from the sum of Fx I have = -300N sin 30° + 200N sin 30° + 250N sin 60°

    -150 + 100 + 216.50=

    Fx=166.5

    To get the magnitude i'm using pithagorm's theorm

    so I have sqrt 166.5^2 + 0 (because the sum in the y dir is zero)

    magnitue= 52.17N

    To get the direction I have

    Tan -1 (0 / 166.5N) = 0 N = Direction

    Am I on the money or way off base?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 7, 2012 #2
    We have a person pulling on a string or rope in the negative y negative x quadrant---**Sorry about that, to be left, I meant positive y and negative x quadrant =-)**
     
  4. Oct 8, 2012 #3
    For this one I have
    wesson= -300cos30 ihat + 300sin30 jhat=-259.80 ihat + 150 j hat
    canola= 200 cos 30 ihat + 200 sin 30 jhat= 173.20 i hat + 100 j hat
    then add.............................................= 433 i hat - 50 j hat

    magnitude= sqrt 433 ihat ^2 + -50 jhat ^2=

    435.87N=magnitude

    Direction= Tan-1 50jhat / 433 i hat=
    Direction=6.58 degrees above the positive x axis or north of the positive x axis

    is this right?
     
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