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Rope Tension in Static Equilibrium

  1. Dec 18, 2006 #1
    1. The problem statement, all variables and given/known data
    A hanging weight with a mass of 25kg is suspended from a rope from a wall. The rope with the weight is wrapped around a pully which is also attached to a different wall. The angle that rope a(rope with mass) makes with the wall is 41 degrees, and the angle of the other rope with the other wall is 69.5 degrees. Find the force of tension in each rope.
    [​IMG]


    2. Relevant equations
    EFx=0
    Efy=o


    3. The attempt at a solution
    EFx=0
    Acos(49)-Bcos(20.5)=0
    A=Bcos(20.5)/cos(49)
    EFy=0
    Asin(49)+Bsin(20.5)=mg
    (Bcos(20.5)/cos(49))sin(49)+Bsin(20.5)=mg
    B=172N
    Substitute B back into A
    A=172cos(20.5)/cos(49)
    A=246

    I got the force of tension in A to be 246N and the tension in B to be 172N. Is this correct or am I making this to easy? Also is it just a conicedence that the tension on the rope with the weight is the force of gravity?
    Thanks for your help.
     
    Last edited: Dec 18, 2006
  2. jcsd
  3. Dec 18, 2006 #2

    cristo

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    Looks good to me, although next time, please provide equations and values you have calculated. It's a lot easier to check your calculations than do them ourselves!
     
    Last edited: Dec 18, 2006
  4. Dec 18, 2006 #3

    cristo

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    Oh, now youve added the picture.. it's not what I thought! I'd imagined both ropes attached to the mass, and the pulley on the left hand wall (for some reason!) Should still be right though.. just draw a free body diagram at the point at which rope A meets rope B and set up the equations from there.
     
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