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Torque tension force problem

  1. Dec 9, 2006 #1
    Torque

    1. The problem statement, all variables and given/known data

    A 160 N child sit on a light swing and is pulled back and held with a horizontal force 100 N. The tension force of each of the two supporting ropes is:

    A uniform plank is supported by two equal 120N forces at X and Y at both ends of the plank.The support at X is then moved to Z (halfway to the plank center). The support of X and Z now have magnitudes of:

    2. Relevant equations

    a) the first one I dont know where to begin. I know the child's weight acts down 160N and that the tension of the ropes would have to support the child's weight. There would also be a 100N force acting up. I cant do this equilibrium way so I am lost.

    b) ok so I said:
    Fx + Fy= 240N
    because both of them were 120N to be in equilibrium. I took the 240N to be the weight of the plank.
    Then I moved Fx in my drawing to the Z part of the plank, halfway between the center of the board and the end. I then picked a length value to represent the distance. I choose 1m. So I said that Z must now be at 2.5m. I picked Fy as the torque point because it did not move and it was at the end.

    so:
    torque rotations about Fy
    counterclockwise = clockwise
    center of board weight down = Fz force up
    (.5 X 240)= (.25 X Fz)
    Fz= 480N (looks wrong)

    then Im stuck.....:eek:
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 9, 2006 #2
    1- you can do it the eqbm way.
    You've got to balance the following forces:
    Tension, weight, pull.

    weight is down, the pull is back, so the tension must act at an angle to counter both of these. The vertical component of the tension is 160N, the horizontal component is 100N
     
  4. Dec 10, 2006 #3

    daniel_i_l

    User Avatar
    Gold Member

    Draw a diagram with the forces and see what the total tension has to be so that the forces cancel out on both the x and the y axes. but remember that the tension of each rope is only 1/2 of the total tension.
     
  5. Dec 10, 2006 #4
    question

    A 160N child sits on a light swing and is pulled back and held with a horizontal force of 100N. The tension force of each of the two supporting ropes is:

    work:
    The tension force of the Ta and Tb must support the weight of the child and the horizontal force of 100N.

    so:
    Ta + Tb= 160N
    Someone commented that I need to have an angle for the tension ropes in order to support the both horizontal and vertical components. I am having trouble in the equation setup. please help still confused :confused:

    How can deal with this as an equilibrium torque problem if no lengths are given?

    Sorry to post again. but no one responds to my replies
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  6. Dec 10, 2006 #5

    berkeman

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    Staff: Mentor

    Do NOT multiple post. I've merged your other thread into this thread, and I'm giving you warning points. If you continue to multiple-post, you will be banned from the PF. That is not how the PF works.

    As to your question -- does the problem diagram show whether the holding force on the swing is horizontal or tangential? What does your free body diagram look like?
     
  7. Dec 10, 2006 #6
    Sorry, I wont double post
    It says that the holding force for the swing is horizontal. I have the two holding forces going up on by free body diagram, the weight force going down, and the 100N force horizontal.
     
  8. Dec 10, 2006 #7
    Could someone help me?
     
  9. Dec 11, 2006 #8

    berkeman

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    Staff: Mentor

    It sounds like you have the FBD correct. Now just write expressions for the vertical and horizontal components of the forces, and set the sum in each axis to zero (nothing is moving). Then solve for the unknown tensions.
     
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