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Homework Help: Torque Equilbrium

  1. Oct 24, 2007 #1
    A uniform 10.0-N picture frame is supported as shown in Figure P8.62. Find the tension in the cords and the magnitude of the horizontal force at P that are required to hold the frame in the position shown.

    T = Fxsintheta

    So I got what's going on in each direction in a separate equation but I need another equation for the third unknown....

    T1sin50 = T2
    T1cos50 = F

    What am I missing?

    ADD: Just to fill in, the whole equations have the distances but if you use the point of gravity as your axis in each case, they are all equidistant from the axis so the distances cancel out, and the torque due to the weight of the picture ends up being 10N(0m) = 0 N*m

    Attached Files:

    Last edited: Oct 24, 2007
  2. jcsd
  3. Oct 24, 2007 #2
    I'm not sure if this is correct but here goes:

    Since the picture is stationary then the total upward force must equal 10.0N. Because the center of mass is in the middle and the distance to each cord is equal, we can assume that the upward force in each cord must equal 5.00N. Therefore the tension in [tex]\vec{T}_2[/tex] = 5.00N

    and the tension in [tex]\vec{T}_{1,y}[/tex] in the vertical direction is 5.00N.
    So to find the total tension in [tex]\vec{T}_1[/tex] we use:
    5.00N / sin(50.0) = 6.53N

    Then, F should be equal to [tex]\vec{T}_{1,x}[/tex] in the horizontal direction so:
    F = cos(50.0) * 6.53N = 4.20N

    [tex]\vec{T}_1[/tex] = 6.53N
    [tex]\vec{T}_2[/tex] = 5.00N
    F = 4.20N

    I hope that's right
  4. Oct 25, 2007 #3
    None of those are right... Seemed right to me. Any more help from anyone?

    This is due 8:30 AM EST on Friday (tomorrow).
  5. Oct 25, 2007 #4


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    You are not doing your torque and translational equilibrium equations correctly. It's sum of forces in x direction = 0; sum of forces in y direction = 0; and sum of torques about any point = 0. That gives you 3 equations with 3 unknowns (T1, T2, and F) which you can solve. I'm not sure if you are trying to sum torques, but if you are, you have not considered the torques from all the forces.
  6. Oct 25, 2007 #5
    Ok so I have:
    T1sin(50) + T2 - 10N = 0
    T1cos(50) - F = 0
    -10N(.15m) + T1cos50(.15m) + T1sin50(.3m) = 0

    Using the bottom left corner of the picture as the origin which makes the torques due to T2 and F 0 because they're on the axes.

    This gave me:
    T1 = 4.59796
    T2 = 6.47776
    F = 2.9555

    None of these are right.........
    Last edited: Oct 25, 2007
  7. Oct 25, 2007 #6
    sigmaFy T2 + T1sin50deg = 10N
    sigmaFx T1cos50deg = F

    sigma M = 0 at point P

    0.30T2 + 0.15T1sin50 +0.15T1cos50 -0.15(10N) = 0 [this is wrong!]

    Now try to make an equation for T2 and sub it in the equation above to solve for T1.
    I guess you will have to use T2 = 10-T1sin50deg.

    EDIT:im getting 81.08N for T1
    looks very high, so this is not a good solution i believe.
    This is the correct moment equation at point P. should be.
    0.30T2 +0.15T1cos50 -0.15(10N) = 0
    by removing 0.15sin50, i now get 22.49N for t1.
    Last edited: Oct 25, 2007
  8. Oct 25, 2007 #7


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    see my note above in red..you missed a minus sign...
  9. Oct 25, 2007 #8
    Why should it be minus? It's going out in the positive direction

    Is my equation right then instead of poofaces? I don't understand why the torques on the axes wouldn't be 0.

    ADD: you are right about the minus (I'm getting right answers now), but will you explain why?
    Last edited: Oct 25, 2007
  10. Oct 25, 2007 #9
    edit:i was wrong
    Last edited: Oct 25, 2007
  11. Oct 25, 2007 #10
    It works with -10(.15) -T1cos50(.15) + T1sin50(.3) = 0 as PhantomJay said.

    I just don't understand why the t1cos is negative.

    ADD: For completion's sake, my answers were:
    T1 = 11.24478
    T2 = 1.385996
    F = 7.228
    Last edited: Oct 25, 2007
  12. Oct 25, 2007 #11
    yea phantomjay how is the t1cos negative?? it is clockwise around the point no? same as 10N it is clockwise

    EDIT I understand why now. He said that because your T1sin50 was positive although conventionally it should be negative because of its counterclockwise direction.

    it doesnt matter as long as you are consistent, if you make counterclockwise positive then all clockwise must be negative.

    I had the right equation and calculations but I messed up on the calculator itself.

    instead of putting 1.5/0.133 for T1, I dont know why I kept putting 3/0.133. that is why i was getting double the T1, 22.49N

    Good job anyways guys and thanks.
    Last edited: Oct 25, 2007
  13. Oct 25, 2007 #12
    Oh... I guess I didn't fully learn the clockwise/counterclockwise thing. I was just working off an X and Y axis and the 10N was going in the -y direction. I understand now. Thanks for the help.
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