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Center of Gravity/Torque problem

  1. Mar 15, 2008 #1
    1. The problem statement, all variables and given/known data
    A uniform plank of length 2.00m and mass 30.0 kg is supported by three ropes. Find the tension in each rope when a 700N person is 5m from the left end.

    Here's the diagram of the problem:
    http://img206.imageshack.us/img206/3752/torquere9.png [Broken]

    2. Relevant equations
    [tex]\tau[/tex] = rfsin[tex]\vartheta[/tex]

    3. The attempt at a solution
    I'm not really sure where to start or what to do except I'm pretty sure that
    [tex]\sum[/tex][tex]\tau[/tex] = 0

    thanks for helping out :)
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Mar 15, 2008 #2


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    Homework Helper

    Start by splitting the Tension at 40 degrees into vertical and horizontal components. Then consider what the forces mean if the object is in equilibrium
  4. Mar 15, 2008 #3
    [tex]\sum[/tex]F = 0 if it is in equilibrium. I'm still lost :(. I don't know how to find the forces except the two I already have.
  5. Mar 15, 2008 #4
    Initially, set your pivot somewhere so that you only have one variable to worry about.
  6. Mar 15, 2008 #5
    Thanks :) I was thinking the pivot was in the middle for some reason =\ . Well I got T1 (501N) I know that's correct because I checked in the back. Now I just need the other two.
  7. Mar 15, 2008 #6
    The beam has to be in rotational equilibrium as well as in static equilibrium, so the sum of the forces has to equal zero.
  8. Mar 15, 2008 #7
    I got T2, 672N, but now I'm not quite sure how to get T3. To find T2 I did:

    [tex]\sum[/tex]T = (1.5)(700) + (30 * 9.8) - 2T[tex]_{2}[/tex] = 0

    I also did that (but a little different) to find T1. The angle for T3 is 180 so to me it seems like there would be no force, but the book says it is 384N

    Edit: I just read the last post, I think I'll be able to find T3 in a minute. :)

    Edit: Yep, I got it.

    [tex]\sum[/tex]F = 0 = 501cos40 - T2

    Thanks for the tips, and thanks for not solving it all for me it really helped me out :)
    Last edited: Mar 15, 2008
  9. Mar 15, 2008 #8
    If T3 was 0, then the horizontal component of the force of T1 would force the beam to the right and the system would not be in static equilibrium.

    You solve rotational equilibrium aspects of the system using torques, and you solve static equilibrium aspects of the system using forces.
  10. Mar 15, 2008 #9
    Snazzy, do you have AIM by any chance? I have another problem that I'm too stupid to figure out and it would be awesome if you could help. I was gone from school for a week but I have to turn this all in on Monday and I just have a couple questions. :)
  11. Mar 15, 2008 #10
    I don't have AIM but you can just post here or PM me.
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