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Circular Plate

  1. Sep 30, 2007 #1
    URGENT QUIZ TOMMOROW - Circular Plate Tension Question

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

    Weight is 600N, and they form 30 degree angles with the vertical
    [​IMG]

    2. Relevant equations

    determine the tension in each wire

    3. The attempt at a solution

    I realize the two wires on the side are both 100 degrees apart from the DB, so they must be the same tension, ive tried a few methods but really havent gotten anywhere, not sure what is the right order to start this
     
    Last edited: Sep 30, 2007
  2. jcsd
  3. Sep 30, 2007 #2
    i also tried splitting it into components along the x and y axes, but it didnt end up helping me get different answers for the tensions
     
  4. Sep 30, 2007 #3
    can i do this? 600=(T1+T2+T3)cos30
    and assume all the tensions are the same?

    then 600/3cos30=T
    then, T= 230.94 N? or 173.205N?
     
    Last edited: Sep 30, 2007
  5. Sep 30, 2007 #4
    im really hoping someone can reply tonight if possible, ive been working on this but nowhere...any help?
     
    Last edited: Sep 30, 2007
  6. Sep 30, 2007 #5

    dynamicsolo

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    That won't necessarily be true: the tension in the middle cord is likely different. You can say that the tensions in the two end cords are equal.

    I'd suggest ignoring the x-y axes given and use the equal 50º angles you described. But, you need to consider the forces in three dimensions, since they must balance in the plane of the disc and in the vertical direction. What would the force components in each dimension look like?

    Have you dealt with torques yet in your course? I ask because they may turn out to be of concern here...
     
  7. Sep 30, 2007 #6
    Well the 600n is simply F(DO)=600k, but im having troubles breaking up the chords into its components

    and yes, we've just begun on Moments/torques as well.

    the reason i first assumed they were equal is due to the 100 degree spacing from the centre chord, im just having trouble relating it to the entire circle
     
  8. Sep 30, 2007 #7

    learningphysics

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    The way I'm seeing for this problem is... 3 equations, 3 unknowns... 1 equation per axis...

    suppose the three tensions are T1, T2 and T3. What is the component of T1 along the z-axis? What is the component of T1 along the x-axis? What is the component of T1 along the y-axis?
     
  9. Sep 30, 2007 #8
    so Tdc along z-axis would be DCcos30, then along y-axis, (CDsin30)(cos120), and x-axis (CDsin30)(sin60) ?
    Tdc=CD(0.433i-0.25j+0.866k)?
     
    Last edited: Sep 30, 2007
  10. Sep 30, 2007 #9

    learningphysics

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    The z-components are all T1cos30, T2cos30, T3cos30...

    For the T1 component... first divide it into two components... one along the z-axis (T1cos30), and one in the x-y plane... so the magnitude in the xy plane is T1sin30.

    Draw in that vector that has the magnitude T1sin30... it lies along that line which is 50 degrees from the x-axis... What is the component of the T1sin30 along the x-axis... along the y-axis? use trig.
     
  11. Sep 30, 2007 #10

    learningphysics

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    yes, exactly right.

    The numbers aren't right... take i to be for the x-axis, j to be for y-axis, k to be for the z-axis...
     
  12. Sep 30, 2007 #11
    Tdc=CD(0.433i-0.25j+0.866k)
    Tdb=DB(-0.32i-0.383j+0.866k)
    Tda=DA(-0.32i+0383j+0.866k)

    so are those all correct?
     
  13. Sep 30, 2007 #12

    learningphysics

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    yup. they look right to me.
     
  14. Sep 30, 2007 #13
    i thought Da and Dc were suppose to be the same so im confused on how to move on from this stage
     
  15. Sep 30, 2007 #14

    learningphysics

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    why do you say Da and Dc are the same?

    Tdc=CD(0.433i-0.25j+0.866k)
    Tdb=DB(-0.32i-0.383j+0.866k)
    Tda=DA(-0.32i+0383j+0.866k)

    Sum of the forces in the x - direction = max = 0

    CD(0.433) + DB(-0.32) + DA(-0.32) = 0 (1)

    Sum of the forces in the y - direction = may = 0

    CD(-0.25) + DB(-0.383) + DA(0.383) = 0 (2)

    Sum of the forces in the z - direction = maz = 0

    CD(0.866) + DB(0.866) + DA(0.866) - 600 = 0 (3)

    solve (1),(2) and (3).
     
  16. Sep 30, 2007 #15

    dynamicsolo

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    You aren't up to calculating the actual force magnitudes yet. (You haven't dealt with the weight of the disc yet.) You are just finding the directions of the vectors AD, BD, and CD, in which case AD and BD will have components symmetrical about the x-axis (and the vectors are pointing toward D, since the x-components of AD and BD are negative).
     
  17. Sep 30, 2007 #16

    dynamicsolo

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    This is in reference to an early post: the magnitudes of the tensions in cords A and C ought to be equal by symmetry. But their direction vectors in the x-y coordinate frame won't be equal (natch)...
     
  18. Sep 30, 2007 #17
    im a little rusty at this but i got Db=452.4N and im pretty sure that isn't right, can you lead me in the right direction solving for 3 variables
     
  19. Sep 30, 2007 #18

    learningphysics

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    Oh I see. Yes, you're right jedisoccer... sorry about that. the magnitudes of DA and DC are the same by symmetry.

    The first two equations:

    CD(0.433) + DB(-0.32) + DA(-0.32) = 0 (1)

    CD(-0.25) + DB(-0.383) + DA(0.383) = 0 (2)

    will also actually give you that CD and DA are equal...

    you can simplify the problem by just using the first and third equations, and using the fact that AD = DC:

    DA(0.433) + DB(-0.32) + DA(-0.32) = 0 (1)

    DA(0.866) + DB(0.866) + DA(0.866) - 600 = 0 (3)

    two equations. two unknowns.
     
  20. Sep 30, 2007 #19

    learningphysics

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    Yeah, i don't think that's right. have a look at my last post... use the two equations with 2 unknowns.
     
  21. Sep 30, 2007 #20
    took me forever but DB is 103.97N and DA=DC=294.4N

    finally!! thanks a lot, took 5 hours but i got it
     
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