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Tension in negative value?

  1. Dec 24, 2015 #1
    1. The problem statement, all variables and given/known data
    Determine the tension cable A and B and the x , y , z reaction of the ball and socket joint at C .

    2. Relevant equations


    3. The attempt at a solution
    I let the reaction of ball and socket joint at C = Fx , Fy and Fz
    Sum of force in z direction :
    TA +TB +FZ= 400

    Sum of force in x direction :
    Fx = 0

    Sum of force in y direction :
    Fy = 0

    Sum of moment about X :
    3TB -1.3(400) = 0
    3TB+1.6FZ= 520
    Sum of moment about Y :
    -2TA-2TB = -1.5(400)

    2TA+2TB= -600

    Sum of moment about z :
    -1.6Fx = 0 , Fx = 0

    Since TA +TB= -300 , so i sub this eq into TA +TB +FZ= 400 ,
    i gt -300+Fz = 400 , Fz = 700N

    3TB +1.6(700) =520 , TB =-200N

    TA -200+700=400 , TA = -100N

    since the question mention the A, B is in tension , but i gt my TB and TA in negative value , i am wondering my ans is correct or not


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  3. Dec 24, 2015 #2

    SteamKing

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    No, it's not correct.

    From the sum of the moments about the y-axis, you get TA + TB = 300 N, which seems reasonable.

    Where you messed up was substituting back into the force equation: TA + TB + FZ = 600

    If TA + TB = 300, how can FZ = 700 N?
     
  4. Dec 24, 2015 #3
    so , after changing TB +TA = 300 , i have TA = 480 N , TB=120N and FZ= 100N , are they correct now ?
     
  5. Dec 24, 2015 #4

    SteamKing

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    Why did you do this? It doesn't make any sense.

    If TA = 480 and TB = 120, how can TA + TB = 300?

    You're not even doing mathematics anymore. You're guessing or something, which I haven't figured out (and don't want to.)

    Look, if
    TA + TB + FZ = 600

    and

    TA + TB = 300,

    then

    300 + FZ = 600

    What's FZ? This is a simple equation which you should be able to solve.
     
  6. Dec 25, 2015 #5
    so , i have my TA = 180N , TB=120n , FZ= 100N , are they correct now ?
     
  7. Dec 25, 2015 #6

    SteamKing

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    These would normally be OK, but you seem to have overlooked that couple of magnitude 200 N-m which wants to rotate the plate around the z-axis.

    The presence of this couple complicates things, and changes the tensions in cables A and B.
     
  8. Dec 25, 2015 #7
    ok , the @00Nm is the moment about z axis , a i right ? So , my -1.6FX +200 = 0 , FX = 125N , the other reading is not affected , right ?
     
  9. Dec 25, 2015 #8

    SteamKing

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    It's not that simple, I'm afraid.

    After calculating FX above, did you sum all of the forces acting in the x-direction? Was that sum equal to zero? If it was not, then the plate ABDE will rotate about the z-axis, and the cables A and B will no longer remain vertical; there will be some x and y components to the tensions in those cables which necessarily develop to keep the plate in static equilibrium.
     
  10. Dec 25, 2015 #9
    can you find the other force in X direction to counter the FX , i couldnt find any
     
  11. Dec 25, 2015 #10

    SteamKing

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    That's why I wrote what I wrote in Post #8.

    The plate can rotate only about the ball and socket joint at C, but it cannot translate in any of the three coordinate directions x, y, or z. If the plate rotates about the joint at C, then cables A and B are no longer vertical, and since each cable has a tension in it, there will be x and y components of the tension created in each cable where it attaches to the plate. The sum of the x and y force components of the reaction at C and cables A and B must both equal zero for the plate to remain in static equilibrium. That's where the other forces come from to counter FX and FY.
     
  12. Dec 25, 2015 #11
    that seems very complicated
     
  13. Dec 25, 2015 #12

    SteamKing

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    It doesn't have to be, not if you write the correct force and moment equations for the plate taking these facts into consideration.
     
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