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Finding tension

  1. Oct 21, 2008 #1
    1. The problem statement, all variables and given/known data
    http://i241.photobucket.com/albums/ff4/alg5045/image546.jpg

    Find the tension in each cord in the figure if the weight of the suspended object is 250N.

    Part A) Find the tension in the cord A for system (a)
    Part B) Find the tension in the cord B for system (a)
    Part C) Find the tension in the cord C for system (a)
    Part D) Find the tension in the cord A for system (b)
    Part E) Find the tension in the cord B for system (b)
    Part F) Find the tension in the cord B for system (b)


    2. Relevant equations

    F=ma

    3. The attempt at a solution

    I found the tension in both Cs by drawing a free body diagram and setting F=mg = 250N for both.

    I've also found the tension in cord A for system b. Since there are two unknowns, I solved the horizontal equation Ax=Bx for B, and then I substituted B into the vertical equation. So the end equation would be A(sin210sin45-cos210sin45) = 250cos45. I got A = 685N.

    Then I tried substituting A into the Asin(thetaA) = Bcos(thetaB). I got B = 484N, which was wrong.

    I've also tried using the same principle to find the tensions in A and B for system a, but I can't get it.
     
  2. jcsd
  3. Oct 21, 2008 #2

    Q_Goest

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    Are you familiar with how to sum forces in the various directions? Let's make the Y direction verticle and the X direction horizontal. Do you know how to set up the equations? Note that the direction of the force each string acts in is along the axis of the string.
     
  4. Oct 21, 2008 #3
    For system a, the horizontal equation is Ax = Bx or Acos(thetaA) = Bcos(thetaB), and the vertical equation is Ay + By = C or Asin(thetaA) + Bsin(thetab) = C. For system b, the horizontal equation is again Ax = Bx or Asin(thetaA) = Bcos(thetaB) because A is below the horizon in this case. The vertical equation By = Ay + C or Bsin(thetaB) = Acos(thetaA) + C.
     
  5. Oct 21, 2008 #4

    Q_Goest

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    Good! You now have 2 equations with 2 unknowns.
    Acos(thetaA) = Bcos(thetaB)
    or
    Acos(thetaA) - Bcos(thetaB) = 0

    and

    Asin(thetaA) + Bsin(thetab) = C

    The two unknows are A and B. Note that you have all the angles and you also have C which you indicated in the OP.

    Now do you know how to solve the two equations for the 2 unknowns?

    The second problem is basically the same.
     
  6. Oct 21, 2008 #5
    Well I already found A for system b = 685N, which was correct. However, when I plugged 685 into the horizontal equation, I got B = 484N, which was incorrect. I've also tried solving for A for system a. For that I got 685N again, but it wasn't right. I know how to solve them. We've done about a million problems in class. I just can't get the right answer.
     
  7. Oct 21, 2008 #6

    Q_Goest

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    Ok, had a look but I don't get 685 for A on problem a. I got 183 N. Try again and just write out each step. The equations you gave are correct for the first problem:
    Acos(thetaA) - Bcos(thetaB) = 0
    Asin(thetaA) + Bsin(thetaB) = C
     
  8. Oct 21, 2008 #7
    I meant 685 for A for system b. My mistake.
     
  9. Oct 21, 2008 #8

    Q_Goest

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    For b, try writing out the two equations as was done for a. The first should be:
    A sin(60) = B cos(45)
    or
    A sin(60) - B cos(45) = 0

    Can you get that?
    What's the second equation?
     
  10. Oct 21, 2008 #9
    The second equation would be B sin(45) = A cos(210) +250
     
  11. Oct 21, 2008 #10

    Q_Goest

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    Close... the force produced by B is vertically up and the force created by A and the 250N load is down as you show, so you have it set up right. But... why A cos(210)? How about A cos(60)?

    Think of it this way, cos(60) = Ay / A
    (cos of an angle = adjacent/hypotaneus)

    so

    Ay (ie: verticle force created by string A) = A cos(60)

    Now take the two equations and see if you can solve. Start with:

    A sin(60) - B cos(45) = 0
    and
    - A cos(60) + B sin(45) = 250
     
  12. Oct 21, 2008 #11
    Ok. I got everything for system b, and I thank you for your help with that. Now I need help with system a. Would the equation be to solve for A for system a:

    A[sin(150)cos(45) + cos(150)sin(45)] = 250cos(45)
     
  13. Oct 21, 2008 #12

    Q_Goest

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    Where do the 150 degree angles come from?
     
  14. Oct 21, 2008 #13
    I figured it out. I don't know why I keep using the wrong angles. That's what was screwing me up I think. Thanks for your help.
     
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