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What are the tensions in the strings?

  1. Oct 19, 2006 #1
    AB is a uniform bar of length 5m and weighs 60N. It is supported in a horizontal position by two vertical strings P and Q. P is 0.5m from A and Q is 1.5m from B.

    (a) What are the tensions in the strings? (Ans= 20N and 40N)

    (b) What weight must be applied to A if the tensions in the strings are to be equal? (Ans= 15N)

    (c) What is the biggest mass which can be hung from B if the bar is to remain horizontal? (Ans= 4.1kg)

    My teacher gave us the answers to these questions. I have gotten part (a), but I truly am lost with (b) and (c). For some reason I keep getting (b) to be 0N... I would really appreciate any help and explanations. Thanks.
     
  2. jcsd
  3. Oct 19, 2006 #2

    Doc Al

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    Staff: Mentor

    Show us how you trying to solve (b) and we can see what's going on.
     
  4. Oct 19, 2006 #3
    Ok... I'm just working based on the only example my teacher gave us, which is as far from this problem as possible, but here it goes. This time I got 58N, uh.

    F*5 - (60*2.5) - (40*3.5) = 0
    5F - 150 -140 = 0
    5F = 140+150
    5F = 290
    F = 58
     
  5. Oct 19, 2006 #4
    Well, now I was just trying something and got 15N but I have no idea if this is the way to do it. I considered the tension to be 30N.

    5*F = 30 * 2.5
    5F = 75
    F = 15
     
  6. Oct 19, 2006 #5

    radou

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    Regarding (b). The first equation is the sum of moments around B. This will give you an equation with two unknowns - F and P, where P is the tension in the strings. The second equation is the equation of equilibrium of all vertical forces. This equation will give another equation with two unknowns - F and P. Now all you have to do is solve this system of 2 equations with 2 unknowns, i.e. elliminate P from the first equation to obtain F. I got the right answer, so this should work.
     
  7. Oct 19, 2006 #6

    Doc Al

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    To solve (b), start by identifying all the forces acting on the bar (including the unknown weight, F). Hint: I count four forces.

    Then you will apply the conditions for equilibrium:
    (1) Sum of the forces must equal zero.
    (2) Sum of the torques (about any point) must equal zero.
     
  8. Oct 19, 2006 #7
    Um... I tried what I thought you were telling me to do, and I got 30N.

    120 -3P + 0.5F
    60 -2P + F

    240 -6P + F
    180 -6P + 3F

    60 = 2F
    F = 30N

    What went wrong?
     
  9. Oct 19, 2006 #8

    Doc Al

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    I don't understand where this comes from. I assume P = tension; F = added weight?

    I assume you mean:
    60 -2P + F = 0

    That's good.

    Now find the torques about point B to get a second equation.
     
  10. Oct 19, 2006 #9
    Wow I got it after all... thanks! I'll try (c) now... can I still post in this thread for help on that one?
     
  11. Oct 19, 2006 #10

    Doc Al

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    Of course!
     
  12. Oct 19, 2006 #11
    Do I take a similar approach as I did with (b)? And do I treat the tension in the strings as equal forces?
     
  13. Oct 19, 2006 #12

    Doc Al

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    You will use the same general principles that I outlined in post #6.
    Absolutely not! That's part of the trick with this question--how will you know if the bar remains level? Try to picture what's going on as you load extra weight on the bar.
     
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