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Static Equilibrium with buoyant force. High difficulty.

  1. Jan 7, 2013 #1
    1. The problem statement, all variables and given/known data
    I attached the original problem and a diagram I made with the variables on it. Basically we have a rod hanging from a wire at an angle theta with tension and the other end floating in water. So there is T=tension, weight=w=mg and buoyant force=F_b. I introduced symbols for total length L and submerged length x along with the height the rod hangs from h. The rod has a density half that of water or 500 kg/m^3.

    2. Relevant equations
    Ʃτ=0, about top of rod Ʃτ= (L/2)mgcosθ-(L-x/2)Fbcosθ=0
    ƩFx=0,
    ƩFy=0=T+Fb-mg

    3. The attempt at a solution

    The real difficulty I am having is that it's unclear what my answer should look like. I think it should depend upon the height and length of the rod. I really don't know what I'm expected to treat as known and unknown, and I only have two equations so I can only solve for two things correct? It asks for the fraction submerged which is x/L, but if those are the unknowns then the forces would have to be known. I keep going in circles on this one and it's driving me mad. Any guidance would be greatly appreciated.
     

    Attached Files:

    Last edited: Jan 7, 2013
  2. jcsd
  3. Jan 7, 2013 #2

    haruspex

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    You should be able to write down the value of Fb in terms of m, g, L and x.
    Just about everything except x and L should then cancel out in your torque equation.
     
  4. Jan 7, 2013 #3
    Thank you! I was getting worried that nobody would have any advice for me. I did come up for an expression for Fb in those terms, it was (2mgx)/L. (because twice the weight of the fraction of the rod submerged). but then the problem becomes that I have solved for x/L....in terms of x and L? Plugging and chugging that expression got me x/L= L/(2x)-1


    I guess my questions still remain about the legitimacy of this. I never used the Fy equation and if L and x were known quantities then why couldn't I have just put fraction=x/L and QED!?
     
  5. Jan 7, 2013 #4

    haruspex

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    That's just a quadratic in x/L. Solve it!
    It's all legitimate and correct. You didn't need the other equations because you don't care about the tension in the string. You could have taken torque about some other point, which would have given you an equation involving that tension. Then you would have needed the Fy equation to eliminate it again. But why do that when you can finesse it by a smart choice of centre of moments?
     
  6. Jan 7, 2013 #5
    Well I'll be. I got everything on one side and multiplied it by 1/L to get it looking right and the solution came to be 1/2. I really expected that the answer would depend on the height it was hung from, I'm really suprised that there is a real solid solution for this problem, can you understand why that confuses me? I figure if it was hanging high enough the fraction would be necessarily smaller, not the same value no matter the length or height.

    Well thank you for sticking with me anyways. Cheers:smile:
     
  7. Jan 8, 2013 #6
    Actually I DO get it. because I think it needs to be balanced that way for the wire to be perfectly vertical as the problem requires! that's awesome.
     
  8. Jan 8, 2013 #7

    haruspex

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    Hmmm... that's not a solution of the equation you posted before: x/L= L/(2x)-1.
    That equation gives x/L = (√3 - 1)/2. Did your equation change?
     
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