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Buoyancy and Spring force problem.

  1. Feb 21, 2012 #1
    1. The problem statement, all variables and given/known data

    A spring with spring constant 35N/m is attached to the ceiling, and a radius=0.025m, 1.0kg metal cylinder is attached to the lower end. The cylinder is held so that the spring is neither stretched nor compressed, then a tank of water is placed under-neath with the surface of the water just touching the bottom of the cylinder. When released, the cylinder will oscillate a few times but, damped by water, reach an equilibrium position. When in equilibrium, what length of the cylinder is submerged?

    2. Relevant equations

    Fspring=-kx
    Vcylinder=∏r^2h
    Fb=rho*Vfluid*g

    THe answer is given, (18.1cm) but I'm kinda confused as to how the solution came about
    3. The attempt at a solution
    Fnet=Fb-(Fg+Fspring)

    rho*Vfluid*g =mg+ kx ?

    Help/hints much appreciated!
     
    Last edited: Feb 22, 2012
  2. jcsd
  3. Feb 21, 2012 #2

    ehild

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    Gold Member

    Make a drawing, and find the correct direction of forces. How is the submerged length of the cylinder related to the change of length of the spring? What is V(fluid) in terms of the submerged length and radius of the cylinder? Note that you wrote the volume of a sphere as relevant equation instead of the volume of a cylinder.

    ehild
     
  4. Feb 22, 2012 #3
    Woops my bad, okay, I changed it to volume of a cylinder.

    So in equilibrium net force is 0, and the forces in the system are Fb, Fspring, Fg
    After drawing,
    Fnet=(Fb+Fspring)-Fg
    Fb+Fspring=Fg


    Fb=rho of water* Vfluid displaced, which would be broken down into ∏r^2 *h thats height of the object, or cylinder? *g

    so Fb=rho*Vf (or ∏r^2h)*g
    Fg=just the cylinder's weight = 1kg(9.8)
    Fspring=(35N/m)(x)


    generally h submerged is= h of object- h object above water level

    So when you mentioned, that the submerged length of the cylinder is related to the change of length of the spring, do you mean the submerged length of the cylinder would factor into
    the equation Area*(Hobj-change of length of spring) above for Fb?
     
  5. Feb 22, 2012 #4

    ehild

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    I do not understand. Could you draw a picture? Or look at my one and say how is the submerged length (y) related to the change of the length of the spring(ΔL). How much water is displaced in terms of y?

    ehild
     

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    Last edited: Feb 22, 2012
  6. Feb 24, 2012 #5
    You have to make a substitution. Usually the questions that say "and then" somewhere require a comparison of before and after. So first find the length of the spring at equilibrium Fg=-kx. 1.0*9.8=-35x. x=.28m. Then note that the tub of water will add a bouyant force Fb, such that the spring will relax a little, and compress. So delta L= .28-x. You don't know the height of the cylinder, but the height of the 'cylinder of water' that is displaced is also going to equal .28-x. So you should end up with:

    Fb+kx= Fg
    rho fluid * Volume fluid displaced * g + 35 * x = mass obj. * g
    1000 * (.025^2 * pi * (.28-x))(9.8) + 35 * (.28-x) = 1.0 * 9.8
    19.24225(.28-x) + 35(.28-x)= 9.8
    (5.3878 - 19.24225x) + (9.8 - 35x) = 9.8
    5.3878 + 9.8 -9.8 = 54.24225x
    x=5.3878/ 54.24225
    x= 0.099328

    delta L = .28- 0.09932
    = .18067m
    = 18.067 cm , 18.1 cm.
     
  7. Feb 24, 2012 #6
    oh, :eek: I was using just 0.28 before, I see now.
    Thanks! :) (especially ehild for drawing the diagram)
     
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