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Buoyancy Equilibrium on a balance

  1. Nov 21, 2004 #1
    A beaker filled with water is balanced on the left pan of a balance. A cube of 4 cm on an edge is attached to a string and lowered into the water so that it is completely submerged. The cube is not touching the bottom of the beaker. A weight of mass m is added to the right pan to restore equilibrium. What is m ?

    Well i know that the volume of the cube is 64 cm^3 or 6.4x10^-5 m^3. And the buoyancy force would be the density of the water times the volume of the cube times g. But thats all i've gotten pretty much. I've tried using summation of torques about the pivot of the balance, but it just gets me more and more lost. All I need is a step in the right direction. Would i take into consideration the beaker's mass or density? It doesnt really specify it, but I would think it is made of glass and so the density of glass is 2.6 x 10^3 kg/m^3. The density of water is 1000kg/m^3 and i dont know if it is needed either but the density of air is 1.293 kg/m^3. Thanks.
     
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  3. Nov 22, 2004 #2

    Galileo

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    Since the cube is completely submerged and not touching the bottom, there is no net force acting on it. So the buoyancy cancels out the gravitational force on the cube.
     
  4. Nov 22, 2004 #3

    Doc Al

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    True, the cube has no net force acting on it. But there are three forces acting on it: its weight, the string tension, and the buoyant force.

    When the cube is lowered into the water, the left pan of the balance will experience an added force equal to the buoyant force on the cube. Choose the mass m accordingly.
     
  5. Nov 22, 2004 #4
    so are you saying the force mg on the right pan is equal to the buyoyant force on the cube in the beaker on the left pan? If this were true, then the density of the water times the volume of the cube would equal that mass m; which would turn out to be 0.064 kg. Is this correct? Or is there something I am missing?
     
  6. Nov 22, 2004 #5

    Doc Al

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    Exactly correct.
     
  7. Nov 22, 2004 #6
    ok i see. thank you, Doc Al, for helping me out. I spent hours on that problem and didnt realize that it would be like that. I know the problem is done and all, but how do you know that the buoyancy force is equal to the weight of the mass on the right pan? I mean to say, what principles or laws can be used to show this?
     
  8. Nov 22, 2004 #7

    Doc Al

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    Excellent question. I'm glad to see you thinking.

    There are several ways to understand what's going on.

    Analyze the forces on the cube. Apply the equilibrium condition to figure out what the tension in the string must be.

    Then analyze the forces on the beaker plus contents as a single system. What are all the forces acting on that system?
     
  9. Nov 23, 2004 #8
    well the summation of forces on the cube shows the tension force is equal to the weight of the cube minus the Buoyant force. The new forces in the beaker system are the Normal force by the balance pan and the weight of the beaker. But isnt the weight of the beaker dependent on the weight of the water inside it as well as the apparent weight of the cube?
     
  10. Nov 23, 2004 #9

    Doc Al

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    Right.
    Don't forget the string pulling up.
    The weight of the beaker system just depends on the mass of its contents: beaker, water, and cube.
     
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