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A rubber ball is plugging a whole in bottom of tank. When will it fail

  1. Mar 17, 2013 #1
    1. The problem statement, all variables and given/known data
    A 3 cm radius rubber ball weighing 5 g is used to plug a 4cm hole in the base of a tank. The tank is in use and is gradually emptying. At approximately what depth will the plug fail by floating up and out of the hole?


    2. Relevant equations
    Density



    3. The attempt at a solution
    Ok so I don't have a numbers and all attempt at a solution :(
    But I do have a logical thought process as to how the solution should be solved, but I'm stumped on finding the volume of a partial sphere when given radius and the radius of the base circle - I know it is enough information I just cannot work out how to calculate it.

    The way I see it the upward force acting on the sphere will be proportional to the mass of the water displaced, whilst the downward force acting on the sphere will be proportional to the whole sphere so theoretically the plug should fail when the water level is at the height corresponding to the radius of the base circle on the opposite side. I am not sure about this though, as according to the figures given the density of the sphere is only 44 kg/m3 so all my common sense is telling me the plug wouldn't even work :confused:

    Am I taking the right approach?
     
  2. jcsd
  3. Mar 17, 2013 #2

    Andrew Mason

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    Homework Helper

    If the ball was just submerged, you would be right because the water at any depth would provide almost equal pressure all around (slightly more on the bottom than top - the difference being the buoyant force/cross-secitonal area of the ball), so there would always be a net buoyant force. But here the ball is not subject to water pressure all around because there is only atmospheric pressure on the part that is covering the hole. You have to find the downward water force on the ball (the downward water pressure over the 4 cm diameter hole x area) as a function of depth and then compare that to the upward buoyant force.

    AM
     
  4. Mar 17, 2013 #3
    Thanks very much for the information AM :) I really appreciate it.
    I'm just about to go to sleep after I watch the West Ham game, but I'll be working on this first thing tomorrow so I will report back my calculations - I am stoked that I don't need to do that partial sphere calculation as my head may have exploded, and how would I ever complete my assignment then!

    But seriously thanks, a nudge in the right direction makes the world of difference.

    DylanW
     
  5. Mar 25, 2013 #4
    Okay so for this one I've done F(UP)=1000.9.8.0.000113 = 1.1N
    F(DOWN)=pghA + mg = 1000.9.8.h.(0.02^2).Pi+0.005.9.8 = 12.315h + 0.049 N
    When F(Down)=F(Up) plug will fail
    h = (1.1-0.049)/12.315 = 8.5 cm.

    Feedback please?
     
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