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Hollow sphere half submerged in a liquid, find density

  1. Aug 5, 2016 #1
    1. The problem statement, all variables and given/known data
    A hollow sphere of inner radius 9.0 cm and outer radius 10.0 cm floats half submerged in a liquid of specific gravity 0.80. (a) Calculate the density of the material of which the sphere is made.
    (b) What would be the density of a liquid in which the hollow sphere would just float completely submerged?

    Density of liquid = 800 kg/m^3
    Outer radius = 0.10 m ; inner radius = 0.09 m

    2. Relevant equations



    3. The attempt at a solution
    Because it's half submerged so the volume would be:

    V = (0.5) (4 pi / 3) * (outer radius^3 ) = 2.09 x 10^-3 m^3

    The force on an object is equal to the weight of displaced fluid. If the sphere is half-submerged, we can find the displaced volume of fluid and then its weight.

    so ρgV = mg which implies that ρV = m therefore

    mass of the sphere = (800 kg/m^3)(2.09 x 10^-3 m^3)= 1.68 kg

    Then V of the sphere, actually the volume of the mass in the sphere so now we take
    (4 pi / 3) * (outer radius^3- inner radius^3 ) = (4 pi / 3) * (0.1^3 )-(0.09)^3) = 1.135 x 10^-3 m^3

    ρ= 1.68 kg / 1.135 x 10^-3 m^3 = 1479 kg/m^3 ?

    This is what I worked out but I'm a little unsure regarding the volumes I've taken?

    For part b, I'd take the volume with just the outer radius then I'd use
    ρgV=mg so ρV=m where V is the volume I just calculated and m is the mass of the sphere from the previous part and solve for ρ?
     
  2. jcsd
  3. Aug 5, 2016 #2

    TSny

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    Looks good to me.

    Yes.
     
  4. Aug 5, 2016 #3
    Great thank you! Can I ask another question?
    A tank containing water is placed on a spring scale, which registers a total weight W. A stone of weight W is hung from a string and lowered into the water without touching the sides or bottom of the tank. What will be the reading on the spring scale in terms of W and the buoyancy B?(Assume the buoyancy on the stone is B<W).

    I didn't get what he meant by placed "on" a spring scale? Isn't a spring scale just a newton meter so it's supposed to be hung from it? Secondly, do we consider tension of the string here? Doesn't say if it's massless or not? B= W of beaker and water + W of stone or will it be T + B = W+W and T= 2W-B which would be the reading on the spring scale? It doesn't work if they actually mean a kitchen scale :/
     
  5. Aug 5, 2016 #4

    Doc Al

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    I didn't check your arithmetic, but your methods are correct.

    Sure, but you should not have to do any real calculation. (By what factor does V increase?)

    Edit: I see TSny beat me to it while I was getting coffee ;)
     
  6. Aug 5, 2016 #5
    It increases by twice the amount! There's another question in post 3 if you're willing to try? :smile:
     
  7. Aug 5, 2016 #6

    Doc Al

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    Right! So how must the density change?
     
  8. Aug 5, 2016 #7
    Half the amount since they're inversely proportional since the mass is constant
     
  9. Aug 5, 2016 #8

    Doc Al

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    Good.
     
  10. Aug 5, 2016 #9

    Doc Al

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    There are several types of "spring scales". Just think of it as a kitchen scale.

    I would.

    Analyze it similarly to the "block hanging from a cord" problem. Do a force analysis of the stone. And then a force analysis of the "system" (Beaker + Water + stone).
     
  11. Aug 5, 2016 #10
    There's going to be Tension, the buoyant force and the weight on the stone so T + B = W
    and on the beaker, water and stone there will be tension, weight, buoyant force, and the forceby the spring scale on the beaker so T+B+N=W+W?
     
  12. Aug 5, 2016 #11

    Doc Al

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    Good.

    Not exactly. Realize that the buoyant force is the water pushing up on the stone, which will be balanced by the force of the stone pushing down on the water. (That's Newton's 3rd law.)

    Just include external forces on the system in your analysis.
     
  13. Aug 5, 2016 #12
    Ohh right, so it will just be T+N = W+W. We're looking to solve for N
    Then if T+B=W so T=W-B then
    (W-B) +N = 2W so
    N = W + B?
     
  14. Aug 5, 2016 #13

    Doc Al

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    Looks good!

    Now that should make some intuitive sense (after awhile). Before the stone was lowered into the water, the scale just read W. And when you lower the stone, the stone pushes down onto the water (and thus the scale) with a force equal to the buoyant force, thus increasing the scale reading by that much.

    But it's better to systematically analyze things, as you did, rather than rely on intuition. (Until you've done hundreds of problems.)
     
  15. Aug 5, 2016 #14
    Yes that makes sense.
    Thank you for your help! :)
     
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