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Fluids Question

  1. Nov 15, 2006 #1
    My Question is " A spherical aluminum ball of mass 1.26kg contains an empty spherical cavity that is concentric with the ball. The ball just barely floats in water. Calculate (a) the outer radius of the ball and (b) the radius of the cavity."
    I'm notsure if it's right, but so far I have:
    the buoyancy force = the gravitational force.
    B = (density(fluid))(g)(V(object))
    Fg = m(object)g = (density(object))(V(object))(g)
    Setting them equal,
    d(fluid)gV(object) = d(object)V(object)g
    I was then thinking that V(object) = 4/3(pi)(r^3), sothat i can solve for the radius of the object.

    I'm unsure of how to solve for the radius because the V of water is unknown, and I don't know any other equations that i can use to solve it. I'm also unsure of how to incorperate the radius of the cavity in part b. Would it just be r(found in part a)-r(cavity)
     
  2. jcsd
  3. Nov 15, 2006 #2

    Doc Al

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    Staff: Mentor

    OK.
    No need to go beyond the second expression, since you are given the mass of the object. Set those equal to solve for V. Then use your volume formula to solve for the radius.
     
  4. Nov 15, 2006 #3
    So,

    d(fluid)gV(object) = m(object)g
    d(fluid) (4/3(pi)(r^3) = m(object)

    Is my idea for part b right at all?
     
  5. Nov 15, 2006 #4

    Doc Al

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    Yes, that will give you the outer radius.

    What idea?

    Hint for part b: The density of aluminum will come in handy.
     
  6. Nov 15, 2006 #5
    Does this mean the radius from the surface to the center (including the cavity)?
     
  7. Nov 15, 2006 #6
    This is what confused me before.
    I have the equation;
    d(fluid)(4/3(pi)(r^3))g = d(object)(4/3(pi)(r^3))g
    d(fluid)(4/3(pi)(r^3)) = d(object)(4/3(pi)(r^3))

    And when I am this far, the term;
    (4/3(pi)(r^3), will cancel.
     
  8. Nov 15, 2006 #7

    Doc Al

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    Yes. (Radius is always from the center of the sphere to some point.)
     
  9. Nov 15, 2006 #8

    Doc Al

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    You are assuming here that the aluminum is in the shape of a solid ball, but it's really a shell--a hollow ball. (Thus, the radius in each expression would be different, so they would not cancel.) But don't go there.

    When you go from Fg = m(object)g to d(object)V(object)g, realize that V(object) is just the volume of the material, not really the volume of the object. You should use m(ball) = d(aluminum)V(aluminum); V(aluminum) is the volume of the aluminum (the shell), not the entire ball.
     
  10. Nov 15, 2006 #9
    Okay, that makes more sense,

    d(fluid)gV(object) = d(aluminum)gV(aluminum)
    d(fluid)(4/3(pi)(r^3)) = d(aluminum)(m(aluminum)/d(alumimun))
    d(fluid)(4/3(pi)(r^3)) = (m(aluminum))

    So, in this case, when I solve for r, this will give me the radius of only the aluminum and I subtract that from the total radius i have already found in part a to find the radius of the cavity?
     
  11. Nov 15, 2006 #10
    Wait, thats not right, It gives me the same r, I'm missing something does the equation V=4/3 (pi)(r^3) have to be modified to represent only the aluminum as well?
     
  12. Nov 15, 2006 #11

    Doc Al

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    You're going in circles. That last equation is what you should have started with:
    d(fluid)gV(object) = m(object)
    d(fluid)g(4/3(pi)(r^3)) = m(object)​
    The radius here is the radius of the object, the sphere of aluminum. (The volume of the object equals the volume of displaced fluid.) m(object) is given; use it to solve for part (a).


    No. Do this:
    (1) Solve for the volume of the object
    (2) Solve for the volume of the aluminum (as discussed earlier)
    (3) Subtract one from the other to find the volume of the space
    (4) Since the space is spherical, find its radius
     
  13. Nov 15, 2006 #12
    I think I have it this time;
    d(fluid)gV(object) = m(aluminum)g
    d(fluid)(4/3(pi)(r^3)) = m(aluminum)
    I solve this for r, and that is the answer for part a.

    Next,
    V(object) = 4/3(pi)(r^3)
    i sub in the radius i found in part a to find V(object)

    Next,
    V(aluminum) = m(aluminum)/ d(aluminum)
    and solve for v aluminum

    I then subtract,
    V(cavity) = V(object) - V(aluminum)

    Finally,
    4/3(pi)(r^3) = V(cavity),
    and solve for r, this is the answer for b.
     
  14. Nov 15, 2006 #13

    Doc Al

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    All good.

    You can also just use your first equation and get V(object) directly.
     
  15. Nov 15, 2006 #14
    Thank you so much for all your help
     
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