Fluids Question

mirandasatterley

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)

Doc Al

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.

mirandasatterley

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?

Doc Al

Yes, that will give you the outer radius.

What idea?

Hint for part b: The density of aluminum will come in handy.

mirandasatterley

Does this mean the radius from the surface to the center (including the cavity)?

mirandasatterley

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.

Doc Al

Yes. (Radius is always from the center of the sphere to some point.)

Doc Al

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.

mirandasatterley

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?

mirandasatterley

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?

Doc Al

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

mirandasatterley

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.

Doc Al

All good.

You can also just use your first equation and get V(object) directly.

mirandasatterley

Thank you so much for all your help