Stupid Buoyancy, this is getting ridiculous.

[Saladsamurai]

[SOLVED] Stupid Buoyancy, this is getting ridiculous.

A hollow sphere of inner readius 8cm and outer radius 9cm floats half submerged in a liquid of density 800kg/m^3. What is the mass of the sphere and the density of the material it is made from.

$$\sum F=0$$
$$\Rightarrow mg-F_b=0$$
$$\Rightarrow m=\rho*\frac{V_s}{2}$$

Now for the Volume do I use $$\frac{4}{3}\pi r^2$$
or

$$\frac{4}{3}\pi (r_2^2-r_1^2)$$ ?

I thought it would just be the outer radius that was important here. But I have tried BOTH ways and both come out wrong...so I am messing something else here...are there only 2 Forces here?

The answer for mass should be 1.2 kg

ahhhhhhhhhhhhhh!

Casey

 

[Kurdt]

The sphere displaces liquid equal to half its volume. You want to take the outer radius to work this out. Now the weight of this displaced liquid is equal to the magnitude of the upthrust supporting the sphere. This means that the weight of the sphere is equal to that of the displaced water. Therefore all you have to do is find the volume of the sphere material and you should be able to find the density. Remember the whole sphere isn't made of that material just a shell.

 

[Saladsamurai]

The sphere displaces liquid equal to half its volume. You want to take the outer radius to work this out. Now the weight of this displaced liquid is equal to the magnitude of the upthrust supporting the sphere. This means that the weight of the sphere is equal to that of the displaced water. Therefore all you have to do is find the volume of the sphere material and you should be able to find the density. Remember the whole sphere isn't made of that material just a shell.

So shouldn't it just be $$m=\rho_l*V=800*\frac{1}{2}*\frac{4\pi}{3}*(.09)^2$$

 

[Kurdt]

To elaborate more to find the mass of the displaced liquid the volume you want to be using is,

$$V=\frac{4}{3} \pi r_1^3$$

then the volume of the sphere material will be given by,

$$V=\frac{4}{3} \pi (r_1^3-r_2^3)$$

where $r_1=9cm$ and $r_2=8cm$.

 

[Kurdt]

So shouldn't it just be $$m=\rho_l*V=800*\frac{1}{2}*\frac{4\pi}{3}*(.09)^2$$

Yes that's correct for the mass of the sphere. If you plug the numbers in you should get the 1.2Kg you are looking for.

 

[Saladsamurai]

Yes that's correct for the mass of the sphere. If you plug the numbers in you should get the 1.2Kg you are looking for.

But I don't. I get 13.57.

Casey

 

[Kurdt]

Oh right just noticed that you're squaring the radius instead of cubing. Amazing how you see what you think you're going to see. Silly me.

 

[Saladsamurai]

Oh right just noticed that you're squaring the radius instead of cubing. Amazing how you see what you think you're going to see. Silly me.

Very true! I think that every mistake that I have made for the last two semesters has been of this nature! What a jerk! I am going to make my avatar a dunce cap!

Thanks Kurdt!
Casey

 

[Kurdt]

Hey no problem. Everyone on this site has made a similar mistake, I guarantee it.