Let ρAl represent the density of aluminum and ρFe that of iron. Find the radius of a solid aluminum sphere that balances a solid iron sphere of radius rFe on an equal-arm balance. (Use any variable or symbol stated above as necessary.)

Question

anonymous · Answer

does anyone know how to get this?

aravindg · Answer

:)

saifoo.khan · Answer

it's a bit tricky.

saifoo.khan · Answer

deals with the formula:

Density = Mass/Volume

anonymous · Answer

im so confused with it..decided to come here and get some help lol

saifoo.khan · Answer

LOL.
sorry i have no idea.

anonymous · Answer

i'm just playing with this in my head at the moment, and i could be completely wrong but i'm thinking; since density = mass/volume

the weight for both spheres has to be the same (to balance) so the mass will be equal. the density is just a ratio of mass to volume so if you divided pFe by pAl you might be able to find the ratio needed in volume. so my working came out to be something like...

$$rAl = (pFe/pAl) * rFe$$

someone care to explain/correct?

anonymous · Answer

and then obviously reverse the sphere volume equation to find radius from a given volume

anonymous · Answer

im not sure..thats all that is given to me to figure out

anonymous · Answer

ok i thought it over a bit more and i made a mistake with my previous calc i think.

basically for the weight to balance both sides need to be equal weight, so massFe will equal massAl. so in the $$pFe = mFe / vFe$$equation mFe = mAl so you can sub in$$pFe = (pAl * vAl) / vFe$$
then rearrange to solve for vAl and then use the volume of aluminium in the sphere volume equation $$vAl = 4/3 * r^3$$ to solve for the radius of the aluminium sphere.

It's a bit tricky with no numbers (I had to put my own numbers in to check) but I THINK this works. Maybe an actual physics student can correct me.

anonymous · Answer

so you think the answer is 4/3* r^3?

anonymous · Answer

There may be an easier way to do it because the way I suggested is a little convoluted, but if I had to put an answer down I'd put $$vAl = pAl*vFe/pAl$$ and then $$3(vAl)/4*\pi=rAl^3$$

anonymous · Answer

unfortunately that isnt the answer..and the answer has still not been given to me