Attached Image is Problem. Having issues with buoyancy.

Question

anonymous · Answer

So what I got so far, hopefully someone can point out my flaw is as follows.
$$m = \rho_w*V_w + \rho_o(V_c-V_w)$$
m = mass being displaced which since it's in equilibrium should be the same. V sub w is amount of cylinder submerged, and V sub c is volume of volume of total cylinder. I got this part because the amount in the oil should be the amount of the whole cylinder minus the amount in the water. 

I keep getting 0.17955m for the answer but it's telling me I'm wrong.

Thanks in advance.

irishboy123 · Answer

http://www.wolframalpha.com/input/?i=%28753.202+-+1000%29%2F%28747-1000%29*.18

anonymous · Answer

Oh no, thinking I've made simple Algebra errors? Well this is embarrassing. In a few minutes I'll work through it again.

anonymous · Answer

Hmm, still not getting it. I am not coming up with the same numbers as you either.

bbypanda16 · Answer

Hey I can't really see the question, if you could type it out or something than I might be able to answer.

irishboy123 · Answer

well, method wise, if the height of the cylinder in the oil is h, as its total height is r we can say: $\large \rho_{o} \pi r^2 h + \rho_{w} \pi r^2 (r-h) = m_c$ $\large h = \dfrac{m_c - \rho_{w} \pi r^3}{\pi r^2(\rho_{o} - \rho_{w} )}$ slightly different way of calculating it, but same answer: http://www.wolframalpha.com/input/?i=%2813.8-%281000%29%28pi%29%280.18%5E3%29%29%2F%28pi%280.18%5E2%29%28747-1000%29%29

vincent-lyon.fr · Answer

I find the same answer : 0.1756 m

$\large h=h_{cyl}\dfrac{\rho _{water}-\rho _{cyl}}{\rho _{water}-\rho _{oil}}$

with

$\large \rho _{cyl}= \dfrac{m_{cyl}}{\pi r^2 h_{cyl}}$

anonymous · Answer

Ah yes thank you! For some reason your response made it click and I realized... I never converted back from volume of cylinder to the height so I was doing something that didn't even make sense in the end. But I got the same answer when I did that final step and we all agree!

Thanks everybody.