# Steel floating in mercury problem

1. Jun 12, 2009

### Liketothink

1. The problem statement, all variables and given/known data
A 17.85 cm diameter, 32.4 cm tall steel cylinder (ρsteel=7900 kg/m3) floats in mercury. The axis of the cylinder is perpendicular to the surface. What length of steel is above the surface?

2. Relevant equations
vsub/vobj=(ro)obj/(ro)liquid

3. The attempt at a solutionA
I canceled the two areas so the equation is d/h=(ro)obj/(ro)liquid and then I solved for d. H equals to 32.4 cm I think. I don't know what I'm doing wrong. Please help. Thank you.

2. Jun 12, 2009

### cepheid

Staff Emeritus
What makes you think you're doing anything wrong? What was your answer? Do you know what the answer is supposed to be?

3. Jun 12, 2009

### Liketothink

I'm using Loncapa for my class. I received .192 m for an answer an it says I'm wrong.

4. Jun 12, 2009

### cepheid

Staff Emeritus
I get a different answer. It's close, but you may have a rounding error. Check your math! Edit, also, what value did you use for the density of mercury?

5. Jun 12, 2009

### Liketothink

13534 kg/m^3

6. Jun 12, 2009

### Liketothink

I also tried .189 m but that was wrong as well.

7. Jun 12, 2009

### cepheid

Staff Emeritus
Last edited: Jun 12, 2009
8. Jun 12, 2009

### cepheid

Staff Emeritus
Pay close attention to what question you're being asked. You are being asked what length of steel is ABOVE the surface. The calculation I posted is NOT for the portion that is above the surface. If you don't know what I mean, then here's a huge hint: what do you think the sub stands for in the formula you posted, vsub/vobj?

9. Jun 12, 2009

### Liketothink

I tried that as well. The .189 m ,. But that didn't work. It's ok though thank you for your help still. The teacher must have done an error in the homework problem as he frequently does.

10. Jun 12, 2009

### cepheid

Staff Emeritus
NO! Read my latest post! You have made a conceptual mistake. But what you have to do to get from 0.189 m to the right answer is very easy. ;)

11. Jun 12, 2009

### Liketothink

Oh! Yeah that's right! It's the length submerged so I have to subtract it from the original length. Thank you so much, I appreciate it.

12. Jun 12, 2009

### cepheid

Staff Emeritus
No problem! I'm glad you got it!