# Steel floating in mercury problem

## Homework Statement

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?

## Homework Equations

vsub/vobj=(ro)obj/(ro)liquid

## The Attempt at a Solution

A
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.

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

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

cepheid
Staff Emeritus
Gold Member
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?

13534 kg/m^3

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

cepheid
Staff Emeritus
Gold Member
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?

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.

cepheid
Staff Emeritus