Steel floating in mercury problem

  • #1

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.
 

Answers and Replies

  • #2
cepheid
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What makes you think you're doing anything wrong? What was your answer? Do you know what the answer is supposed to be?
 
  • #3
I'm using Loncapa for my class. I received .192 m for an answer an it says I'm wrong.
 
  • #4
cepheid
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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?
 
  • #6
I also tried .189 m but that was wrong as well.
 
  • #8
cepheid
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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
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
cepheid
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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
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
cepheid
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No problem! I'm glad you got it!
 

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