Finding density of an object

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Homework Statement


A piece of driftwood floating in the ocean has 35 percent of its volume above water. What is the density of the driftwood if the ocean has a density of 1,025 kg/m^3.


Homework Equations


ρvol(driftwood)=ρvol(ocean)


The Attempt at a Solution



Well based on a problem we did in class I attacked it this way.

Since 35% is above, I would assume 65% is below, giving me ρ.35=1025(.65).

Then ρ.35=666.25 so ρ=1903.57 kg/m^3
 

Answers and Replies

  • #2
phinds
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The density you calculated is higher than the stated density of the water. Does that seem right to you?
 
  • #3
No and that is what has me worried. I know for buoyancy, an object whose density is greater than that of the fluid in which it is submerged tends to sink and if the object is either less dense than the liquid or is shaped appropriately (as in a boat), the force can keep the object afloat. I am wondering if maybe it fits the shape part. (no picture was given for the problem as well)

I am not sure how to go about this problem otherwise.
 
  • #4
phinds
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Well, I seriously doubt that driftwood is going to be shaped like a boat, especially since that is not stated in the problem.

Since your math is giving you an obviously wrong answer, you need to re-examine your math (and I dont' mean your arithemetic)
 
  • #5
Ok how about this...Archimedes' principle!

We have ρocean=1025kg/m^3 ρwood=? Vol(below)=65% Vol(wood)=100% our g's cancel out though.

This would give us ρoceanVol(below)=ρwoodVol(wood) 1025(65)=ρwood(100) ρwood=666.25kg/m^3 I really hope this is it. I know this is similar to what I did before but for some reason I made our vol(wood) .35 instead of the 100 and converted our vol(below) to .65.
 
  • #6
phinds
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Yep, I think you were making a really simple problem difficult by not immediately gloming onto the obvious:

If half an item is under water, then it seems pretty obvious that it has half the density of water, yes? Similarly, if 65% of an object is under water then it has 65% of the density of water. It's all that simple.
 
  • #7
Thanks!

I have been having that problem all semester. What you said makes sense as well. Is it safe to assume then for most problems like this it should be the density of the liquid*volume of liquid(submerged part)*g=density of object*density total of object*g
 
  • #8
phinds
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Thanks!

I have been having that problem all semester. What you said makes sense as well. Is it safe to assume then for most problems like this it should be the density of the liquid*volume of liquid(submerged part)*g=density of object*density total of object*g

Looking at that just makes my head hurt. It THIS simple: if an object is X% submerged, then it has a density that is X% of the fluid density.
 

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