SOLUTION: </i>I've been having some problems acquiring a formula for finding how much of an object is submerged in water. I have an object that is 10 cm wide, 10 cm, and 4 cm high. Its densi

Question 388646: I've been having some problems acquiring a formula for finding how much of an object is submerged in water. I have an object that is 10 cm wide, 10 cm, and 4 cm high. Its density is 1.000 g/cm³.
It is being placed in salt water, which has a density of 1.025 g/cm³. So I guess what I'm asking is for a formula that will solve for s. I know that if I use a number substituting s, for example .50
(%50 submerged) that that will give me a buoyancy force of 205g. I'm trying to figure out how far the object will be submerged before the buoyancy force is equal to mass of the object.
B = p_fV_os
Where
B is buoyancy force.
p_f is the density of the liquid the object is being placed in.
V_o is the volume of the object.
s is the percent of the object submerged in the liquid.
p_f = 1.025 g/cm³
V_o = 400 cm³
Found 2 solutions by stanbon, Alan3354:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Please have a look at the following site:
http://en.wikipedia.org/wiki/Buoyancy
============
Cheers,
Stan H.
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
I have an object that is 10 cm wide, 10 cm, and 4 cm high. Its density is 1.000 g/cm³.
It is being placed in salt water, which has a density of 1.025 g/cm³. So I guess what I'm asking is for a formula that will solve for s. I know that if I use a number substituting s, for example .50
(%50 submerged) that that will give me a buoyancy force of 205g. I'm trying to figure out how far the object will be submerged before the buoyancy force is equal to mass of the object.
---------------------------
The object have a volume of 400 cc, so it weighs 400 grams.
It will displace 400 grams of water.
400/1.025 = 390.24 cc
------------------
You say it's 4 cm high, implying that the 4 cm is the vertical dimension. If it's submerged in that orientation, its depth will be 390.24/100 cm
=~ 3.9 cm submerged.

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