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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 = pfVos
Where
B is buoyancy force.
pf is the density of the liquid the object is being placed in.
Vo is the volume of the object.
s is the percent of the object submerged in the liquid.
pf = 1.025 g/cm3
Vo = 400 cm3
Found 2 solutions by stanbon, Alan3354:
Answer by stanbon(75887)   (Show Source):
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.
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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
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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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