Calculating Minimum Volume for Floating Ice Slab - AP Physics B Problem

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To determine the minimum volume of an ice slab that can support a 70 kg man without submerging, the relationship between the mass of the ice, the man's weight, and the buoyant force must be established. The equation combines the mass of the ice and the person, equating it to the buoyant force, which is the product of the density of water and the volume submerged. By expressing the mass of the ice in terms of its volume and density, the equation can be simplified to isolate the volume of the ice slab. The final equation reveals that the volume of ice must compensate for the weight of the man while considering the differences in density between ice and water. This approach leads to a solution for the required volume of the ice slab.
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AP physics B problem- Buoyant Forces?


A slab of ice floats on a freshwater lake. What is the minimum volume of the slab such that a 70 kg man can stand on it without getting his feet wet? The density of ice is 0.922 kg/m^3 and the density of freshwater is 1,000 kg/m^3

I have gotten to mass(ice) + mass(person) = Density of water x volume submerged

From there I have two unknown variables I am at a loss
 
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Your unknown variables are the mass of the ice and the volume of the ice, right? What other equation relates them to each other?
 
Hint: How does the volume submerged relate to the volume of ice?
 
density = mass/volume , but I only know the density. Also the volume submerged would equal the total volume, but I don't know where to go with that...
 
llama0lover said:
density = mass/volume , but I only know the density. Also the volume submerged would equal the total volume, but I don't know where to go with that...
Good. Express the mass of the ice in terms of its volume.
 
so density= m/v so m=density x volunme submerged, but how would that help you if you don't know the volume submerged.
 
The volume is what you are trying to find.
 
so volume submerged = m/density (sorry if I'm a little slow at this I'm not really great at physics) how would you solve that if you only know the density... You still have two unknown variables
 
llama0lover said:
so volume submerged = m/density (sorry if I'm a little slow at this I'm not really great at physics) how would you solve that if you only know the density... You still have two unknown variables
Use the definition of density to rewrite the equation you gave in post #1 completely in terms of volume, not mass. You'll only have one variable.
 
  • #10
so you would end up with mass of ice + mass of person = density of water x (mass of ice/ density of ice)

so you would just simplify it from there, but how would you do that

[ m(i) + m(p) ] / density of water= m/ density of ice

how would you get both onto one side?
 
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  • #11
thank you by the way :-) You're a huge help
 
  • #12
llama0lover said:
so you would end up with mass of ice + mass of person = density of water x (mass of ice/ density of ice)

so you would just simplyfy it from there, but how would you do that
Start with your original equation:
llama0lover said:
I have gotten to mass(ice) + mass(person) = Density of water x volume submerged
Replace each mass with its equivalent in terms of density and volume.
 
  • #13
densityof ice(volume ice) + 70 = Density of water x volume submerged

I can't separate the mass into volume and density of the person because I don't know either of those
 
  • #14
llama0lover said:
densityof ice(volume ice) + 70 = Density of water x volume submerged
Perfect. Now you can solve for the volume. (Remember your answer to my question in post #3.)

I can't separate the mass into volume and density of the person because I don't know either of those
Skip that one--my mistake. :rolleyes: You already know the person's mass.
 
  • #15
so [ density of ice(volume ice) + 70 ] / density of water= volume of ice
how would you divide it with the seventy still there
 
  • #16
llama0lover said:
so [ density of ice(volume ice) + 70 ] / density of water= volume of ice
No. Go back to the previous equation and collect all terms with volume to one side. Then you can isolate the volume and solve for it.
 
  • #17
70 = Density of water (volume ice) + density of ice(volume ice)

70= volume of ice(density of water +density of ice)

would that be right?
 
  • #18
llama0lover said:
70 = Density of water (volume ice) + density of ice(volume ice)

70= volume of ice(density of water +density of ice)

would that be right?
Almost. But you messed up a sign when you moved a term from left to right.
 
  • #19
70= volume of ice(density of water - density of ice)

so from there you would just simplify

ohhhhh I get it.

Thanks you so much you were a huge help
 
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