- #1
mwalter
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Homework Statement
A slab of ice floats on a freshwater lake. What minimum volume must the slab have for a 49 kg person to be able to stand on it without getting his or her feet wet?
(use 920kg/m^3 as density of ice and 1000 kg/m^3 for the density of freshwater)
Homework Equations
Archimedes Principle:
Fg = Fb
mg = pVg
The Attempt at a Solution
Hi guys! First post. I have the solution to my problem, however I don't understand how I came to it.
What I did: to set up the problem, I found the person's weight, and then set it equal to the weight of the ice to try to solve for volume. This didn't work, but I'm still not really sure what the quantity I found means.
mg = rho_ice*V_ice*g
49kg*9.8m/s^2 = 920 kg/m^3*V_ice*9.8m/s^2
Solve for V_ice:
V_ice = (49kg*9.8m/s^2)/(920kg/m^3*9.8m/s^2)
V_ice = 0.053 m^3
Since my physics class uses MasteringPhysics for homework input, I tried this answer but it was wrong, so I googled the question and found these forums, and this post: https://www.physicsforums.com/showthread.php?t=288259
So although LunarJK set his initial problem up differently, I tried to use the second portion of his solution, as here:
piceVice = pwaterVwater
(920 kg/m^3)(0.053 m^3 + x) = (1000 kgm3)x
0.053 m^3*920 kg/m^3+920 kg/m^3*x = 1000x
48.76 = 1000x - 920x
48.76 = x (1000-920)
x = 0.6095 m^3 --> volume of ice under water (which I believe is the total volume of ice)
The total volume of ice ended up being 0.61 m^3 according to MasteringPhysics, but I can't seem to wrap my mind around what my first equation found, and what the proper method of solving this should have been. I've been having a lot of trouble in this class with analysis and setting up problems properly. Any help explaining or pointers/links to help with analysis on future problems would be a huge life-saver. Thanks,
Matthew