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zzoon
Apr8-08, 11:19 PM
This has been bugging me. Any help will be much appreciated. Thanks.

1. The problem statement, all variables and given/known data

A 355 mL soda can is 6.2 cm in diameter and has a mass of 20 g. Such a soda can half full of water is floating upright in water. What length of the can is above the water level?

The answer is 5.22 cm but I can't reach that answer.

2. Relevant equations

Bouyant force = density(given by rho) * g * volume displaced

Density of water = 1000 kg/ cubic meter


3. The attempt at a solution

mass of water inside the can: 355/2 mL * 1kg/L = 0.1775 kg
mass of water inside the can plus the can itself = 0.1975

From the bouyant force equation, mg = rho * g * volume displaced
m = rho * volume displaced
volume displaced = m/rho = .1975 kg /(1000 kg/(M^3)) = 0.0001975 M^3

The area of the can's top is (3.1 cm)^2 * pi = 30.19 cm^2 = 0.003019 M^2

volume = area * x where x is the depth to which the can has sunk into the water. Which I reason should also be the height of the part sticking out of the water since the can is half full.

0.0001975 M^3 = 0.003019 M^2 * x
x = 0.0654 M = 6.54 cm

luben
Apr9-08, 04:25 AM
you almost got it. you found the immersion depth. how high is the can?

zzoon
Apr9-08, 04:05 PM
Thanks for letting me know I was on the right track. With that hint, I solved it.