Fluid dynamics - hydorstatic

1. Dec 14, 2008

rhodium

Hi everyone,

I hope you can help me out with this question as my exam is tomorow.

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

A 355 mL soda can with diameter=6.2 cm has a mass of 20g. If it is half full with water, and it is floating upright on water, what length of the can is abover the water level?

2. Relevant equations
$$\rho$$ = m/V

F of buoyancy = $$\rho$$ of fluid*V of fluid displaced*g = $$\rho$$ of object*V of object*g

Since V= Area*height, and the area is the same, then the Area cancel out and we are left with:
$$\rho$$ of fluid*h of fluid displaced = $$\rho$$ of object*h of object

Isolating for height of fluid displaced

height of fluid displaced = $$\rho$$ of object*h of object/$$\rho$$ of fluid

$$\rho$$ is density

so height above water level = (H of object) - (H of fluid displaced)

3. The attempt at a solution

m of object = 0.02 kg + (355 mL/2)(1 g/1 mL)
= 177.52 g

density of object = m of object/ volume
= 0.17752 kg / 0.000355 m^3
= 500 kg/m^3

h =V / ($$\pi$$ * radius^2)
= 11.76 cm
= 0.1176 m

Now, subbing into the buoyancy eqation:

height of fluid displaced = ($$\rho$$ of object) (h of object) / ($$\rho$$ of fluid)
= (50)(0.1176 )/1000
= 0.0588 m

height above water level = (H of object) - (H of fluid displaced)
= 0.1176 -0.0588
= 5.88 cm

2. Dec 14, 2008

Staff: Mentor

3. Dec 14, 2008

rhodium

hey,

God...Thank you sooo much. It took me a while to notice it actually. :)

4. Dec 15, 2008

trotter

i got a similar problem with fluid dynamics , i got a test later in the week havent been prepped for it at all just been handed some sheets wondering if any one could help with an example i have got , as it will help me see what i need to do for future reference

QUESTION

A cube of 0.25m length sides which has a weight of 50N is immersed in a tank of fluid .

If the specific gravity of teh fluid is 0.8, calculate
a) the density of the cube material
b) the density of the fluid
c) the resultant up-thrust of the cube

ATTEMPTS

w=mg
50=mg
50/0.8 = m
m = 62.5 kg

p=m/v
p=62.5/0.25x0.25x0.25
p=4000 kg/m3

Thats all i can manage i havent really been taught this yet so any help would be appreciated

5. Dec 16, 2008

rhodium

Hey,

I'll try to help with what I know.