## Saturday, August 23, 2008

### 2nd common force: Buoyancy

Buoyancy

My last topic about forces (but don't be scared, there is more to life than just that) is about buoyancy.

One day, Archimedes was taking a bath. He had already spent some years on finding out why an object in water experiences a lower gravitational force. Suddenly, he noticed he kept floating in the water, and that hes bath had overflowed when he stepped in it. He found the explanation, and completely naked, he ran on to the streets screaming "eureka" (I found it) (running naked on the streets wasn't such a big deal back then).

What was he so happy about? He had discovered that an object immersed in a fluid, experiences an upward force, equal to the gravitation on the displaced mass of water.

Knowing that, we can easily producea formula for the so called buoyancy.
All matter has a density, P (The Greek letter 'rho'). This is the ratio of the mass to the volume (for water, this is 1000kg/m³ ). So, the mass of an amount of fluid is the product of P and the volume of the fluid. But since the volume of the displaced fluid equals the volume of the object in the water, we can say:

m = V . P

So the gravitation on this mass of displaced fluid is:

FA = m . g => FA = P . g . V

If the buoyancy on an object is bigger than the gravitation on the object, the object will rise and come out of the water, until the gravitation and the buoyancy are equal. The opposite happens when the gravitation is bigger than the buoyancy. Then the object will keep on sinking until it hits the bottom. When the gravitation equals the buoyancy, the object floats on or in the water.

Now, we can calculate we whether a small submarine of 10 kg with a volume of 0,05 m³ (under water) will sink, float or rise out of the water.

FG = 10kg. 9,81N/kg = 9,81. 10 N

FA = P.g.V => FA = 0,05m³ . 9,81N/kg . 1000kg/m³ = 50. 9,81N

Since FA = 5 . FG => FA > FG. So, the object will rise (disregarding atmospheric pressure).

We can also calculate how much of the sub will remain under water. To do this, we'll need an equitation, since the sub will only stop rising when it is in rest. This is, when the resultant force ( = the sum of all force vectors) equals zero. In this case:

FG + FA = 0 <=> FG = FA
=> m.g = V.g.P

We know almost everything, so, same as before:

10kg . g = V . g . 1000kg/m³
=> V = (10kg . m³)/1000kg = 0,001 m³ = 10 dm³

So, the sub will keep rising until only 10 dm³ is still under water and 40dm³ is above (disregarding atmospheric pressure).

So, that's what I have to say about buoyancy. Hope you understand it. I think next, I'll start with pressure or work, we'll see.
Bye.
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