# Math Help - Buoyancy Force Question

1. ## Buoyancy Force Question

Hi,
I am looking to investigate the buoyancy force of a cuboid. The dimensions of the cuboid are 30cm x 20cm x 15 cm. When the cuboid is placed in the water (the face 30cm x 20cm faces the water), 4cm of the cuboid will be submerged. If I push the cuboid so that the top face is 15mm below the surface of the water, I will have pushed the cuboid (15 - 4) + 1.5 = 12.5 cm. Therefore, using some notes that I have I have worked out the buoyancy force to be:

F = p*A*y*g

where p = pressure of the water (998.2 kg/m^3, pressure of tap water at 20oC), A is the cross sectional area to face the water (0.06 m), y is the overall displacement (0.125 m) and g = 9.8m/s, acceleration due to gravity. If I work this out, I get: 73.37 Newtons.

First things first, is this correct? Please note that this is NOT homework, just a problem that I am looking into. I have looked at some notes where I was placing a bottle filled with sand in water, and then pushing it to a new depth. We used F = p*A*y*g to work out the buoyancy force in that case, however I dont believe that we considered the case if the bottle was pushed so far into the water that it was fully submerged (as in this case). Therefore, I think that the equation is correct so far as I push the cuboid down to a depth of 11cm so that the top face of the cuboid will be level with the top of the water, however I am unsure if I can carry the above reasoning through to when the object is fully submerged in water. I then want to consider pushing the cuboid so that the top face is 15cm and then 30cm deep, if I know that I can apply the above principle at 15mm then I know that I can extend it to these two cases.

What I am really confused with (the reason i dont think that this works past 11cm) is that Archimedes said the Buoyancy Force is equal to the mass of the water displaced. However, when fully submerged, surely there wont be any more water displaced if we push the cuboid deeper and deeper?!?

Any help on this would be greatly appreciated!

Si

2. [quote=Abelian;206947]Hi,
I am looking to investigate the buoyancy force of a cuboid. The dimensions of the cuboid are 30cm x 20cm x 15 cm. When the cuboid is placed in the water (the face 30cm x 20cm faces the water), 4cm of the cuboid will be submerged. If I push the cuboid so that the top face is 15mm below the surface of the water, I will have pushed the cuboid (15 - 4) + 1.5 = 12.5 cm. Therefore, using some notes that I have I have worked out the buoyancy force to be:

F = p*A*y*g

where p = pressure of the water (998.2 kg/m^3, pressure of tap water at 20oC), A is the cross sectional area to face the water (0.06 m), y is the overall displacement (0.125 m) and g = 9.8m/s, acceleration due to gravity. If I work this out, I get: 73.37 Newtons.

[\quot]

What you have written as p is usualy $\rho$ and is the density of the water (as you should be able to see from the units)

The boyancy force is equal (and opposite if we are being carefull about direction) to the weight of the displaced water. In this case the cuboid is completley submerged so the boyancy force is:

$F=\rho A h = \rho Vg$

where $h$ is the dimension of the cuboid normal to a face of area $A$.

What you had is only applicable if the cuboid is not completly submerged.

What I am really confused with (the reason i dont think that this works past 11cm) is that Archimedes said the Buoyancy Force is equal to the mass of the water displaced. However, when fully submerged, surely there wont be any more water displaced if we push the cuboid deeper and deeper?!?
That is not what Archimedes principle is, it is that the boyancy force is equal to the weight of the fluid displaced.

CB