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buoyant force

 Definition/Summary Buoyant force (or buoyancy) is the net force on a body caused by the pressure differences in the surrounding medium caused by gravity. Buoyant force acts through the centre of gravity of the displaced fluid (the centre of buoyancy). If the centre of buoyancy is not on the same vertical line as the centre of gravity, there will be a torque (a turning force). Archimedes' Principle: The buoyant force on a body is vertical, and is equal and opposite to the weight of the fluid displaced by the body.

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 Scientists Archimedes (APXIMH∆H∑) of Syracuse (287-212 BC)

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 Breakdown Physics > Classical Mechanics >> Fluid Statics

 Extended explanation Archimedes' principle: From Archimedes' principle we see that the buoyant force on an object is equal to the weight of the fluid it displaces. Why is this so? Let us imagine a cube of any substance of side length $h$. Fluid around the cube exerts pressure on each side of the cube. On the perpendicular faces the force cancels out, but on the top and bottom faces there is a pressure differential due to the variation of pressure with depth in fluid. If we call the depth of the top face $x_t$ and the depth of the bottom face $x_b$, then the pressure differential across the cube can be written as: $$\Delta P = \rho_f g(x_b-x_t) = \rho_f gh$$ Since, $$x_b-x_t = h$$ Since pressure is a force per unit area the pressure differential must be the same as the buoyant force divided by the area of one of the cube's faces. In other words we can write the buoyant force as being the pressure differential multiplied by the area of one face of the cube. $$F_B = (\Delta P) h^2$$ $$F_B = (\rho_f g h) h^2 = \rho_f V g = M_f g$$ This shows that by considering pressure differentials we have arrived at the same result Archimedes' principle tells us we should arrive at. Alternatively, consider the forces which would be on the displaced fluid if it were still there … the only forces are the weight of the displaced fluid, and the pressure from the surrounding fluid … if the fluid is in equilibrium, then these forces must be equal and opposite … and the pressure from the surrounding fluid on the body is the same as it would be on the displaced fluid. Centre of buoyancy: The buoyant force acts through the centre of buoyancy. The centre of buoyancy of a body is at the centre of gravity of the fluid displaced (for a fluid of uniform density, that is at the centroid of the displaced volume). Water, even at great depths, may be taken to be of uniform density. If the body is totally immersed in a fluid of uniform density, it is at the same point, whatever the orientation of the body. Note that the centre of gravity of the body may change if the body is not rigid (for example, if it contains shifting cargo). If the body also is of uniform density, it is at the centre of gravity (the centre of mass) of the body. If the body is floating at the surface, the shape of the volume of fluid displaced will change as the orientation changes, and so the centre of buoyancy will change. For stability, the centre of buoyancy should be above the centre of gravity. Total immersion ("underwater"): If the body is totally immersed, then the volume of fluid displaced is equal to the volume of the body, and so the buoyant force (measured vertically upward) is that volume times the density of the fluid minus the average density of the body: $$F_b\ =\ Vg\left(\rho_{fluid}\ -\ \rho_{body}\right)\ =\ mg\left(1\ -\ \frac{\rho_{body}}{\rho_{fluid}}\right)$$ and the acceleration upward will be: $$g\left(1\ -\ \frac{\rho_{body}}{\rho_{fluid}}\right)$$ Obviously, if the body has lesser average density than the fluid, that acceleration will be positive, and the body will rise. And if it has greater average density, that acceleration will be negative, and the body will sink. Partial immersion ("floating") of a uniform body: If a body of uniform density is floating, in equilibrium, only partially immersed, at the surface of the fluid, then the volume of fluid displaced is less than the volume of the body, and so the proportion submerged equals the ratio of the densities. A ship is not of uniform density (its volume is mostly filled with air), and so there is no such simple relationship. Strictly, a ship is totally immersed, in water and air, but the weight of the displaced air is so small, compared with the weight of the displaced water, that it can be ignored. If such a floating body is pushed down by an external force, the volume displaced will increase, and so the body will feel an increased buoyant force. This increased force, together with the weight of the body, provides a restoring force, tending to bring the body back to its equilibrium position. If the body has a uniform vertical cross-section, this restoring force will be proportional to depth, and so the body will bob up and down in simple harmonic motion (shm). If a floating body is pushed sideways by an external force, the volume displaced will depend on the shape of the body … the stability of the body will be greater if the cross-section is nearer an arc of a circle.