It states that a body that is partially or fully submerged in a fluid at rest experiences an upward force equal in magnitude to the weight of fluid displaced by it.

or, `B = V rho g`

`B =` Buoyant force, `V =` volume of fluid displaced, `rho =` densiry of fluid

Consider a cylindrical body submerged in a fluid of density `'rho'` .

The top of cylinder is at depth `h_1` and bottom at `h_2`.

The horizontal forces acting on cylinder at any depth cancels out. As for each force there is a force diametrically opposite, which is equal in magnitude and opposite in direction.

The vertical components are

`F_1 = P_1 A = (P_0 +h_1 rho g )A, F_2 = P_2 A = (P_0 +h rho g) A `

`A = ` Area of cross-section of cylinder

Net buoyant force acts upward on the cylinder is

`B = F_1 -F_2 = (h_2-h_2) rho g A, B = A(h_2 -h_1) rho g`

`A(h_2 - h_1) = ` volume of cylinder

V= Volume of fluid displaced

`B = V rho g =(V rho) g`

`V rho = ` mass of fluid displaced

`B = m_fg` where , `m_f = ` mass of fluid displaced

This relation is called Archimede's Principle.
Buoyant force arises because of pressure difference in the fluid.

Maximum buoyant force `B_(max)` that can act on the body is equal to the weight of maximum volume of fluid displaced.

`B_(max) = V rho g `

1. If `B_(max) >= W_(block)`, then block will float in fluid.

(i) `B_(max) = W_(block)` , it floats just being fully submerged
`V rho g = V sigma g`

Or, `rho =sigma , sigma =` density of material of block.

(ii) `B_(max) > W_(block)` , then it floats being partially submerged `Vrho g > V sigma g, rho > sigma`

2. If `B_(max) < W_(block)` , then block will sink in the fluid.
If there is no viscous force then equation of motion of block is given by

`W_(block) - B_(max) =ma`
`m =` mass of block; `a =` acceleration of block

Note:

If the density of material `sigma` is greater than fluid `'rho'`. then object can be made to float provided it is not a uniform solid.

The resultant buoyant force acts at a fixed point located within the body called centre of buoyancy.

This point is centre of mass of the displaced fluid.

The weight of body acts at its centre of gravity.

If the centre of buoyancy and centre of gravity lies on the same vertical line as in figure (a), then the body floats in equilibrium.

If centre of buoyancy lies above the centre of gravity as in figure (a), then equilibrium is stable otherwise it is unstable.

Consider a fluid in a tank moving horizontally with acceleration 'a'. The free surface fluid makes an angle `theta` with the horizontal.

Consider a cylindrical section of fluid of length `l` and cross section area 'A'.

`F_1 =` Force on left face of cylinder

`F_1 = P_1 xx A`

`F_2 =` Force on right face of cylinder

`F_2 = P_2 xx A`

`F_1-F_2 =ma, m= ` mass of cylinder

`P_1A - P_2A = ma `

`(rho g y_1- rho g y_2)A = ma`

`rho g(y_1 - y_2)A = rho A l a`

`(y_1-y_2)/l =a/g`

`(y_1-y_2)/l = tan theta, tan theta =a/g`

Consider a beaker filled with fluid of density `rho` accelerating vertically with acceleration `rho` .

Imagine a cylindrical element of fluid of cross section 'A' with upper face at depth `h_1` and lower face at depth `h_2` .

If `P_1` and `P_2` be pressures at upper and lower face of element then, from Newton's second law,

`P_2A -(P_1A +mg) = ma`

`m = ` mass of cylinder `= A(h_2-h_1)rho`

`(P_1-P_2)A =A(h_2-h_1) rho (g+a)`

`P_2-P_1 = (h_2-h_1)rho(g+a)`

`(g+a) = g_(eff) =` Effective gravity

`P_2 -P_1 =(h_2-h_1) rho g_(eff)`

If fluid is accelerating vertically downward with acceleration a, then

`P_2-P_1 = (h_2-h_1)rho(g+a)`

or `P_2 -P_1 =(h_2-h_1) rho g_(eff)`

where, `g_(eff) = (g-a)`

If fluid is subjected to combined horizontal and vertical acceleration.

`tan theta = a_H/g_(eff)`

Where, `theta = ` angle of inclination of surface with horizontal

`a_H =` horizontal acceleration

`g_(eff) = ` Effective gravity `= g pm a`

`a_v = ` Vertical acceleration