Archimede’s Principle , Buoyancy

What is Archimede’s Principle ?

When a body is partially or fully immersed in a fluid at rest, then the body experiences an upward force called buoyancy. Buoyancy is equal to the weight of the fluid displaced by the body.

Suppose that a uniform cylindrical block of mass M and of uniform cross-sectional area A is kept submerged in a liquid of density ρ. The material of the block has density ρo .

Suppose that the total volume of the block is V and the submerged volume of the block is VS (say).

The weight of the block is Wb = Mg = Vρog

The volume of the fluid displaced is equal to the submerged volume of the block = VS.

Therefore, the weight of the displaced fluid is

FB = Vsρg    , acting vertically upward.

The maximum possible buoyant force FB/max that can act on the block is equal to the weight of maximum volume of the fluid displaced, which occurs when the body is completely immersed within the fluid (by force if necessary).

FB/max = Vρg

(a) If FB/max ≥ Wb , then the block will float in the fluid.

(i) More specifically when FB/max = Wb, it just floats just submerged,

(ii) When FB/max > Wb , then it floats partially submerged.

(b) If FB/max <  Wb

then, the block will sink in the fluid. If there is no fluid friction (viscosity) then the equation of motion of the block is given by

Wb – FB/max = Ma

Solved Example : A body of mass 2 kg and volume .001/m3 is in a fluid of density (4/3) gm/cm3. The fluid is at rest. Determine whether the body will float or sink in the fluid. If it floats, then, what fraction is immersed of volume in the fluid? If sinks, then find out the acceleration.

Solution : The force acting on the body

Fb = Vρg = (0.001) (4/3) × 1000 (9.8)

$\large F_b = \frac{4\times 9.8}{3}$  = 13.06 N

⇒ Weight of the body (W = mg = 2 × 9.8 = 19.6 N ) is greater than Fb

⇒  The body sinks completely.

Its acceleration $\large = \frac{W-F_b}{m}$

$\large = \frac{19.6- \frac{4\times 9.8}{3}}{2}$

= 3.3 m/s2

Solved Example : A rubber ball of mass 10 gm and volume 15 cm3 is dipped in water to a depth of 10 m. Assuming density of water uniform throughout the depth, find

(a) the acceleration of the ball, and

(b the time taken by it to reach the surface if it is released from rest  (take g = 980 cm/s2).

Solution:  The maximum buoyant force on the ball is

FB = V ρwg = 15 × 1 × 980 dyne = 14700 dyne.

The weight of the ball is mg = 10 × g = 10 × 980 dyne.

The net upward force, F = (15 × 980 – 10 × 980) dyne

= 5 × 980 dyne = 4900 dyne.

(a) Therefore, acceleration of the ball upward

$\large a = \frac{F}{m} = \frac{5 \times 980}{10}$

a = 490 cm/s2 = 4.9 m/s2.

(b) Time taken by it reach the surface is

$\large t = \sqrt{\frac{2 h}{a}}$

$\large t = \sqrt{\frac{2 \times 10}{4.9}}$

= 2.02 sec

What is the Centre of Buoyancy & Equilibrium ?

The resultant force of buoyancy acts at a fixed point within the body and is known as the centre of buoyancy.

If the centre of gravity and centre of buoyancy lie on the same vertical line then the body floats in fluid in equilibrium.

If the centre of buoyancy is above the centre of gravity the equilibrium is stable, otherwise it is unstable.

Solved Example :  A vessel contains oil of density 0.8 gm/cm3 over mercury of density 13.6 gm/cm3. A homogeneous sphere floats with half of its volume immersed in mercury and other half in oil. What is the density of the material of the sphere ?

Solution:  Let the volume of the sphere be 2 V and density  ρ .

According to Archimedes’ principle,  2 V. ρ g = V ρmercury g + V ρoil g.

Therefore, $\large \rho = \frac{\rho_{mercury} + \rho_{oil}}{2}$

$\large \rho = \frac{13.6 + 0.8}{2}$

= 7.2 g/cm3

Solved Example: A uniform wooden bar of length L and mass M is hinged on a vertical wall of a container containing water, at one end [in figure]. 3/5th part of the bar is submerged in water. Find the ratio of densities of the liquid and the bar.

Solution: Let ρ = density of liquid & σ = density of bar .

Taking the moment of W = mg and Fb = V ρ g  about O, we obtain

$\large V \rho g (\frac{7}{10} l sin\theta) = ( \frac{5}{3}V ) \sigma g (\frac{l}{2} sin\theta)$

$\large \frac{7 \rho}{10} = \frac{5 \sigma}{6}$

$\large \frac{\rho}{\sigma} = \frac{25}{21}$

Solved Example: A cube of 1 kg is floating at the interface of two liquids of density ρ1 = 3/4 gm/cm3 and ρ2 = 2.5 gm/cm3. The cube has 3/5 part in ρ2 and 2/5 part in ρ1. Find the density of the cube.

Solution: The weigh of the cube = (force acting by the liquid of density ρ1) + (force acting by the liquid of density of ρ2)

Let mass of the block = m

$\large m g = V_1 \rho_1 g + V_2 \rho_2 g$

$\large m = V_1 \rho_1 + V_2 \rho_2$

$\large A L \rho = A(\frac{2}{5}L)\rho_1 + A(\frac{3}{5}L)\rho_2$

$\large \rho = \frac{2\rho_1 + 3 \rho_2}{5}$

$\large \rho = \frac{2(3/4) + 3 \times 2.5 }{5}$

$\large \rho = \frac{1.5 + 7.5}{5} = \frac{9}{5}$

= 1.8 g/cm3