Buoyancy Force

The tendency for an immersed body to be lifted up in a fluid, due to an upward force that acts opposite to the action of gravity is called buoyancy.

Example: When an empty plastic bottle is pushed into the water, an upward push is felt. When it is pushed further down, it is observed that it is difficult to push deeper and deeper. This indicates that water exerts a force on the bottle in the upward direction. 

An upward force (called 'buoyant force') acts on an object which is immersed partially or completely in a liquid. This force increases when the volume of the object immersed in the liquid increases and reaches a maximum value when the object is completely immersed in the liquid.

It is an upward force that is exerted by a fluid on any object immersed partly or wholly in the fluid.

1.0Cause of the Buoyant Force

The buoyant force is caused by the pressure that is exerted by a fluid on an object in the fluid. Figure shows a cube shaped object submerged in a glass of water. The water exerts pressure everywhere over the surface of the object. The direction of the pressure on a surface is always perpendicular to the surface. Also, the pressure exerted by a fluid increase as you go deeper into the fluid. In figure, the bottom of the cube is deeper in the water. Therefore, the pressure that is exerted by the water at the bottom of the cube is greater than it is at the top of the cube. The higher pressure near the bottom means that the water exerts a net upward force on the cube. This net upward force is the buoyant force.

A buoyant force acts on all objects that are placed in a fluid, whether they are floating or sinking.

2.0Archimedes’ Principle 

According to Archimedes’ principle, any object completely or partially submerged in fluid experiences an upward buoyant force equal in magnitude to the weight of the fluid displaced by the liquid. 

Buoyant force = Weight of displaced fluid = $m_{l}g$

$m_l$ = mass of displaced fluid

$m_l={\rho }_l\times v_l$

F_B = ${\mathrm{P}}_{l}V\times g$

$V_l$ = volume of displaced liquid

Apparent weight : Because of an upward force acting on a body immersed in a fluid, either wholly or partially, there occurs an apparent loss in weight of the body. The net weight of an object immersed in a fluid is called apparent weight.

Apparent weight, W' = Weight of body in air – Buoyant force 

Archimedes’ was a Greek mathematician who was born in Syracuse. According to legend, the king of Syracuse suspected that a certain golden crown was not pure gold. While bathing, Archimedes figured out how to test the crown’s authenticity when he discovered the buoyancy principle. He is reported to have then exclaimed, “Eureka!” meaning “I’ve found it!”

3.0Sinking and Floating

The buoyant force pushes an object in a fluid upward, but gravity pulls the object downward. 

(1) If the weight of the object is greater than the buoyant force, the net force on the object is downward and it sinks [see figure (a)]

Let an object of density ρ_s be immersed in a liquid of density ρ_L.

If ρ_s > ρ_L, the body will sink to the bottom. 

(2) If the buoyant force is equal to the object’s weight, the forces are balanced and the object floats. 

(a) If ρ_s = ρ_L, apparent weight = 0 i.e., weight of the body in air = buoyant force   

This means, the body will just float or remain hanging at whatever height it is left inside the liquid [see figure (b)].

(b) If ρ_s < ρ_L, apparent weight = 0  

i.e., weight of the body in air = buoyant force   

This means the body will float, but it is immersed partly in the liquid [see figure (c)].

4.0Changing the Buoyant Force

An object sinks or floats depends on whether the buoyant force is smaller than or equal to its weight. The fluid exerts upward pressure on the entire lower surface of the object that is in contact with the fluid. If this surface is made larger, then upward pressure is exerted on larger surface of the object and the buoyant force becomes large enough to float the object. e.g. If an aluminium sheet is crumpled, the buoyant force on it is less than the weight, so the aluminium sheet sinks. When the aluminium is flattened into a thin curved sheet, the buoyant force is large enough so that the sheet floats.

Metal ships are able to float because the metal is formed into a curved sheet that has a large surface area in contact with the water, which is called the hull of the ship. 

The contact area of the hull with the water is much greater than if the metal were a solid block. As a result, the buoyant force on the hull is greater than it would be on a metal block and thus, the ship floats.

Finding buoyant force on an object immersed in a liquid

When a stone tied to a rubber string is gradually lowered in the water, elongation of the string decreases. 

But, no further change is observed once the stone gets fully immersed in the water.

Since the extension decreases once the stone is lowered in water, it means that some force acts on the stone in upward direction. As a result, the net force on the string decreases and hence the elongation also decreases. This upward force exerted by water is called buoyant force or force of buoyancy.

If W₁ be the weight of an object in air and W₂ be its weight (apparent weight) when it is completely immersed in a liquid, then,

W₂ = W₁ – F_B where, F_B is the buoyant force acting on it.  or $F_B=W_1-W_2$

Thus, 'buoyant force is the loss of weight of an object when it is immersed in a liquid'.

Measuring relative density of an object using Archimedes' principle

Let us take an object having weight W₁ in air. The weight W₁ is measured by using a spring balance . 

Now take a vessel filled with water and immerse the object completely in water and again note the spring balance reading. This reading gives the weight (W₂) of the object in water. Thus, the buoyant force on the object due to water is given by,

F_B= W₁ – W₂ …(1)

Now, relative density is given by,

$\text{R.D.}=\frac{\text{Density of object}}{\text{Density of water}}=\frac{\rho }{{\rho }_w}=\frac{\rho Vg}{{\rho }_{w}Vg}$ ...(2)

(Multiplying and dividing by Vg)

Where, V_s= V_L = V = volume of object = volume of water displaced, as the object is immersed completely, and 'g' is the acceleration due to gravity.

Now, weight of object in air, W₁ = ρVg  ...(3) 

Buoyant force, F_B = ρ_wVg  ……… (4)

From (2), (3) and (4), we get, R.D. = $\frac{W_1}{F_B}$

From (1) and (4), we get,

$R.D=\frac{W_1}{W_1-W_2}$

Melting of ice that is floating on water

When a piece of ice floating on water in a beaker completely changes (melts) to liquid state, the level of water in the beaker remains unchanged. Let us understand it using mathematics. 

Let M = mass of ice ; ρ = density of water ; V = volume of water displaced

Now,   W = F_B   

or Mg = ρVg    

or M = 1 × V    (ρ_water = 1 g/cm³)

or M = V … (1)

Now, volume V' of water is formed on melting, 

$V^{\prime }=\frac{M}{\rho }=\frac{M}{1}$

or V' = M … (2)

From (1) and (2), we get, V' = V, i.e., there is no change in the volume of the contents of the beaker. Hence, level of water will not change.

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