Law of Flotation

Law of Flotation

When a body is allowed to immerse in a liquid, the following forces act on it:

1. The weight of the body acting vertically downward and
2. The upthrust on the body acting vertically upward.

Depending on the magnitude of these forces, the following three cases can be studied:

Case I: When the weight of the body is greater than the upthrust acting on it, the resultant force will be vertically downwards. As a result, the body gets sunk in the liquid.

Mathematically,

 weight of the body (w₁) ˃ weight of the liquid displaced

or, m₁g ˃ m₂g [\(\because\) w = m × g]

or, m₁ ˃ m₂

or, d₁v₁ ˃ d₂v₂ [\(\because\) d = \(\frac{m}{v}\) or m = d × v]

\(\therefore\) d₁ ˃ d₂ [\(\because\) v₁ = v₂]

Thus, a body sinks in a liquid only if the density of the body (d₁) is greater than the density of the liquid (d₂) in which the body is kept. Due to this reason, a piece of iron, being denser than water, sinks completely in water.

 Case II: When the weight of the body is just equal to the upthrust acting on it, the resultant force acting on it is zero. As a result, the body floats in the liquid.

Mathematically,

W₁ = w₂

or, m₁g = m₂g

or, m₁ = m₂ [\(\because\) w = m × g]

or,  d₁v₁ = d₂v₂ [d = \(\frac {m}{v}\) or m = d × v]

Since, the volume of body (v₁) is equal to the volume of displaced liquid (v₂) i.e., v₁ = v₂, d₁ is also equal to d₂.

\(\therefore\) d₁ = d₂

Thus, a body just floats on a liquid when the density of the body (d₁) is equal to the density of the liquid (d₂) in which the body is kept.

Case III: When the weight of the body is less than the upthrust acting on it, the resultant force will be directed vertically upwards. As a result, a body will come at rest with its some part in air and some part in water. In this case, the weight of the body will be equal to the weight of the liquid displaced by the immersed portion of the body.

Mathematically,

Weight of the body (w₁) = weight of the liquid displaced (w₂)

or, m₁g = m₂g [\(\because\) w = m × g]

or, m₁ = m₂

or, d₁v₁ ˂ d₂v₂ [\(\because\) d = \(\frac {m}{v}\) or m = d × v]

Since the body is partially immersed in the liquid, the volume of the body ˃ the volume of the liquid displaced i.e. v₁ ˃ v₂ and hence we have, d₁ ˂ d₂.

\(\therefore\) d₁ ˂ d₂

Thus, when the density of the body (d₁) is lesser than that of the liquid (d₂), a body floats keeping some of its parts out from the liquid.

From the above discussion, we can conclude that:

A body sinks in a liquid if the density of the body is greater than that of the liquid. In this case, the weight of liquid displaced is lesser than the weight of the body. But, if we shape the body in such a way that the weight of the body is equal to the weight of the liquid displaced by the body, it floats on the liquid. For example: a floating ship.

The law of floatation states that “weight of a floating body is equal to the weight of the fluid displaced by the body”.

i.e. weight of floating body (w₁) = weight of the displaced fluid (w₂)

Experimental Verification of Law of Floatation

Verification:

Materials Needed:

• Eureka can
• Beaker
• Top Pan Balance
• Dry Wooden Block
• Water
• Tripod Stand

Process:

1. An overflow can is kept over a top pan balance.
2. The can is filled with water up to its spout and an empty beaker is kept below the spout of the can.
3. The weight of water with overflow can [Eureka can] is noted.
4. Now, a dry wooden block is kept in the water of the can.

Conclusion:

The water is displaced into the beaker but the top pan balance does not show any change. It proves that the weight of the displaced water is equal to the weight of the block. It verifies the law of floatation.