5.7 Archimedes' Principle and Floatation

5.7 Archimedes' Principle and Floatation

An air-filled balloon immediately shoots up to the surface when released under the surface of water. The same would happen if a piece of wood is released under water. We might have noticed that a mug filled with water feels light under water but feels heavy as soon as we take it out of water.

More than two thousand years ago, the Greek scientist, Archimedes noticed that there is an upward force which acts on an object which is kept inside a liquid. As a result, an apparent reduction in weight of the object is observed. This upward force acting on the object is called the upthrust of the liquid. Archimedes' principle states that:

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When an object is totally or partially immersed in a liquid, an upthrust acts on it equal to the weight of the fluid it displaces.

Consider a solid cylinder of cross-sectional area $A$ and height $h$ immersed in a liquid as shown in Fig. 5.8. Let $h_{1}$ and $h_{2}$ be the depths of the top and bottom faces of the cylinder respectively from the surface of the liquid. Then:

If $P_{1}$ and $P_{2}$ are the liquid pressures at depths $h_{1}$ and $h_{2}$, respectively and $\rho$ is its density, then using equation $P = \rho gh$ of liquid pressure at height $h$:

Let the force $F_{1}$ be exerted at the top of cylinder by the liquid due to pressure $P_{1}$ and the force $F_{2}$ be exerted at the bottom of the cylinder by the liquid due to $P_{2}$. Then:

$F_{1}$ and $F_{2}$ are the forces acting on the opposite faces of the cylinder. Therefore, the net force $F$ will be equal to the difference of these forces. This net force $F$ on the cylinder is called the upthrust of the liquid. Hence:

or Upthrust of liquid $= \rho gAh$

Here $Ah$ is the volume $V$ of the cylinder and is equal to the volume of the liquid displaced by the cylinder, therefore, $\rho gV$ is the weight of the liquid displaced. This equation shows that an upthrust acts on a body immersed in a liquid and is equal to the weight of liquid displaced, which is Archimedes's principle.

Example 5.2

A wooden cube of sides $10\,\mathrm{cm}$ each has been dipped completely in water. Calculate the upthrust of water acting on it.

Solution

Given:

• Length of side $L = 10\,\mathrm{cm} = 0.1\,\mathrm{m}$
• Volume $V = L^{3} = (0.1\,\mathrm{m})^{3} = 1 \times 10^{-3}\,\mathrm{m^{3}}$
• Density of water $\rho = 1000\,\mathrm{kg/m^{3}}$
• Upthrust $F = ?$

Using Archimede's principle:

Thus, upthrust of water acting on the wooden cube is $9.8\,\mathrm{N}$.