Fluid Mechanics
4.0 Buoyant force
4.0 Buoyant force
Buoyant force is an upward force exerted by a liquid that opposes the weight of an immersed object.Note:
- $V'$ is the submerged volume
- Buoyant force depends on,
- density of the liquid
- volume of the body submerged inside the liquid
- acceleration due to gravity
- Buoyant force does not depend on,
- density of the body
- total volume of the body
Buoyant force exerted by a liquid of density $\rho $ is, $$\begin{equation} \begin{aligned} {\text{Buoyant force}} = {\text{(Density of liquid)}} \times {\text{(volume of the body submerged)}} \times {\text{g}} \\ F = \rho V'g \\\end{aligned} \end{equation} $$
4.1 Archimedes principle
Archimedes principle states that any object wholly or partially immersed in a fluid is buoyed up by a force equal to the weight of the fluid displaced by the object.
Law of flotation
Consider a body of density ${\rho _b}$ and volume $V$ floating in a liquid of density ${\rho _L}$.
Let $V$ be the volume of the body immersed in the liquid as shown in the figure.
At equilibrium, $$\begin{equation} \begin{aligned} F = mg \\ {\rho _L}V'g = {\rho _b}{V_g} \\ \frac{{{\rho _L}}}{{{\rho _b}}} = \frac{V}{{V'}} \\\end{aligned} \end{equation} $$ or $$\frac{{{\rho _b}}}{{{\rho _L}}} = \frac{{V'}}{V}$$
Note:
If ${\rho _b} < {\rho _L}$, then the body will float partially immersed in the liquid
If ${\rho _b} = {\rho _L}$, then the body will be completely immersed in the liquid. The body remains floating in the liquid wherever it is left
If ${\rho _b} > {\rho _L}$, then the body will sink
4.2 Apparent weight of a body inside a liquid
When a body is immersed in a liquid, the effective weight of the body is decreased.
The decrease in weight of the body is due to the upthrust on the body as shown in the FBD
At equilibrium, $$\begin{equation} \begin{aligned} {W_{apparent}} = {W_{actual}} - {F_{buoyant}} \\ {W_{apparent}} = mg - BF \\ {W_{apparent}} = {\rho _b}Vg - {\rho _L}V'g \\\end{aligned} \end{equation} $$
When the body is completely immersed in the liquid then, $$V = V'$$ So, $${W_{apparent}} = \left( {{\rho _b} - {\rho _L}} \right)Vg$$
4.3 Buoyant force in accelerating fluids
A body immersed in a liquid of density ${\rho _L}$ is kept in a lift which is moving upwards with an acceleration $a$ as shown in the figure.
From FBD, $$\begin{equation} \begin{aligned} BF - W = ma \\ BF = mg + ma \\ BF = m(g + a) \\ BF = {\rho _b}V(g + a) \\\end{aligned} \end{equation} $$
Similarly when the lift is descending downwards,
From FBD, $$\begin{equation} \begin{aligned} W - BF = ma \\ BF = W - ma \\ BF = m(g - a) \\ BF = {\rho _b}V(g - a) \\\end{aligned} \end{equation} $$ Here from the above two cases we can formulate a general equation of buoyant force as, $$BF = {\rho _b}V{\overrightarrow g _{eff}}$$ where, ${\overrightarrow g _{eff}} = \left| {\overrightarrow g - \overrightarrow a } \right|$
