When a body is fully or partially immersed in a fluid, it experiences an upthrust which is equal to weight of the fluid displaced by the body.
Suppose a body has weight of W newton in air and volume V. If the body is immersed in a liquid, let the weight of the body in it be W_{1}. Then
Loss in weight of the body in liquid or buoyancy = W-W_{1 }
According to Archimedes’ principle thrust weight of displaced fluid
upthrust= W-W_{1}
Experimental Verification
Take a body which is not soluble in a liquid and weight it in air with a spring balance as shown in the figure. Let its weight be W_{1}. Weigh the body again by immersing the body in the liquid in a beaker without touching the sides and bottom of the beaker. Let the weight be W_{2} in the liquid.
The difference in weight = W_{1} – W_{2}
After noting the initial level of a liquid in a measuring cylinder, immerse the body in the liquid and then note the final reading level in it. The difference in liquid levels given in the volume of displaced liquid which is equal to the volume of the body. The weight of displaced liquid = Vρg where ρ is the density of the fluid. It can be observed that
W_{1 -}W_{2}= V ρ g
In this way, Archimedes’ principle is verified.
If a body is floating in a liquid, its weight W acts vertically downward and the upthrust U due to displaced liquid acts vertically upward. If the weight of the body of volume V is greater than its upthrust, the body will sink in the liquid and will lie at the bottom of the container. So,
\begin{align*} W &> U \\ \text {or,} \: \rho Vg &> V\rho _l g \\ \text {or,} \: \rho &>\rho _l \\\end{align*}
Hence, a body will sink in a liquid as shown in the figure. If the density of the body is greater than the density of the liquid. In case, W=U, the body just sinks and remains inside the liquid with its upper surface near the liquid surface as shown in the figure. So,
\begin{align*} W &=U \\ \text {or,} \: \rho Vg &= V\rho _l g \\ \text {or,} \: \rho &=\rho _l \\\end{align*}
If the weight of the body is smaller than the upthrust, W<U, the body will float on the surface of liquid. So,
\begin{align*} W &< U \\ \text {or,} \: \rho Vg &< V\rho _l g \\ \text {or,} \: \rho &<\rho _l \\\end{align*}
A body will float in a liquid if its density is smaller than the density of the liquid. For example, the density of cork is smaller than that of water, and a cork floats in water. This is shown in figure. So, a floating body displaces the liquid of its own weight. Then,
\begin{align*} mg &= U \\ \text {or,} \: V\rho g &= V_1\rho _1 g \\ \text {or,} \: V\rho &= V_l\rho _1 \\ \text {or,} \: V_1/V &= \rho \rho _l \\ \end{align*}
Where V_{l} is volume of the displaced liquid = the volume of the body inside the liquid. So, for the floating body,
When a body is floating in a liquid, its weight acting vertically downward is equal to the upthrust acting vertically upward.
At the equilibrium, the centre of gravity C.G. of the body and centre of buoyancy C.B. of the displaced liquid both lie on the vertical axis.
If the floating body is slightly titled from its equilibrium position, then the C.G and C.B will not lie on the same vertical line, as C.B. shifts away.
The point of intersection of the vertical line passing through C.B. and original vertical line is called the Meta Centre, M.C. of the floating body on the liquid.
Condition that the body regains its equilibrium position or falls in the liquid depends upon the position of M.C. and C.G. of the body. These two possible cases are discussed below.
Three Possible Cases
Examples of Floatation
Suppose a body has weight of W newton in air and volume V. If the body is immersed in a liquid, let the weight of the body in it be W_{1}. Then
Loss in weight of the body in liquid or buoyancy = W-W_{1 }
_{ }According to Archimedes’ principle thrust weight of displaced fluid
upthrust= W-W_{1}
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ASK ANY QUESTION ON Archimedes’ Principle, Principle of Flotation and Equilibrium of Floating bodies
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