Using a scale, a piece of alloy has a measured mass of 86 g...

Using a scale, a piece of alloy has a measured mass of $86 \mathrm{~g}$ in air and $73 \mathrm{~g}$ when immersed in water. Find its volume and its density. The apparent change in measured mass is due to the buoyant force of the water. Figure $13-7$ shows the situation when the object is in water. From the figure, $F_{B}+F_{T}=m g$, so $$F_{B}=(0.086)(9.81) \mathrm{N}-(0.073)(9.81) \mathrm{N}=(0.013)(9.81) \mathrm{N}$$ But $F_{B}$ must be equal to the weight of the displaced water. or $$ \begin{aligned} F_{B} &=\text { Weight of water }=(\text { Mass of water })(g) \\ &=(\text { Volume of water })(\text { Density of water })(g) \\ (0.013)(9.81) \mathrm{N} &=V\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right) \end{aligned} $$ from which $V=1.3 \times 10^{-5} \mathrm{~m}^{3}$. This is also the volume of the piece of alloy. Therefore, $$ \rho \text { of alloy }=\frac{\text { Mass }}{\text { Volume }}=\frac{0.086 \mathrm{~kg}}{1.3 \times 10^{-5} \mathrm{~m}^{3}}=6.6 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3} $$

Using a scale, a piece of alloy has a measured mass of $86 \mathrm{~g}$ in air and $73 \mathrm{~g}$ when immersed in water. Find its volume and its density.
The apparent change in measured mass is due to the buoyant force of the water. Figure $13-7$ shows the situation when the object is in water. From the figure, $F_{B}+F_{T}=m g$, so
$$F_{B}=(0.086)(9.81) \mathrm{N}-(0.073)(9.81) \mathrm{N}=(0.013)(9.81) \mathrm{N}$$
But $F_{B}$ must be equal to the weight of the displaced water.
or
$$
\begin{aligned}
F_{B} &=\text { Weight of water }=(\text { Mass of water })(g) \\
&=(\text { Volume of water })(\text { Density of water })(g) \\
(0.013)(9.81) \mathrm{N} &=V\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)
\end{aligned}
$$
from which $V=1.3 \times 10^{-5} \mathrm{~m}^{3}$. This is also the volume of the piece of alloy. Therefore,
$$
\rho \text { of alloy }=\frac{\text { Mass }}{\text { Volume }}=\frac{0.086 \mathrm{~kg}}{1.3 \times 10^{-5} \mathrm{~m}^{3}}=6.6 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}
$$

Key Concepts

Archimedes' Principle

Archimedes' Principle states that any object fully or partially immersed in a fluid experiences an upward buoyant force equal to the weight of the fluid displaced by the object. This principle is fundamental for understanding how the displaced fluid can be used to determine the volume of an object, as well as for explaining the difference in measured weights when an object is in air versus when it is submerged in a fluid.

Buoyant Force

Buoyant force is the upward force exerted by a fluid on an immersed object. It is equal to the weight of the fluid that the object displaces. This force accounts for the difference between an object's weight in air and its apparent weight in water, and it serves as a practical tool for measuring the volume of the object, since the magnitude of the buoyant force is directly related to the displaced fluid's volume and density.

Density

Density is a measure of mass per unit volume and is a crucial property of materials. In problems like this, once the volume of the object is determined from the buoyant force, the density can be calculated by dividing the object's mass by its volume. This concept is widely used in material science, fluid mechanics, and engineering to characterize substances and predict their behavior under various conditions.

Volume Determination via Fluid Displacement

The method of determining an object's volume through fluid displacement involves measuring the volume of fluid displaced by the object, which is directly proportional to the weight of the displaced fluid due to the buoyant force. This technique relies on the understanding that the displaced fluid's weight, when divided by its known density and gravitational acceleration, yields the object's volume. It is a practical application of the principles of buoyancy and Archimedes' Principle.

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