Question

A beaker of mass ${\mathit{m}}_{\mathbf{b}}$ containing oil of mass ${\mathit{m}}_{\mathbf{o}}$ and density $\mathit{\rho }$rests on a scale. A block of iron of mass ${\mathit{m}}_{\mathbf{F}\mathbf{e}}$ suspended from a spring scale is completely submerged in the oil as shown in Figure P14.71. Determine the equilibrium readings of both scales.

Short answer

The equilibrium readings of both scales is$n=\left(m_b+m_o+\left(\frac{{\rho }_0}{{\rho }_{Fe}}\right)m_{Fe}\right)g$

Step-by-step solution

A concept:

The pressure in a fluid at rest is given by:

$\mathit{P}\mathbf{=}\mathit{\rho }\mathit{g}\mathit{h}$

….. (1)

Where, $P$ is the pressure of the fluid, $\rho$ is the Density of fluid, $g$ is the gravitational constant, and $h$ is the high attained by the fluid.

When an object is partially or fully submerged in a fluid, the fluid exerts on the object an upward force called the buoyant force.

According to Archimedes’s principle, the magnitude of the buoyant force is equal to the weight of the fluid displaced by the object:

$B={\rho }_{fluid}Vg$

….. (2) Here, $\mathit{V}$ is the volume and $\mathit{g}$is gravity.

Determine the equilibrium readings of both scales:

Looking first at the top scale and the iron block, you have

$T+B=F_{g,iron}$

….. (1)

Where, $T$ is the tension in the spring scale, $B$ is the buoyant force, and $F_{g,Fe}$ is the weight of the iron block.

The buoyant force exerted on the iron block by the oil is,

$$\begin{gathered} B={\rho }_0{V}_{Fe}g \\ ={\rho }_0\left(\frac{m_{Fe}}{{\rho }_{Fe}}\right)g \end{gathered}$$

Rearrange equation (1) and substitute known equation.

$$\begin{gathered} T=F_{g,Fe}-B \\ =m_{Fe}g-{\rho }_0\left(\frac{m_{Fe}}{{\rho }_{Fe}}\right)g \end{gathered}$$

Or

$T=\left(1-\frac{{\rho }_o}{{\rho }_{Fe}}\right)m_{Fe}g$

This is the reading on the top scale.

Now, consider the bottom scale, which exerts an upward force of n on the beaker-oil-iron combination.

$$\begin{gathered} \underset{}{\overset{}{\sum F=0}} \\ T+n-F_{g,beaker}-F_{g,oil}-F_{g,Fe}=0 \\ n=(m_{beaker}+m_{oil}+m_{iron})g-\left(1-\frac{{\rho }_o}{{\rho }_{Fe}}\right)m_{Fe}g \end{gathered}$$

Or the reading of the bottom scale is,

$n=\left(m_b+m_o+\left(\frac{{\rho }_0}{{\rho }_{Fe}}\right)m_{Fe}\right)g$