Chapter 9, Problem 43P

A 1.00-kg beaker containing 2.00 kg of oil (density = 916 kg/m³) rests on a scale. A 2.00-kg block of iron is suspended from a spring scale and is completely submerged in the oil (Fig. P9.31). Find the equilibrium readings of both scales.

Figure P9.31

To determine

The equilibrium readings of both scales.

Answer to Problem 43P

The equilibrium readings of both scales are $17.3\text{N}$ and $31.7\text{N}$.

Explanation of Solution

Section1:

To determine: The equilibrium reading of the upper scale.

Answer: The equilibrium reading of the upper scale is $17.3\text{N}$.

Explanation: The volume of the iron block is $V=m_{\text{iron}}/{\rho }_{\text{iron}}$ and the buoyancy force on the block is $B={\rho }_{\text{oil}}Vg={\rho }_{\text{oil}}\left(m_{\text{iron}}/{\rho }_{\text{iron}}\right)g$. Using Newton’s equations of motion at equilibrium condition, the upper scale reading would be ${\displaystyle \sum F_y}=0=F_{\text{upper}}+B-m_{\text{iron}}g\Rightarrow F_{\text{upper}}=m_{\text{iron}}\left[1-\left({\rho }_{\text{oil}}/{\rho }_{\text{iron}}\right)\right]g$.

Given info: The mass of iron block is $2.00\text{kg}$, acceleration due to gravity is $9.80{\text{m/s}}^{\text{2}}$, density of iron is $7.86\times {10}^3{\text{kg/m}}^{\text{3}}$, and the density of oil is $916{\text{kg/m}}^{\text{3}}$.

The formula for the upper scale reading is,

$F_{\text{upper}}=m_{\text{iron}}\left[1-\frac{{\rho }_{\text{oil}}}{{\rho }_{\text{iron}}}\right]g$

• $m_{\text{iron}}$ is mass of iron.
• $g$ is acceleration due to gravity.
• ${\rho }_{\text{iron}}$ is density of iron.
• ${\rho }_{\text{oil}}$ is density of oil.

Substitute $2.00\text{kg}$ for $m_{\text{iron}}$, $9.80{\text{m/s}}^{\text{2}}$ for $g$, density of oil is $7.86\times {10}^3{\text{kg/m}}^{\text{3}}$, $7.86\times {10}^3{\text{kg/m}}^{\text{3}}$ for ${\rho }_{\text{iron}}$, and $916{\text{kg/m}}^{\text{3}}$ for ${\rho }_{\text{iron}}$ to find $F_{\text{upper}}$.

$\begin{array}{c}{F}_{\text{upper}}=\left(2.00\text{kg}\right)\left(1-\frac{916{\text{kg/m}}^{\text{3}}}{7.86\times {10}^3{\text{kg/m}}^{\text{3}}}\right)\left(9.80{\text{m/s}}^{\text{2}}\right)\\ =17.3\text{N}\end{array}$

Thus, the equilibrium reading of the upper scale is $17.3\text{N}$.

Section2:

To determine: The equilibrium reading of the lower scale.

Answer: The equilibrium reading of the lower scale is $31.7\text{N}$.

Explanation: Use Newton’s equation of motion as $F_{\text{lower}}-B-\left(m_{\text{oil}}+m_{\text{beaker}}\right)g=0$ and it is rearranged for the lower scale reading is $F_{\text{lower}}=\left[{\rho }_{\text{oil}}\left(m_{\text{iron}}/{\rho }_{\text{iron}}\right)+\left(m_{\text{oil}}+m_{\text{beaker}}\right)\right]g$.

Given info: The mass of the beaker is $1.00\text{kg}$, mass of the oil is $2.00\text{kg}$, acceleration due to gravity is $9.80{\text{m/s}}^{\text{2}}$, mass of iron block is $2.00\text{kg}$, density of oil is $916{\text{kg/m}}^{\text{3}}$, and the density of iron is $7.86\times {10}^3{\text{kg/m}}^{\text{3}}$.

The formula for the lower scale reading is,

$F_{\text{lower}}=\left[m_{\text{iron}}\left({\rho }_{\text{oil}}/{\rho }_{\text{iron}}\right)+\left(m_{\text{oil}}+m_{\text{beaker}}\right)\right]g$

• $m_{\text{iron}}$ is mass of iron.
• $m_{\text{oil}}$ is mass of the oil.
• $m_{\text{beaker}}$ is mass of the beaker.
• $g$ is acceleration due to gravity.
• ${\rho }_{\text{iron}}$ is density of iron.
• ${\rho }_{\text{oil}}$ is density of oil.

Substitute $2.00\text{kg}$ for $m_{\text{iron}}$, $2.00\text{kg}$ for $m_{\text{oil}}$, $1.00\text{kg}$ for $m_{\text{beaker}}$, $9.80{\text{m/s}}^{\text{2}}$ for $g$, $7.86\times {10}^3{\text{kg/m}}^{\text{3}}$ for ${\rho }_{\text{iron}}$, and $916{\text{kg/m}}^{\text{3}}$ for ${\rho }_{\text{iron}}$ to find $F_{\text{lower}}$.

$\begin{array}{c}{F}_{\text{lower}}=\left[\left(2.00\text{kg}\right)\left(916{\text{kg/m}}^{\text{3}}/7.86\times {10}^3{\text{kg/m}}^{\text{3}}\right)+\left(2.00\text{kg}+1.00\text{kg}\right)\right]\left(9.80{\text{m/s}}^{\text{2}}\right)\\ =31.7\text{N}\end{array}$

Thus, the equilibrium reading of the lower scale is $31.7\text{N}$.

Conclusion:

Therefore, the equilibrium readings of both scales are $17.3\text{N}$ and $31.7\text{N}$.