Chapter 13, Problem 109A

(a)

To determine

To Find: The percentage of the sphere that would be submerged if the sphere were released to float freely.

Explanation of Solution

Given:

The amount of force required to completely submerge the sphere, $F=700\text{N}$

Density of the sphere, $\rho =95{\text{kg/m}}^{\text{3}}$

Formula Used:

  $F_B={\rho }_{f}Vg$

Here, $F_B$ is the buoyant force, ${\rho }_f$ is the density of the fluid, V is the volume of the submerged part and g is the acceleration due to gravity.

Calculations:

Let the volume of sphere be $V_s$ and the volume of submerged portion be $V_{sub}$ .

Density of water, ${\rho }_w=1000{\text{kg/m}}^{\text{3}}$

For the sphere to float freely over water, the buoyant force must be equal to the weight of the sphere.

  $\begin{array}{l}{F}_B=w\\ {\rho }_f{V}_{sub}g=\rho V_{s}g\\ \Rightarrow \frac{V_{sub}}{V_s}=\frac{\rho }{{\rho }_f}\\ \Rightarrow \frac{V_{sub}}{V_s}=\frac{95}{1000}\\ \Rightarrow \frac{V_{sub}}{V_s}\%=\frac{95}{1000}\times 100\%=9.5\%\end{array}$

Conclusion:

Thus, the percentage of the sphere that would be submerged if the sphere were released to float freely is $9.5\%$ .

(b)

To determine

To Find: The weight of the sphere in the air.

Explanation of Solution

Given:

The amount of force required to completely submerge the sphere, $F=700\text{N}$

Density of the sphere, $\rho =95{\text{kg/m}}^{\text{3}}$

Formula Used:

  $F_B={\rho }_{f}Vg$

Here, $F_B$ is the buoyant force, ${\rho }_f$ is the density of the fluid, V is the volume of the submerged part and g is the acceleration due to gravity.

Calculations:

Let the mass of the sphere be m .

700 N force is required to keep the sphere fully submerged. This can be expressed in the form of following equation:

  $70\text{0}\text{N}+w=F_B$

  $\begin{array}{l}{F}_B={\rho }_w\times V_s\times g\\ \because V_s=\frac{m}{\rho }\\ \therefore F_B={\rho }_w\times \frac{m}{\rho }\times g\\ w=mg\\ \Rightarrow F_B=w\times \frac{{\rho }_w}{\rho }\end{array}$

  $\begin{array}{l}70\text{0}\text{N}+w=F_B\\ \Rightarrow 70\text{0}\text{N}+w=w\times \frac{{\rho }_w}{\rho }\\ \Rightarrow w=\frac{700\text{N}}{\left(\frac{{\rho }_w}{\rho }-1\right)}\\ \Rightarrow w=\frac{700\text{N}}{\left(\frac{1000}{95}-1\right)}=73.5\text{N}\end{array}$

Conclusion:

Thus, the weight of the sphere in the air is $73.5\text{N}$ .

(c)

To determine

To Find: The volume of the sphere.

Explanation of Solution

Given:

Weight of the sphere in the air, $w=73.5\text{N}$ .

Density of the sphere, $\rho =95{\text{kg/m}}^{\text{3}}$

Formula Used:

Volume of the sphere can be obtained by:

  $\begin{array}{l}{V}_s=\frac{m}{\rho }\\ m=\frac{w}{g}\end{array}$

Calculations:

Mass of the sphere is:

  $\begin{array}{l}m=\frac{w}{g}\\ m=\frac{73.5\text{N}}{9.8{\text{m/s}}^{\text{2}}}\\ m=7.\text{5}\text{kg}\end{array}$

Volume of the sphere can be calculated as follows:

  $\begin{array}{l}{V}_s=\frac{m}{\rho }\\ V_s=\frac{7.5\text{kg}}{95{\text{kg/m}}^{\text{3}}}\\ V_s=0.079{\text{m}}^{\text{3}}\end{array}$

Conclusion:

Thus, the volume of the sphere is $0.079{\text{m}}^{\text{3}}$ .