Chapter 5, Problem 74P

To determine

The ratio of the weight of the twin in the boat to the weight of his brother on solid ground.

Answer to Problem 74P

The ratio of the weight of the twin in the boat to the weight of his brother on solid ground is $\underset{\_}{0.9999}$.

Explanation of Solution

Write the expression for the density of the lake.

    ${\rho }_{\text{lake}}=\frac{m_{\text{lake}}}{V}$        (I)

Here, ${\rho }_{\text{lake}}$ is the density of lake, $m_{\text{lake}}$ is the mass of lake, $V$ is the volume.

Write the expression for the density of Earth.

    ${\rho }_{\text{Earth}}=\frac{m_{\text{Earth}}}{V}$        (II)

Here, ${\rho }_{\text{Earth}}$ is the density of Earth, $m_{\text{Earth}}$ is the mass of Earth.

Write the expression for the force due to lake on the twin.

    $F_{\text{lake}}=\frac{G{m}_{\text{lake}}{m}_{\text{twin}}}{r^2}$        (III)

Here, $F_{\text{lake}}$ is the force due to the lake acting on twin, $G$ is the universal gravitation constant, $m_{\text{twin}}$ is the mass of twin, $r$ is the distance.

Write the expression for the force due to Earth on the twin.

    $F_{\text{Earth}}=\frac{G{m}_{\text{Earth}}{m}_{\text{twin}}}{r^2}$        (IV)

Here, $F_{\text{Earth}}$ is the force due to Earth acting on the twin.

Use equation (II) and (I) to solve for the ratio of force of gravity due to the lake on the twin to the Earth acting on the twin.

    $\begin{array}{c}\frac{F_{\text{lake}}}{F_{\text{Earth}}}=\frac{\frac{G{m}_{\text{lake}}{m}_{\text{twin}}}{r^2}}{\frac{G{m}_{\text{Earth}}{m}_{\text{twin}}}{r^2}}\\ =\frac{m_{\text{lake}}}{m_{\text{Earth}}}\\ =\frac{{\rho }_{\text{lake}}}{{\rho }_{\text{Earth}}}\end{array}$        (V)

Use equation (V) to solve for the $F_{\text{Earth}}$.

    $F_{\text{Earth}}=\left(\frac{{\rho }_{\text{lake}}}{{\rho }_{\text{Earth}}}\right)F_{\text{lake}}$        (VI)

Write the expression for $r$.

    $r=\frac{d}{2}$        (VII)

Here, $d$ is the distance between the deepest point on the lake on the twin.

Use equation (VII) to write the expression for the $V$.

    $V=\frac{4}{3}\pi {\left(\frac{d}{2}\right)}^3$        (VIII)

Use equation (VIII) in (I) to solve for $m_{\text{lake}}$.

    $m_{\text{lake}}={\rho }_{\text{lake}}\frac{4}{3}\pi {\left(\frac{d}{2}\right)}^3$        (IX)

Use equation (IX) and (VII) in (IV) to solve for $F_{\text{Earth}}$.

    $\begin{array}{c}{F}_{\text{Earth}}=\frac{G{\rho }_{\text{lake}}\frac{4}{3}\pi {\left(\frac{d}{2}\right)}^3{m}_{\text{twin}}}{{\left(\frac{d}{2}\right)}^2}\\ =G{\rho }_{\text{lake}}\frac{4}{3}\pi \left(\frac{d}{2}\right)m_{\text{twin}}\end{array}$        (X)

Write the expression for the difference in the force due to Earth to the lake acting on the twins.

    $\Delta F=F_{\text{Earth}}-F_{\text{lake}}$        (XI)

Here, $\Delta F$ is the difference between the force due to Earth to the lake acting on the twins.

Write the expression for the ratio of total weight experienced by the twins.

    $\frac{W_{\text{water}}}{W_{\text{Earth}}}=\frac{m_{\text{twin}}g-\Delta F}{m_{\text{twin}}g}$        (XII)

Use equation (XI) in (XII) to solve for $\frac{W_{\text{water}}}{W_{\text{Earth}}}$.

    $\frac{W_{\text{water}}}{W_{\text{Earth}}}=\frac{m_{\text{twin}}g-\left(F_{\text{Earth}}-F_{\text{lake}}\right)}{m_{\text{twin}}g}$        (XIII)

Conclusion:

Substitute $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^2$ for $G$, $1000\text{kg}/{\text{m}}^3$ for ${\rho }_{\text{lake}}$, $1600\text{m}$ for $d$, $70\text{kg}$ for $m_{\text{twin}}$ in equation (X) to find $F_{\text{Earth}}$.

    $\begin{array}{c}{F}_{\text{Earth}}=\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^2\right)\left(1000\text{kg}/{\text{m}}^3\right)\frac{4}{3}\pi \left(\frac{1600\text{m}}{2}\right)\left(70\text{kg}\right)\\ =0.016\text{N}\end{array}$

Substitute $1000\text{kg}/{\text{m}}^3$ for ${\rho }_{\text{lake}}$, $6000\text{kg}/{\text{m}}^3$ for ${\rho }_{\text{Earth}}$, $0.016\text{N}$ for $F_{\text{lake}}$ in equation (VI) to solve for $F_{\text{Earth}}$.

    $\begin{array}{c}{F}_{\text{Earth}}=\left(\frac{1000\text{kg}/{\text{m}}^3}{6000\text{kg}/{\text{m}}^3}\right)\left(0.016\text{N}\right)\\ =\left(\frac{1}{6}\right)\left(0.016\text{N}\right)\\ =0.094\text{N}\end{array}$

Substitute $0.094\text{N}$ for $F_{\text{Earth}}$, $0.016\text{N}$ for $F_{\text{lake}}$, $70\text{kg}$ for $m_{\text{twin}}$, $9.8\text{m}/{\text{s}}^2$ for $g$ in equation (XIII) to find $\frac{W_{\text{water}}}{W_{\text{Earth}}}$.

    $\begin{array}{c}\frac{W_{\text{water}}}{W_{\text{Earth}}}=\frac{\left(70\text{kg}\right)\left(9.8\text{m}/{\text{s}}^2\right)-\left(0.094\text{N}-0.016\text{N}\right)}{\left(70\text{kg}\right)\left(9.8\text{m}/{\text{s}}^2\right)}\\ =0.9999\end{array}$

Therefore, the ratio of the weight of the twin in the boat to the weight of his brother on solid ground is $\underset{\_}{0.9999}$.