Chapter 3, Problem 96P

(a)

To determine

The force exerted by the wind on the sail.

Answer to Problem 96P

The force exerted by the wind on the sail is $730\text{N}$.

Explanation of Solution

Write the equation for the force exerted by the wind on the sail.

    $F_{\text{wind}}=\frac{1}{2}\rho A{v}^2$        (I)

Here, $F_{\text{wind}}$ is the force exerted by the wind on the sail, $\rho$ is the density of air, $A$ is the area of the triangular sail and $v$ is the velocity of the air with respect to the sail.

Write the equation for the area of the triangular sail.

    $A=\frac{1}{2}bh$        (II)

Here, $b$ is the breadth of the triangular sail and $h$ is the height of the triangular sail.

 Substitute equation (II) in equation (I).

    $F_{\text{wind}}=\frac{1}{2}\rho \left(\frac{1}{2}bh\right)v^2$

Conclusion:

Substitute $1.3{\text{kg/m}}^{\text{3}}$ for $\rho$, $5.0\text{m}$ for $b$, $10\text{m}$ for $h$ and $15\text{mi/h}$ for $v$.

     $\begin{array}{c}{F}_{\text{wind}}=\frac{1}{2}\left(1.3{\text{kg/m}}^{\text{3}}\right)\left(\frac{1}{2}\left(5.0\text{m}\right)\left(10\text{m}\right)\right){\left(\text{6}\text{.7m/s}\right)}^2\\ =730\text{N}\end{array}$ 

Therefore, the force exerted by the wind on the sail is $730\text{N}$.

(b)

To determine

The drag force due to water.

Answer to Problem 96P

The drag force due to water is $-730\text{N}$.

Explanation of Solution

Write the equation for the net force acting using Newton’s second law.

    $\begin{array}{c}\Sigma F=F_{\text{wind}}+F_{\text{water}}=0\\ F_{\text{wind}}=-F_{\text{water}}\end{array}$        (III)

Here, $F_{\text{water}}$ is the drag force due to water and $F_{\text{wind}}$ is the force exerted by the wind on the sail.

Conclusion:

Substitute $730\text{N}$ for $F_{\text{wind}}$.

    $F_{\text{water}}=-730\text{N}$

Therefore, the drag force due to water is $-730\text{N}$.

(c)

To determine

The increase in speed of the boat.

Answer to Problem 96P

The speed of the boat increases by $4$ times when the wind speed is doubled.

Explanation of Solution

Combine equation (I) and (III).

    $F_{\text{wind}}=-F_{\text{water}}=\frac{1}{2}\rho A{v}_{\text{wind}}^2$        (IV)

Write the equation for the drag force of the water using Stokes’ equation.

    $F_{\text{water}}=-Cr{v}_{\text{boat}}$        (V)

Combine equation (IV) and (V).

    $\frac{1}{2}\rho A{v}_{\text{wind}}^2=Cr{v}_{\text{boat}}$        (VI)

Rewrite equation (VI) when the speed of the wind is doubled.

      $\frac{1}{2}\rho A{\left(2v_{\text{wind}}\right)}^2=Cr\left(v_{\text{boat,new}}\right)$        (VII)

Conclusion:

Find the ratio of equation (VI) to equation (VII).

    $\begin{array}{c}\frac{\frac{1}{2}\rho A{\left(2v_{\text{wind}}\right)}^2}{\frac{1}{2}\rho A{v}_{\text{wind}}^2}=\frac{Cr\left(v_{\text{boat,new}}\right)}{Cr{v}_{\text{boat}}}\\ \frac{v_{\text{boat,new}}}{v_{\text{boat}}}=4\end{array}$

Therefore, the speed of the boat increases by $4$ times when the wind speed is doubled.