Solution

Engineering Mechanics (Dynamics)

Chapter 12

Exercise 220

Given: $v_{B/w} = 16$ km/h, $\vec{v}_W = (-4 \hat{j})$ km/h. Boat's velocity is:

$$\begin{align*} \vec{v}_B &= \vec{v}_W + \vec{v}_{B/W} \\ v_B \cos70\text{\textdegree} \hat{i} + v_B \sin70\text{\textdegree} \hat{j} &= - 4 \hat{j} + 16\sin\theta \hat{i} + 16\cos\theta \hat{j} \\\\ v_B \cos70\text{\textdegree} &= 16 \sin\theta \quad \Rightarrow \quad v_B = \frac{16\sin\theta}{\cos70\text{\textdegree}}\\ v_B\sin70\text{\textdegree} &= -4 + 16\cos\theta\\\\ \frac{16\sin\theta}{\cos70\text{\textdegree}} \cdot \sin70\text{\textdegree} &= -4 +16\cos\theta \\ 43.96 \sin\theta &= -4 + 16\cos\theta \\ 43.96 \cdot \sqrt{1-cos^2\theta} &= -4 + 16\cos\theta \bigg/^2 \\ 1932.5 (1-cos^2\theta) &= 16 - 128 \cos\theta + 256 \cos^2\theta \\ 2188.5 \cos^2\theta -128 \cos\theta& -1916.5 = 0 \bigg/ : 2188.5 \\ \cos^2\theta - 0.0585 \cos\theta& - 0.876 = 0\\ \cos\theta_{1,2} &= \frac{0.0585 \pm \sqrt{0.00342+3.5}}{2} \\ &= \frac{0.0585\pm1.881}{2}\\ \cos\theta_1 &= 0.97 \qquad \theta_1 = 14.13 \text{\textdegree} \\ \cos\theta_2 &= -0.911 \qquad \theta_2 = 155.68 \text{\textdegree} \end{align*}$$

The bearing angle $\theta$ needed for the boat to stay on course is $14.1\text{\textdegree}$

$\text{Answer}$

$\theta =14.1\text{°}$.