## Physics (10th Edition)

$F_w=18.45N$ Its direction is $68^o$ north of east.
1) First, we find the acceleration of the boat $(a_b)$ Taking east to be $+x$ and north to be $+y$ direction, we rewrite the initial and final velocity in unit-vector notation: $$\vec{v}_i=(2\cos15m/s)i+(2\sin15m/s)j$$ $$\vec{v}_i=(1.93m/s)i+(0.52m/s)j$$ $$\vec{v}_f=(4\cos35m/s)i+(4\sin35m/s)j$$ $$\vec{v}_f=(3.28m/s)i+(2.29m/s)j$$ So, $$\Delta \vec{v}=\vec{v}_f-\vec{v}_i=(1.35m/s)i+(1.77m/s)j$$ As $\Delta t=30s$, we have $$\vec{a_b}=\frac{\Delta\vec{v}}{\Delta t}=(0.045m/s^2)i+(0.059m/s^2)j$$ 2) The total net force acting on the boat $\sum F$ The mass of the boat $m_b=325kg$. According to Newton's 2nd Law: $$\sum \vec{F}=m_b\vec{a_b}=(14.63N)i+(19.18N)j$$ 3) We rewrite the three forces acting on the boat in unit-vector notation: $$\vec{F_1}=(31\cos15N)i+(31\sin15N)j=(29.94N)i+(8.02N)j$$ $$\vec{F_2}=-(23\cos15N)i-(23\sin15N)j=-(22.22N)i-(5.95N)j$$ (its direction is $15^o$ south of west, opposite from positive directions) Since $\sum \vec{F}=\vec{F_1}+\vec{F_2}+\vec{F_w}$, $$\vec{F_w}=\sum \vec{F}-\vec{F_1}-\vec{F_2}$$ $$\vec{F_w}=(6.91N)i+(17.11N)j$$ - Magnitude: $F_w=\sqrt{6.91^2+17.11^2}=18.45N$ - Direction: take $\theta$ to be the angle with respect to due east. $$\tan\theta=\frac{F_{w, j}}{F_{w, i}}=2.476$$ $$\theta=68^o$$