两个力 $\mathbf{P}$ 和 $\mathbf{Q}$ 作用在一个质点上。

• $\mathbf{P}$ 的大小为 $10\,\mathrm{N}$，方向为正西。
• $\mathbf{Q}$ 的大小为 $8\,\mathrm{N}$，方向的方位角为 $330^\circ$。

已知 $\mathbf{F}=\mathbf{P}+\mathbf{Q}$，求 $\mathbf{F}$ 的大小。

方位角 $330^\circ$ 表示北偏西 $30^\circ$，所以 8 N 力有向西 4 N、向北 $4\sqrt3$ N 的分量，再与向西的 10 N 相加。

Take east and north as the positive coordinate directions. Since a bearing of $330^\circ$ is $30^\circ$ west of north, the components of $\mathbf Q$ are

Adding $\mathbf P$, the resultant has components $-14$ N eastwards and $4\sqrt3$ N northwards. Its magnitude is

两力同起点时夹角为 $60^\circ$。首尾相接的力三角形内角为其补角 $120^\circ$，可用余弦定理求合力。

The bearing of due west is $270^\circ$, so the angle between the two force directions is $330^\circ-270^\circ=60^\circ$.

In the head-to-tail force triangle, the sides of lengths 10 N and 8 N include the interior angle $120^\circ$. Therefore

Hence $\boxed{|\mathbf F|=2\sqrt{61}\ \mathrm N}$.