题目

Problem

Two forces $\mathbf{F}_1$ and $\mathbf{F}_2$ act on a particle. The resultant of the two forces is $\mathbf{R}$ .

Given that $\mathbf{R}$ has the same direction as the vector $(3\mathbf{i}-2\mathbf{j})$ ,

(a) find the size of the angle between $\mathbf{R}$ and the vector $\mathbf{j}$ .

(2)

Given that $\mathbf{F}_1=(a\mathbf{i}+3\mathbf{j})\,\mathrm{N}$ and $\mathbf{F}_2=(-4\mathbf{i}+b\mathbf{j})\,\mathrm{N}$ ,

(b) show that $2a+3b+1=0$ .

(4)

题目中文翻译

两个力 $\mathbf{F}_1$ 和 $\mathbf{F}_2$ 作用在一个质点上，它们的合力为 $\mathbf{R}$ 。

已知 $\mathbf{R}$ 与向量 $(3\mathbf{i}-2\mathbf{j})$ 的方向相同，

(a) 求 $\mathbf{R}$ 与向量 $\mathbf{j}$ 之间的夹角大小。

(2)

已知 $\mathbf{F}_1=(a\mathbf{i}+3\mathbf{j})\,\mathrm{N}$ ， $\mathbf{F}_2=(-4\mathbf{i}+b\mathbf{j})\,\mathrm{N}$ ，

(b) 证明 $2a+3b+1=0$ 。

(4)

解答

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合力向东南，与向北的单位向量夹角应是钝角。先求合力在东方向以下的角，再加上 $90^\circ$ 。

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The vector $3\mathbf{i}-2\mathbf{j}$ is directed south-east. Let $\theta$ be its angle below east. Then

\tan\theta=\frac23, \qquad \theta=33.690\ldots^\circ.

The angle between $\mathbf{R}$ and the northward vector $\mathbf{j}$ is

90^\circ+\theta=\boxed{124^\circ}\quad\text{(3 s.f.)}.

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合力是两力的和。同方向表示合力是给定向量的正倍数，不能把它们直接写成相等。用两个分量方程消去倍数。

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Adding the forces gives

\mathbf{R}=(a-4)\mathbf{i}+(3+b)\mathbf{j}.

Since $\mathbf{R}$ has the same direction as $3\mathbf{i}-2\mathbf{j}$ , there is a constant $k>0$ such that

a-4=3k,\qquad 3+b=-2k.

Multiplying the first equation by $2$ and the second by $3$ , then adding,

2(a-4)+3(3+b)=6k-6k=0.

Expanding and collecting terms,

2a-8+9+3b=0 \quad\Longrightarrow\quad \boxed{2a+3b+1=0}.