大小分别为 $17\,\mathrm{N}$、$51\,\mathrm{N}$ 和 $34\,\mathrm{N}$ 的共面力沿图示方向作用于点 $O$，其中 $\tan\alpha=\frac{15}{8}$。

(a) 求这三个力的合力的大小和方向。

将大小为 $51\,\mathrm{N}$ 的力替换为方向相同、大小为 $P\,\mathrm{N}$ 的力。此时三个力的合力沿 $x$ 轴正方向作用。

(b) 求 $P$ 的值。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

Where a numerical value for the acceleration due to gravity ($g$) is needed, use $10\,\mathrm{m\,s^{-2}}$.

需要使用重力加速度 $g$ 的数值时，取 $10\,\mathrm{m\,s^{-2}}$。

由8、15、17直角三角形确定正弦余弦。特别注意17N的力同时有向左和向下的分量。

Since $\tan\alpha=15/8$ and $\alpha$ is acute,

Taking rightwards and upwards as positive, the resultant components are

Therefore the magnitude is

Both components are positive. The angle above the positive $x$-axis is

也可沿34N方向及与之垂直、反向于17N的方向分解。最后必须把所得相对角换回与水平正向的夹角。

Resolve along the $34\,\mathrm{N}$ force and along the direction opposite to the $17\,\mathrm{N}$ force. These directions are perpendicular.

Since $\sin\alpha=15/17$ and $\cos\alpha=8/17$,

The components along the chosen directions are

Thus

The resultant makes an anticlockwise angle $\delta=\tan^{-1}(431/95)$ with the $34\,\mathrm{N}$ force. That force is $\alpha=\tan^{-1}(15/8)$ below the positive $x$-axis. Hence the resultant direction is

The resultant is $15.6^\circ$ above the positive $x$-axis.

合力沿水平正向，先令竖直分量为零，再检查水平分量确实为正。

For a horizontal resultant,

Thus $15P=510+136=646$, giving

The horizontal component is $8P/17+16-15=8P/17+1>0$, so the resultant acts in the positive $x$-direction.