题目

Problem

A light elastic string $AB$ has modulus of elasticity $2mg$ and natural length $ka$ , where $k$ is a constant.

The end $A$ of the elastic string is attached to a fixed point. The other end $B$ is attached to a particle of mass $m$ . The particle is held in equilibrium, with the elastic string taut, by a force that acts in a direction that is perpendicular to the string. The line of action of the force and the elastic string lie in the same vertical plane. The string makes an angle $\theta$ with the downward vertical at $A$ , as shown in Figure 2.

Given that the length $AB=\dfrac{21}{10}a$ and $\tan\theta=\dfrac34$ , find the value of $k$ .

(6)

题目中文翻译

弹性轻绳 $AB$ 的弹性模量为 $2mg$ ，原长为 $ka$ ，其中 $k$ 为常数。

弹性绳的 $A$ 端连接于固定点，另一端 $B$ 连接于质量为 $m$ 的质点。一个方向垂直于绳的力使质点保持平衡，弹性绳绷紧。力的作用线与弹性绳位于同一竖直平面内。如图 2 所示，绳与 $A$ 点处竖直向下方向成 $\theta$ 角。

已知长度 $AB=\dfrac{21}{10}a$ 且 $\tan\theta=\dfrac34$ ，求 $k$ 的值。

(6)

解答

解法一

思路

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沿绳方向分解可直接消去垂直于绳的外力。胡克定律的分母是原长 $ka$ ，分子中的伸长为实际长度减原长。

答题过程

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Let $T$ be the tension. The applied force is perpendicular to the string and therefore has no component parallel to it. Resolving towards $A$ along the string gives

T=mg\cos\theta.

The string makes an acute angle with the downward vertical, and $\tan\theta=3/4$ , so $\cos\theta=4/5$ . Hence

T=\frac45mg.

The extension of the string is

e=AB-ka=\left(\frac{21}{10}-k\right)a.

Using Hooke’s law with modulus $2mg$ and natural length $ka$ ,

\frac45mg=\frac{2mg}{ka}\left(\frac{21}{10}-k\right)a.

Cancel $mg$ and multiply by $k$ :

\frac45k=\frac{21}{5}-2k, \qquad \frac{14}{5}k=\frac{21}{5}.

Therefore

\boxed{k=\frac32.}

This gives extension $e=3a/5>0$ , consistent with a taut elastic string.

M3 2023 June Q2

题目

解答

解法一

思路

答题过程