题目

Problem

A light elastic string $AB$ has natural length $l$ and modulus of elasticity $2mg$ .

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 and horizontal, by a force of magnitude $F$ . The line of action of the force and the elastic string lie in the same vertical plane. The direction of the force makes an angle $\alpha$ , where $\tan\alpha=\dfrac34$ , with the upward vertical, as shown in Figure 2.

Find, in terms of $l$ , the length $AB$ .

(6)

题目中文翻译

弹性轻绳 $AB$ 的原长为 $l$ ，弹性模量为 $2mg$ 。

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。大小为 $F$ 的力使质点保持平衡，弹性绳绷紧且水平。该力的作用线与弹性绳位于同一竖直平面内。如图 2 所示，力的方向与竖直向上方向成 $\alpha$ 角，其中 $\tan\alpha=\dfrac34$ 。

用 $l$ 表示长度 $AB$ 。

(6)

解答

解法一

思路

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外力的角度从竖直方向量起，因此其竖直分量为 $F\cos\alpha$ 。先由两方向平衡求张力，再用胡克定律求伸长；最后还要加回原长。

答题过程

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Let $T$ be the tension in the string. Resolving vertically and horizontally gives

F\cos\alpha=mg, \qquad F\sin\alpha=T.

Dividing the second equation by the first,

\frac{T}{mg}=\tan\alpha=\frac34, \qquad T=\frac34mg.

Let $e=AB-l$ be the extension. By Hooke’s law,

T=\frac{2mg}{l}e.

Consequently

\frac{2mg}{l}e=\frac34mg, \qquad e=\frac{3l}{8}.

The required length, rather than the extension, is

\boxed{AB=l+e=l+\frac{3l}{8}=\frac{11l}{8}.}

M3 2023 Jan Q2

题目

解答

解法一

思路

答题过程