题目

Problem

A particle, $P$ , of mass $km$ lies on a fixed rough plane. The plane is inclined to the horizontal at an acute angle $\theta$ . A horizontal force of magnitude $X$ acts on $P$ , as shown in Figure 1. The line of action of the force lies in the vertical plane which contains the line of greatest slope of the inclined plane that passes through $P$ . The coefficient of friction between $P$ and the inclined plane is $\mu$ .

When $X=mg$ , the particle $P$ is in equilibrium and on the point of sliding down the plane.

(a) Show that

\mu=\frac{k\tan\theta-1}{k+\tan\theta}

(10)

(b) Deduce that, when $k=1$ , $\theta$ must be greater than $45^\circ$ .

(2)

题目中文翻译

一个质量为 $km$ 的质点 $P$ 位于固定粗糙斜面上。斜面与水平面成锐角 $\theta$ 。如图 1 所示，一个大小为 $X$ 的水平力作用在 $P$ 上。该力的作用线位于包含经过 $P$ 点的斜面最大坡度线的竖直平面内。 $P$ 与斜面之间的摩擦系数为 $\mu$ 。

当 $X=mg$ 时，质点 $P$ 处于平衡状态，且恰好要沿斜面向下滑动。

(a) 证明

\mu=\frac{k\tan\theta-1}{k+\tan\theta}

(10)

(b) 推断：当 $k=1$ 时， $\theta$ 必须大于 $45^\circ$ 。

(2)

解答

(a)

解法一

思路

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将要向下滑，所以摩擦力沿斜面向上且达到极限值。水平推力不仅有沿斜面向上的分量，也有压向斜面的分量。消去支持力与摩擦力后，再把正弦余弦之比化成正切。

答题过程

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Let the normal reaction be $R$ and the limiting friction be $F$ . Since the particle is about to slide down, $F$ acts up the plane and $F=\mu R$ .

Resolving perpendicular to the plane, with $X=mg$ ,

R=kmg\cos\theta+mg\sin\theta.

Resolving parallel to the plane,

F+mg\cos\theta=kmg\sin\theta,

F=mg(k\sin\theta-\cos\theta).

Using $F=\mu R$ and cancelling $mg$ gives

\mu(k\cos\theta+\sin\theta)=k\sin\theta-\cos\theta.

As $\theta$ is acute, $\cos\theta>0$ . Dividing numerator and denominator by $\cos\theta$ ,

\begin{align*} \mu=&\,\frac{k\sin\theta-\cos\theta}{k\cos\theta+\sin\theta}\\[4mm] =&\,\frac{k\sin\theta/\cos\theta-1}{k+\sin\theta/\cos\theta}\\[4mm] =&\,\boxed{\frac{k\tan\theta-1}{k+\tan\theta}}. \end{align*}

解法二

思路

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也能沿水平和竖直方向分解。支持力指向左上方，摩擦力沿斜面指向右上方；用极限摩擦关系先消去两个力，再整理。

答题过程

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Let $R$ be the normal reaction and $F=\mu R$ the limiting friction up the plane. Horizontal and vertical equilibrium give

mg+F\cos\theta=R\sin\theta,

R\cos\theta+F\sin\theta=kmg.

Substituting $F=\mu R$ ,

mg=R(\sin\theta-\mu\cos\theta),

kmg=R(\cos\theta+\mu\sin\theta).

Multiply the first equation by $k$ and compare with the second. Cancelling the positive reaction $R$ gives

k\sin\theta-k\mu\cos\theta=\cos\theta+\mu\sin\theta.

Collecting terms in $\mu$ ,

\mu(k\cos\theta+\sin\theta)=k\sin\theta-\cos\theta.

Dividing through and then dividing numerator and denominator by $\cos\theta>0$ ,

\mu=\frac{k\sin\theta/\cos\theta-1}{k+\sin\theta/\cos\theta} =\boxed{\frac{k\tan\theta-1}{k+\tan\theta}}.

(b)

解法一

思路

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粗糙斜面有 $\mu>0$ 。代入 $k=1$ 后，先确认分母为正，再推出分子必须为正；锐角区间内正切单调递增。

答题过程

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Putting $k=1$ in (a),

\mu=\frac{\tan\theta-1}{1+\tan\theta}>0.

Since $0<\theta<90^\circ$ , the denominator is positive. Therefore

\tan\theta-1>0, \qquad \tan\theta>1=\tan45^\circ.

Hence $\boxed{\theta>45^\circ}$ .

M1 2019 Oct Q4

题目

解答

(a)

解法一

思路

答题过程

解法二

思路

答题过程

(b)

解法一

思路

答题过程