题目

Problem

A particle of weight $W$ newtons lies at rest on a rough horizontal surface, as shown in Figure 3.

A force of magnitude $P$ newtons is applied to the particle.

The force acts at an angle $\alpha$ to the horizontal, where $\tan\alpha=\frac{4}{3}$ .

The coefficient of friction between the particle and the surface is $\frac{1}{4}$ .

Given that the particle does not move, show that

P\leq\frac{5W}{8}

(7)

题目中文翻译

如图 3 所示，一个重 $W$ 牛顿的质点静止在粗糙水平面上。

对质点施加一个大小为 $P$ 牛顿的力。

该力与水平方向成 $\alpha$ 角，其中 $\tan\alpha=\frac{4}{3}$ 。

质点与平面之间的摩擦系数为 $\frac{1}{4}$ 。

已知质点不运动，证明

P\leq\frac{5W}{8}

(7)

解答

展开

原图中的外力斜向下压，所以支持力大于重力。静止时实际摩擦力等于水平分力，但它只能小于或等于最大静摩擦力，不能直接假设达到极限。

展开

Since $\tan\alpha=4/3$ , $\sin\alpha=4/5$ and $\cos\alpha=3/5$ .

Let $R$ be the normal reaction and $F$ the frictional force. Vertical and horizontal equilibrium give

R=W+P\sin\alpha=W+\frac{4P}{5},\qquad F=P\cos\alpha=\frac{3P}{5}.

For the particle to remain at rest, $F\leq\mu R=R/4$ . Therefore

\begin{align*} \frac{3P}{5}\leq&\,\frac14\left(W+\frac{4P}{5}\right),\\ \frac{2P}{5}\leq&\,\frac W4,\\ P\leq&\,\boxed{\frac{5W}{8}}, \end{align*}

as required.