题目

Problem

A parcel of mass $20\,\mathrm{kg}$ is at rest on a rough horizontal floor. The coefficient of friction between the parcel and the floor is $0.3$ .

Two forces, both acting in the same vertical plane, of magnitudes $200\,\mathrm{N}$ and $T\,\mathrm{N}$ are applied to the parcel. The line of action of the $200\,\mathrm{N}$ force makes an angle of $15^\circ$ with the horizontal and the line of action of the $T\,\mathrm{N}$ force makes an angle of $25^\circ$ with the horizontal, as shown in Figure 1. The parcel is modelled as a particle $P$ .

Find the smallest value of $T$ for which $P$ remains in equilibrium.

(9)

题目中文翻译

一个质量为 $20\,\mathrm{kg}$ 的包裹静止在粗糙水平地板上。包裹与地板之间的摩擦系数为 $0.3$ 。

两个大小分别为 $200\,\mathrm{N}$ 和 $T\,\mathrm{N}$ 的力作用在包裹上，两个力均位于同一竖直平面内。如图 1 所示， $200\,\mathrm{N}$ 的力的作用线与水平方向成 $15^\circ$ 角， $T\,\mathrm{N}$ 的力的作用线与水平方向成 $25^\circ$ 角。将包裹视为质点 $P$ 。

求使 $P$ 保持平衡的 $T$ 的最小值。

(9)

解答

解法一

思路

展开

$T$ 最小时，包裹恰好要被左侧的力拉向左，摩擦力向右并达到极限。两个斜向上的力都会减小支持力，所以必须把竖直平衡与水平平衡联立。

答题过程

展开

Let the normal reaction be $R$ and the friction be $F$ . At the minimum $T$ , the parcel is on the point of moving left, so friction acts right with $F=0.3R$ .

Vertical equilibrium gives

R+200\sin15^\circ+T\sin25^\circ=20g.

Horizontal equilibrium gives

200\cos15^\circ=T\cos25^\circ+F.

Eliminating $R$ and $F$ ,

\begin{align*} 200\cos15^\circ=&\,T\cos25^\circ\\[4mm] &\,+0.3\big(196-200\sin15^\circ-T\sin25^\circ\big). \end{align*}

Rearranging,

\begin{align*} T(\cos25^\circ-0.3\sin25^\circ) =&\,200\cos15^\circ-58.8\\[4mm] &\,+60\sin15^\circ. \end{align*}

Hence

T=\frac{200\cos15^\circ-58.8+60\sin15^\circ} {\cos25^\circ-0.3\sin25^\circ} =192.31\ldots\,\mathrm{N}.

The normal reaction from the vertical equation is positive, so contact is maintained. The minimum force is $\boxed{192\,\mathrm{N}}$ (3 s.f.).

M1 2021 Jan Q3

题目

解答

解法一

思路

答题过程