题目

Problem

A uniform beam $AB$ , of mass $15\,\mathrm{kg}$ and length $6\,\mathrm{m}$ , rests with end $A$ on rough horizontal ground. The end $B$ of the beam rests against a rough vertical wall.

The beam is inclined at $75^\circ$ to the ground, as shown in Figure 2.

The coefficient of friction between the beam and the wall is $0.2$

The coefficient of friction between the beam and the ground is $\mu$

The beam is modelled as a uniform rod which lies in a vertical plane perpendicular to the wall.

The beam rests in limiting equilibrium.

(a) Find the magnitude of the normal reaction between the beam and the wall at $B$ .

(5)

(b) Find the value of $\mu$

(6)

题目中文翻译

质量为 $15\,\mathrm{kg}$ 、长为 $6\,\mathrm{m}$ 的均匀梁 $AB$ ，其 $A$ 端位于粗糙水平地面上， $B$ 端靠在粗糙竖直墙上并静止。

如图 2 所示，梁与地面成 $75^\circ$ 角。

梁与墙之间的摩擦系数为 $0.2$ 。

梁与地面之间的摩擦系数为 $\mu$ 。

将梁视为位于垂直于墙面的竖直平面内的均匀杆。

梁处于极限平衡状态。

(a) 求梁与墙在 $B$ 点之间的法向反作用力大小。

(5)

(b) 求 $\mu$ 的值。

(6)

解答

(a)

解法一

思路

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梁有向下滑动的趋势，所以墙面摩擦向上、地面摩擦向墙。对 $A$ 取矩消去地面两项力；墙面的支持力和摩擦力都有力矩。

答题过程

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Let the wall’s normal reaction be $R_B$ , acting horizontally away from the wall, and its frictional force be $F_B$ upwards. In limiting equilibrium,

F_B=0.2R_B.

The weight $15g$ acts at the midpoint of the beam. Taking moments about $A$ ,

15g(3\cos75^\circ)=R_B(6\sin75^\circ)+F_B(6\cos75^\circ).

Substituting $F_B=0.2R_B$ ,

R_B=\frac{15g(3\cos75^\circ)}{6\sin75^\circ+1.2\cos75^\circ}.

Therefore the normal reaction at the wall is

\boxed{R_B=18.7\text{ N}}.

(b)

解法一

思路

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地面支持力竖直向上，摩擦水平向墙。先利用平衡求两项，再用地面处的极限摩擦关系；不要把墙与地面的两个摩擦系数混用。

答题过程

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Let the ground’s normal reaction be $R_A$ upwards and its frictional force be $F_A$ towards the wall.

Resolving horizontally,

F_A=R_B.

Resolving vertically,

R_A+F_B=15g,

R_A=15g-0.2R_B.

In limiting equilibrium at the ground,

F_A=\mu R_A.

Hence

\mu=\frac{R_B}{15g-0.2R_B}.

Using the unrounded value of $R_B$ from (a),

\boxed{\mu=0.130}.

M2 2023 June Q5

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程