题目

Problem

A uniform beam $AB$ , of weight $5W$ and length $12a$ , rests with end $A$ on rough horizontal ground. A package of weight $W$ is attached to the beam at $B$ . The beam rests in equilibrium on a smooth horizontal peg at $C$ , with $AC=9a$ , as shown in Figure 5.

The beam is inclined at an angle $\theta$ to the ground, where $\tan\theta=\frac5{12}$

The beam is modelled as a rod that lies in a vertical plane perpendicular to the peg. The package is modelled as a particle.

The normal reaction between the beam and the peg at $C$ has magnitude $kW$

Using the model,

(a) show that $k=\frac{56}{13}$

(4)

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

Given that the beam is resting in limiting equilibrium,

(b) find the value of $\mu$

(6)

题目中文翻译

重为 $5W$ 、长为 $12a$ 的均匀梁 $AB$ ，其 $A$ 端位于粗糙水平地面上并静止。一个重为 $W$ 的包裹连接于梁的 $B$ 点。如图 5 所示，梁在 $C$ 点支承于光滑水平销钉上并处于平衡状态，其中 $AC=9a$ 。

梁与地面成 $\theta$ 角，其中 $\tan\theta=\frac5{12}$ 。

将梁视为位于垂直于销钉的竖直平面内的细杆。将包裹视为质点。

梁与销钉在 $C$ 点之间的法向反作用力大小为 $kW$ 。

利用此模型，

(a) 证明 $k=\frac{56}{13}$ 。

(4)

$A$ 与地面之间的摩擦系数为 $\mu$ 。

已知梁处于极限平衡状态，

(b) 求 $\mu$ 的值。

(6)

解答

(a)

解法一

思路

展开

光滑销钉的反力垂直于梁，力臂为 $9a$ 。梁重为 $5W$ 且在中点作用，端点另有 $W$ ；这里 $W$ 已是重量，不再乘 $g$ 。

答题过程

展开

From $\tan\theta=5/12$ ,

\sin\theta=\frac5{13},\qquad \cos\theta=\frac{12}{13}.

The peg reaction $kW$ is perpendicular to the beam. Taking moments about $A$ ,

5W(6a\cos\theta)+W(12a\cos\theta)=kW(9a).

Cancelling $Wa$ and substituting $\cos\theta=12/13$ ,

30\left(\frac{12}{13}\right)+12\left(\frac{12}{13}\right)=9k.

Thus

9k=\frac{504}{13},

and hence

\boxed{k=\frac{56}{13}}.

(b)

解法一

思路

展开

销钉反力向左上方，地面摩擦因此向右。水平、竖直分解后用极限摩擦 $F=\mu R$ ；地面支持力不是总重 $6W$ ，还要扣除销钉反力的竖直分量。

答题过程

展开

Let the ground reaction at $A$ be $R$ vertically upwards and the frictional force be $F$ horizontally to the right.

Resolving horizontally,

F=kW\sin\theta=\frac{56}{13}W\frac5{13}=\frac{280}{169}W.

Resolving vertically,

R+kW\cos\theta=5W+W.

Hence

R=6W-\frac{56}{13}W\frac{12}{13}=\frac{342}{169}W.

In limiting equilibrium, $F=\mu R$ , so

\mu=\frac{F}{R}=\frac{280/169}{342/169}=\frac{280}{342}.

Therefore

\boxed{\mu=\frac{140}{171}}.

M2 2024 Oct Q6

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程