题目

Problem

A uniform rod $AB$ has length $4\,\mathrm{m}$ and weight $50\,\mathrm{N}$ .

The rod has its end $A$ on rough horizontal ground. The rod is held in equilibrium at an angle $\alpha$ to the ground by a light inextensible cable attached to the rod at $B$ , as shown in Figure 2. The cable and the rod lie in the same vertical plane and the cable is perpendicular to the rod. The tension in the cable is $T$ newtons.

Given that $\sin\alpha=\frac35$

(a) show that $T=20$

(3)

Given also that the rod is in limiting equilibrium,

(b) find the value of the coefficient of friction between the rod and the ground.

(6)

题目中文翻译

均匀杆 $AB$ 长 $4\,\mathrm{m}$ ，重 $50\,\mathrm{N}$ 。

杆的 $A$ 端位于粗糙水平地面上。如图 2 所示，一条连接于杆的 $B$ 点的不可伸长轻缆绳，使杆与地面成 $\alpha$ 角并保持平衡。缆绳与杆位于同一竖直平面内，且缆绳垂直于杆。缆绳中的张力为 $T$ 牛顿。

已知 $\sin\alpha=\frac35$ ，

(a) 证明 $T=20$ 。

(3)

又知杆处于极限平衡状态，

(b) 求杆与地面之间的摩擦系数。

(6)

解答

(a)

解法一

思路

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张力垂直于杆，关于 $A$ 的力臂就是整根杆长。重量已给出牛顿值，不要再乘 $g$ ；其作用点在中点。

答题过程

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The weight acts at the midpoint of the uniform rod, 2 m from $A$ . Since the cable is perpendicular to the rod, taking moments about $A$ gives

4T=50(2\cos\alpha).

As $\sin\alpha=3/5$ and $\alpha$ is acute, $\cos\alpha=4/5$ . Therefore

4T=50(2)\left(\frac45\right)=80,

and hence

\boxed{T=20\text{ N}}.

(b)

解法一

思路

展开

张力向左上方，地面摩擦力应向右。地面支持力为竖直分量，须扣除张力提供的向上分量。最后使用极限平衡条件。

答题过程

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Let the frictional force at $A$ be $F$ to the right and the normal reaction be $R$ upwards. Resolving horizontally,

F=T\sin\alpha=20\left(\frac35\right)=12\text{ N}.

Resolving vertically,

R+T\cos\alpha=50,

R=50-20\left(\frac45\right)=34\text{ N}.

In limiting equilibrium, $F=\mu R$ . Therefore

12=34\mu,

and

\boxed{\mu=\frac6{17}}.

M2 2022 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