两个质点 $A$ 和 $B$ 的质量分别为 $m$ 和 $3m$，连接在一根轻且不可伸长的细绳两端。质点 $A$ 被保持静止在粗糙水平桌面上。

质点 $A$ 与桌面之间的摩擦系数为 $\frac{1}{5}$。细绳沿桌面放置，并绕过固定在桌子边缘的一个光滑轻质小滑轮。如图 4 所示，质点 $B$ 静止在与水平面成 $\alpha$ 角的粗糙斜面上，其中 $\tan\alpha=\frac{4}{3}$。质点 $B$ 与斜面之间的摩擦系数为 $\frac{1}{3}$。细绳位于包含滑轮及斜面一条最大坡度线的竖直平面内。系统从静止释放，细绳保持绷紧，$B$ 沿斜面下滑。已知 $A$ 未到达滑轮，

(a) 求细绳中的张力，

(b) 指出在计算的哪一步使用了“将细绳视为轻绳”这一假设，

(c) 求细绳对滑轮施加的力的大小。

绳绷紧且不可伸长，两质点加速度大小相同。分别求支持力和摩擦力，再沿运动方向列牛顿第二定律；摩擦力均与各自运动方向相反。

Since $\tan\alpha=4/3$ and $\alpha$ is acute,

Let the common acceleration magnitude be $a$ and the tension be $T$. For $A$, the normal reaction and friction are

For $B$, resolving perpendicular to the plane gives

$A$ accelerates towards the pulley, while $B$ accelerates down the plane. Their equations of motion are

Adding (1) and (2),

Substituting into (1),

也可以先将两个质点作为整体，沿各自运动方向合并方程，消去内力张力；再单独研究 $A$ 求张力。支持力和摩擦力的计算与解法一相同。

Using $F_A=mg/5$ and $F_B=3mg/5$ as obtained in method 1, the equation for the connected system is

Therefore

For $A$ alone, $T-F_A=ma$, so

对应到前面的计算，指出哪一个未知量因轻绳模型而能在两端共用。

The same tension $T$ was used in the equations for both particles: the tension is uniform throughout the light string passing over the smooth pulley.

绳对滑轮的两个拉力沿两段绳指向外侧，夹角为 $180^\circ-\alpha$。先取水平、竖直分量再合成；不要把两个张力的大小直接相加。

The horizontal string pulls the pulley to the right with force $T$. The inclined string pulls downwards and to the left, with components $T\sin\alpha$ downwards and $T\cos\alpha$ to the left.

Thus the resultant components are

Its magnitude is

两个大小相等的力，其合力沿夹角平分线。将每个张力向平分线分解，垂直于平分线的分量互相抵消。

The angle between the two forces of magnitude $T$ is $180^\circ-\alpha$. Hence

Since $\alpha$ is acute,

Using (a),