Solving Connected Bodies Problems | 连接体问题解法

📚 Solving Connected Bodies Problems | 连接体问题解法

Connected bodies problems appear frequently in A-Level Mechanics. They involve two or more objects linked by a light string, a rod, or direct contact, moving under shared constraints. The key is to apply Newton’s second law carefully to each body and then combine the equations.

连接体问题是 A-Level 力学中的高频考点。题目通常涉及两个或多个物体通过轻绳、轻杆或相互接触相连，在共同约束下运动。解题的核心是对每个物体分别应用牛顿第二定律，再联立方程求解。

1. What Is a Connected Bodies Problem? | 什么是连接体问题？

A connected bodies problem describes a system where objects influence each other’s motion through tension, thrust, or contact forces. Common examples include a mass hanging from a pulley and pulling a block across a table, or a car towing a trailer.

连接体问题是指物体之间通过张力、推力或接触力相互影响运动的系统。常见例子包括：悬挂物绕过滑轮拉动桌面上的木块，或者汽车牵引拖车。

In all these situations, the motion of one object is constrained by the motion of another. If the string is inextensible, the magnitudes of their accelerations are equal. If the string passes over a smooth pulley, the tension is the same on both sides.

在这些情形中，一个物体的运动受到另一个物体的约束。如果绳子不可伸长，则各物体的加速度大小相等。如果绳子绕过光滑滑轮，则两侧张力相同。

The aim is usually to find the acceleration of the system, the tension in the string, or the force transmitted between objects.

解题目标通常是求系统的加速度、绳中的张力，或者物体之间传递的作用力。

2. Assumptions and Idealisation | 假设与理想化

A-Level problems rely on standard idealisations that simplify the maths while preserving the physics.

A-Level 题目依赖一些标准理想化假设，这些假设简化了数学计算，但不失物理本质。

Light string: the string has negligible mass, so tension is the same throughout it.
轻绳：绳子质量忽略不计，因此绳上张力处处相等。
Inextensible: the length of the string does not change, so all objects connected by it have equal acceleration magnitudes.
不可伸长：绳长不变，因此所有由绳连接的物体加速度大小相等。
Smooth pulley / peg: no friction at the pulley, so tension is unchanged when the string bends around it.
光滑滑轮 / 光滑钉：滑轮处无摩擦，绳子绕过时张力不改变。
Air resistance ignored: motion is treated as ideal unless the question states otherwise.
忽略空气阻力：除非题目另有说明，运动被视为理想状况。

Always state these assumptions when you define variables, and check whether the question introduces extra forces such as friction or resistance.

在设变量时应当说明这些假设，并注意题目是否额外给出摩擦力或阻力等条件。

3. Drawing Free-Body Diagrams | 受力分析

For each object, draw a separate diagram showing all forces acting on it. Label the weight \(W = mg\), normal reaction \(R\), tension \(T\), friction \(F\), and any applied force \(P\).

对每个物体分别画受力图，标出重力 \(W = mg\)、法向反力 \(R\)、张力 \(T\)、摩擦力 \(F\) 以及外加力 \(P\)。

Choose a positive direction for the whole system before writing equations. Often it is easiest to take the direction of acceleration as positive for each body, but be aware that different bodies may have different positive directions if they are on opposite sides of a pulley.

在列方程前先为整个系统规定正方向。通常取每个物体的加速度方向为正方向，但要注意：如果物体位于滑轮两侧，它们的正方向可能不同。

For a mass on a horizontal table connected to a hanging mass, the forces on the table block are: its weight, the normal reaction, the tension, and possibly friction. The forces on the hanging mass are its weight and the tension.

例如，水平桌面上的木块通过绳子连接一个悬挂物。木块受重力、支持力、张力和可能的摩擦力；悬挂物受重力和张力。

Once the diagrams are complete, write Newton’s second law for each direction.

受力图完成后，沿每个方向列出牛顿第二定律方程。

4. Applying Newton’s Second Law | 应用牛顿第二定律

For each body, the resultant force in the direction of motion is equal to mass times acceleration: \(F = ma\).

对每个物体，运动方向上的合力等于质量乘以加速度：\(F = ma\)。

Consider a block of mass \(m_1\) on a smooth horizontal table connected by a light string over a pulley to a hanging mass \(m_2\). The hanging mass accelerates downward; the block accelerates horizontally toward the pulley.

考虑一个质量 \(m_1\) 的木块在光滑水平桌上，通过轻绳绕过滑轮与悬挂质量 \(m_2\) 相连。悬挂物向下加速，木块水平向滑轮加速。

For the hanging mass \(m_2\), if we take downward as positive:

对悬挂物 \(m_2\)，取向下为正：

\(m_2g – T = m_2a\)

For the block \(m_1\), taking motion toward the pulley as positive:

对木块 \(m_1\)，取向滑轮方向为正：

\(T = m_1a\)

Notice that the acceleration \(a\) has the same magnitude for both bodies because the string is inextensible. The two equations contain two unknowns: \(a\) and \(T\).

由于绳子不可伸长，两物体的加速度大小 \(a\) 相同。两个方程包含两个未知量：\(a\) 和 \(T\)。

5. Solving for Acceleration and Tension | 求解加速度与张力

Add the two equations to eliminate \(T\):

将两方程相加消去 \(T\)：

\(m_2g = (m_1 + m_2)a\)

Hence the acceleration is:

因此加速度为：

\(a = \frac{m_2g}{m_1 + m_2}\)

Substitute \(a\) back into \(T = m_1a\) to find the tension:

将 \(a\) 代回 \(T = m_1a\) 求张力：

\(T = \frac{m_1m_2g}{m_1 + m_2}\)

As a check: if \(m_2\) is very large compared with \(m_1\), then \(a \approx g\) and \(T \approx m_1g\). This makes physical sense: the hanging mass falls almost freely, and the block’s acceleration is limited by the tension.

检验：若 \(m_2\) 远大于 \(m_1\)，则 \(a \approx g\)，\(T \approx m_1g\)。这符合物理直觉：悬挂物几乎自由下落，木块加速度受到张力限制。

6. Systems Moving in the Same Direction | 同方向运动的连接系统

Sometimes the connected bodies move in the same straight line, such as a car towing a trailer. In that case, the tension or thrust in the tow bar is internal to the system, but it appears when isolating one body.

有时连接体沿同一直线同向运动，例如汽车拖着拖车。此时拖杆中的张力或推力是系统内力，但隔离其中一个物体时就会出现。

Suppose a car of mass \(M\) tows a trailer of mass \(m\) with constant acceleration \(a\). The engine provides a driving force \(F\), and the tow bar exerts a tension \(T\) on the trailer. Air resistance is neglected.

设质量 \(M\) 的汽车以加速度 \(a\) 牵引质量 \(m\) 的拖车。发动机驱动力为 \(F\)，拖杆对拖车施加张力 \(T\)。忽略空气阻力。

For the system as a whole:

对整个系统：

\(F = (M + m)a\)

For the trailer alone:

对拖车单独分析：

\(T = ma\)

Thus the tension in the tow bar depends only on the trailer’s mass and the common acceleration. If friction or resistance is present, write separate equations including those forces.

因此拖杆拉力只取决于拖车质量和共同加速度。若存在摩擦或阻力，需在各自方程中加入这些力。

7. Pulley Systems and Atwood Machines | 滑轮系统与阿特伍德机

When two masses \(m_1\) and \(m_2\) hang vertically on opposite sides of a smooth pulley, the system is called an Atwood machine. If \(m_2 > m_1\), the heavier mass moves downward.

当两个质量 \(m_1\) 和 \(m_2\) 分别挂在光滑滑轮两侧竖直下垂时，称为阿特伍德机。若 \(m_2 > m_1\)，较重物体向下运动。

For \(m_2\), taking downward as positive:

对 \(m_2\)，取向下为正：

\(m_2g – T = m_2a\)

For \(m_1\), taking upward as positive:

对 \(m_1\)，取向上为正：

\(T – m_1g = m_1a\)

Adding the two equations gives:

两式相加得：

\((m_2 – m_1)g = (m_1 + m_2)a\)

Hence:

因此：

\(a = \frac{(m_2 – m_1)g}{m_1 + m_2}\)

Substitute back to obtain \(T\). Notice that if \(m_1 = m_2\), then \(a = 0\) and \(T = m_1g = m_2g\), so the system is in equilibrium.

代回可求得 \(T\)。注意若 \(m_1 = m_2\)，则 \(a = 0\)，\(T = m_1g = m_2g\)，系统平衡。

8. Inclined Planes and Friction | 斜面与摩擦力

Connected bodies may sit on inclined planes. In that case, resolve forces parallel and perpendicular to the plane.

连接体也可能位于斜面上。此时需要沿斜面方向和垂直于斜面方向分解力。

Consider a block of mass \(m_1\) on a rough plane inclined at angle \(\theta\), connected by a string over a pulley at the top to a hanging mass \(m_2\). The block tends to move up the plane as \(m_2\) descends.

设质量 \(m_1\) 的木块放在倾角 \(\theta\) 的粗糙斜面上，通过绳子绕过斜面顶端滑轮与悬挂质量 \(m_2\) 相连。当 \(m_2\) 下降时，木块沿斜面向上运动。

For the block, the weight component down the plane is \(m_1g\sin\theta\), the normal reaction is \(R = m_1g\cos\theta\), and friction is \(F = \mu R = \mu m_1g\cos\theta\).

对木块：重力沿斜面向下的分量为 \(m_1g\sin\theta\)，法向反力 \(R = m_1g\cos\theta\)，摩擦力 \(F = \mu R = \mu m_1g\cos\theta\)。

Taking up the plane as positive for the block:

对木块取沿斜面向上为正：

\(T – m_1g\sin\theta – F = m_1a\)

For the hanging mass, downward as positive:

对悬挂物，取向下为正：

\(m_2g – T = m_2a\)

Be careful: friction always opposes the direction of motion. If the block might move down the plane instead, the friction direction changes and the equation must be adjusted.

注意：摩擦力总是阻碍相对运动方向。若木块可能沿斜面向下运动，摩擦力的方向应反向，方程需相应调整。

9. Multi-Body Chains | 多物体链式连接

When three or more bodies are connected in a line, the force between any two bodies can be found by isolating the appropriate section of the chain.

当三个或更多物体沿直线相连时，求任意两个物体之间的作用力，需要隔离相应的一段链条。

Imagine three blocks \(A\), \(B\), and \(C\) in contact on a frictionless horizontal surface, pushed by a force \(P\) applied to \(A\). The common acceleration is \(a = P/(m_A + m_B + m_C)\).

设想光滑水平面上有三个紧挨着的木块 \(A\)、\(B\)、\(C\)，力 \(P\) 作用在 \(A\) 上推动它们。共同加速度为 \(a = P/(m_A + m_B + m_C)\)。

To find the contact force between \(B\) and \(C\), treat \(C\) alone:

求 \(B\) 与 \(C\) 之间的接触力，可单独隔离 \(C\)：

\(F_{BC} = m_C a\)

To find the force between \(A\) and \(B\), treat \(B\) and \(C\) together as a composite body:

求 \(A\) 与 \(B\) 之间的力，可将 \(B\) 和 \(C\) 看作一个整体：

\(F_{AB} = (m_B + m_C)a\)

This “section method” avoids the need to solve many simultaneous equations for internal forces.

这种“隔离段法”可以避免为内部力解大量联立方程。

10. Combining Dynamics with Kinematics | 动力学与运动学结合

Once acceleration is found, you can use SUVAT equations to determine displacement, velocity, or time. For uniform acceleration \(a\), with initial velocity \(u\), final velocity \(v\), displacement \(s\), and time \(t\):

求出加速度后，可以用运动学公式计算位移、速度或时间。对于匀加速度 \(a\)，初速度 \(u\)，末速度 \(v\)，位移 \(s\)，时间 \(t\)，有：

\(v = u + at\)

\(s = ut + \frac{1}{2}at^2\)

\(v^2 = u^2 + 2as\)

Example: in a pulley system, if the heavier mass starts from rest and we need the speed after it descends 2 m, use \(v^2 = u^2 + 2as\) with \(u = 0\), \(s = 2\), and the acceleration found from the dynamics equations.

例如：在滑轮系统中，较重物体从静止开始下落，求下落 2 m 后的速度。用 \(v^2 = u^2 + 2as\)，其中 \(u = 0\)，\(s = 2\)，加速度由动力学方程求得。

Remember that if the string becomes slack or a body hits the ground, the motion changes. You must then re-analyse the system with new forces or treat the remaining motion separately.

记住：如果绳子松弛或某一物体落地，运动将改变。此时必须重新分析受力，或对剩余运动单独处理。

11. Checking Signs and Limits | 检查符号与极限情形

A common source of errors is inconsistent sign conventions. Always decide the positive direction for each body and stick to it when substituting into equations.

常见错误来源是符号约定不一致。务必为每个物体明确正方向，并在代入方程时保持一致。

For a hanging object, if you choose downward as positive, then \(mg\) is positive and \(T\) is negative. If you choose upward as positive, the signs reverse. The final acceleration magnitude must be positive in the physical direction of motion.

对悬挂物体，若取向下为正，则 \(mg\) 为正、\(T\) 为负；若取向上为正，则符号相反。最终加速度大小必须在实际运动方向上为正。

Use limiting cases to verify results. If a mass becomes zero, the tension should tend to zero. If friction becomes huge, motion may cease and acceleration should become zero. These checks catch algebraic mistakes.

利用极限情形检验结果。若某质量趋于零，张力应趋于零；若摩擦力很大，运动可能停止，加速度应变为零。这些检验能发现代数错误。

Also check that tension is never greater than the weight of a hanging object that accelerates downward; otherwise the object would accelerate upward.

还要检查：向下加速的悬挂物所受张力不可能大于其重力；否则物体将向上加速。

12. Common Mistakes and Exam Tips | 常见错误与考试技巧

Many students lose marks because they forget to include all relevant forces or they apply the same acceleration direction to bodies moving in opposite coordinate directions.

许多学生丢分是因为漏掉某个力，或者对不同运动方向的物体误用了相同的坐标方向。

Here are the most important tips for solving connected bodies problems:

以下是解连接体问题最重要的建议：

Draw clear diagrams: separate diagrams for each body, with all forces labelled.
画清晰受力图：每个物体单独画图，标出所有力。
Use the same symbol \(a\) for the magnitude of acceleration of all bodies connected by an inextensible string.
使用相同符号 \(a\) 表示不可伸长绳连接的所有物体的加速度大小。
Write equations in terms of resultant force, not individual forces unless isolated.
按合力列方程，不要只写单个力，除非是隔离体。
Check whether the question asks for internal forces; if so, use the isolation method, never the whole-system equation.
注意题目是否要求内力；如果要求，必须用隔离法，不能只用整体方程。
Substitute units at the end to ensure answers are dimensionally correct.
最后代入单位，确保答案量纲正确。

Finally, practise with past paper questions. Connected bodies problems follow predictable patterns, and mastering the method will save you time and improve accuracy in the exam.

最后，多练习历年真题。连接体问题模式固定，掌握方法后既能节省时间，又能提高考试准确率。

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Solving Connected Bodies Problems | 连接体问题解法