📚 Pulleys: Connected Particles, Tension and Acceleration | 滑轮：连接体、张力与加速度
Pulleys appear throughout Edexcel A-Level Mechanics, often as the core of connected-particles problems. By routing a light inextensible string over a smooth pulley, you link the motions of two bodies so that their accelerations and the string tension become deeply related. Mastering pulleys means grasping how Newton's laws apply simultaneously to each mass, using the constraint that the string length is constant. This article walks you through the key assumptions, standard problem types — horizontal surfaces, vertical Atwood setups, inclined planes — and the exam technique needed to score full marks on pulley questions.
滑轮在爱德思 A-Level 力学中反复出现，常作为连接体问题的核心。将一根轻质不可伸长的绳子跨过光滑滑轮，就把两个物体的运动联系在了一起，它们的加速度和绳中张力因此密切相关。掌握滑轮意味着理解如何对每个质量同时应用牛顿定律，同时利用绳子长度不变这一约束条件。本文带你梳理关键假设、常见题型——水平桌面、竖直阿特伍德机、斜面——以及确保滑轮题目拿满分的考试技巧。
1. What Makes Pulley Problems Tick | 滑轮问题的底层逻辑
At the heart of any pulley problem sits the principle that a light, inextensible string passing over a smooth pulley transmits tension unchanged and forces the connected particles to move with the same magnitude of acceleration. The pulley itself merely changes the direction of the tension force. This allows us to write two separate equations of motion — one for each mass — and then solve them simultaneously to find the common acceleration a and the string tension T. The direction of acceleration must be chosen consistently; usually you let the heavier side accelerate downwards and the lighter side accelerate upwards, or for a horizontal table, the hanging mass accelerates vertically downwards while the table mass accelerates horizontally.
任何滑轮问题的核心都基于这样一个原理：一根轻质、不可伸长的绳子跨过光滑滑轮时，张力大小不变地传递，并且迫使连接的各物体以相同大小的加速度运动。滑轮本身只改变拉力的方向。这使得我们可以对每个质量分别写出运动方程，然后联立求解共同的加速度 a 和绳中张力 T。加速度的方向必须前后一致地选定：通常让较重的一侧向下加速、较轻的一侧向上加速；如果是水平桌面，则悬挂的重物竖直向下加速，而桌面上的物体水平加速。
2. Key Assumptions You Must Memorise | 必须记住的关键假设
Edexcel pulley problems are always set under a specific set of modelling assumptions. Breaking any of these assumptions invalidates the simple relationships you rely on. Here are the five you will see in every mark scheme:
爱德思的滑轮题目总是基于一套特定的建模假设。违背其中任何一个，你所依赖的简单关系就不成立了。以下是每个评分标准中都会看到的五条：
| Assumption | What it means |
|---|---|
| Light string | Mass of string is zero → tension is uniform throughout |
| Inextensible string | String does not stretch → both particles move with the same magnitude of acceleration |
| Smooth pulley | No friction at the pulley → tension is the same on both sides |
| Light pulley | Rotational inertia of pulley is neglected → no energy lost to spinning the pulley |
| Particles | Bodies are treated as point masses → no rotational effects, only translation |
中文对应：轻绳——绳子质量为零，因此张力处处相等；不可伸长——绳子不拉伸，因此两物体加速度大小相等；光滑滑轮——滑轮处没有摩擦，两边张力相等；轻滑轮——忽略滑轮转动惯量，无能量损耗于旋转；质点——物体视为质点，只考虑平动。
3. The Universal Equation Block: Two Masses, One String | 通用方程框架：两个质量，一根绳
For any system with masses m₁ and m₂ connected by a light string over a smooth pulley, once you have chosen the positive direction of acceleration, applying Newton's second law to each mass gives a pair of equations. If m₂ is the heavier hanging mass and accelerates downwards, we write:
m₂g – T = m₂a
and for the other mass moving upwards:
T – m₁g = m₁a
Adding these eliminates T and yields the common acceleration a = (m₂ – m₁)g / (m₁ + m₂). Tension then follows by substituting a back into either equation.
对于任何由轻绳跨过光滑滑轮连接的质量 m₁ 和 m₂ 的系统，一旦选定了正加速度方向，对每个质量应用牛顿第二定律就可以得到一对方程。若 m₂ 是较重的悬挂质量并向下加速，则写出：m₂g – T = m₂a；另一个质量向上加速时：T – m₁g = m₁a。两式相加消去 T，得到共同加速度 a = (m₂ – m₁)g / (m₁ + m₂)。再将 a 代回任一方程即可求出张力。
4. Case I: One Mass on a Table, One Hanging | 情况一：一物在桌，一物悬挂
able, connected by a string passing over a pulley at the edge to a hanging mass B. Here, A' anced by the normal reaction, so the only horizontal force is T. For mass A we write T = mAa. For mass B, which moves downwards, mBg – T = mBa. Solving yields a = mBg / (mA + mB) and T = mAmBg / (mA + mB). If the table is rough, an extra term –μR appears in A's equation, where R = mAg.
经典装置是将物体 A 放在光滑水平桌面上，通过跨过桌面边缘滑轮的绳子连接悬挂的物体 B。此时 A 的重力被法向反作用力平衡，唯一的水平力是 T。对 A 有 T = mAa。对向下运动的 B 有 mBg – T = mBa。解得 a = mBg / (mA + mB)，T = mAmBg / (mA + mB)。若桌面粗糙，A 的方程中要加入 –μR 项，其中 R = mAg。
5. Case II: Atwood Machine – Both Masses Hanging Vertically | 情况二：阿特伍德机——两物均竖直悬挂
ften forget that both sides feel tension upwards, regardless of motion direction. The cleanest method is to draw a free-body diagram for each mass, label all forces, and assign a consistent acceleration direction. For masses m₁ (lighter) and m₂ (heavier), the equations are T – m₁g = m₁a (up positive) and m₂g – s less than the weight of the heavier mass and greater than the weight of the lighter mass — a quick self-check in exams.
当两个质量自由悬挂在滑轮两侧的绳子两端时，较重的一侧向下加速，较轻的一侧向上加速。学生常忘记无论运动方向如何，绳子对两个物体的拉力都是向上的。最清晰的方法是画出每个质量的受力图，标出所有力，并统一加速度方向。对于较轻的质量 m₁ 和较重的质量 m₂，方程是 T – m₁g = m₁a（向上为正）和 m₂g – T = m₂a（向下为正）。可以求出张力 T = 2m₁m₂g / (m₁ + m₂)。注意 T 总是小于较重物体的重量，大于较轻物体的重量——这在考试中是个快速自查的好方法。
6. Case III: Inclined Plane with a Pulley | 情况三：滑轮加上斜面
Edexcel loves combining pulleys with inclined planes. Suppose mass A sits on a smooth slope inclined at angle θ to the horizontal, connected via a pulley at the top of the slope to a hanging mass B. The component of A's weight down the slope is mAg sin θ, and you must decide which way the system accelerates. If B is heavy enough, it accelerates downwards, pulling A up the slope. A's equation up the slope: T – mAg sin θ = mAa. B's equation downwards: mBg – T = mBa. If friction is present, include –μR where R = mAg cos θ. Always resolve perpendicular to the slope to find R first.
爱德思考试局喜欢将滑轮与斜面结合起来。假设物体 A 放在与水平面成 θ 角的光滑斜面上，通过斜面顶端的滑轮连接悬挂物体 B。A 所受重力沿斜面的分量为 mAg sin θ，你必须判断系统朝哪个方向加速。若 B 足够重，它将向下加速，拉动 A 沿斜面向上。A 沿斜面向上的方程：T – mAg sin θ = mAa。B 向下的方程：mBg – T = mBa。如果有摩擦，要加上 –μR 项，其中 R = mAg cos θ。务必先沿垂直斜面方向分解求出 R。
7. The Towing Analogy: Horizontal Pulley Systems | 拖车类比：水平滑轮系统
via a horizontal rope that passes over a pulley. While the pulley may not be physically present, the idea is identical: the tension acts as an internal force linking the accelerations of two connected bodies. For a car of mass M pulling a trailer of mass m with a rope tension T, the system' ole system gives the common acceleration; isolating the trailer gives T. This is structurally the same as a table-and-hanging-mass problem, except gravity is replaced by an engine force.
有时候滑轮会出现在水平方向上，比如汽车通过一根水平绳索经滑轮牵引拖车。虽然滑轮不一定实际画出来，但其思路完全一样：张力作为连接两个物体加速度的内力。对于质量为 M 的汽车以绳索张力 T 拉动质量为 m 的拖车，系统所受的净力是驱动力减去阻力。对整个系统应用 F = ma 得到共同加速度；隔离拖车则可求出 T。这与桌面—悬挂质量问题的结构相同，只不过发动机力代替了重力。
8. The Importance of Free-Body Diagrams | 受力图的重要性
Examiners consistently report that students who skip drawing clear free-body diagrams lose marks by misassigning signs. A good diagram shows each particle separately, with all forces drawn as arrows originating from the particle's centre. Include tension T acting away from the mass, weight mg acting downwards, normal reaction R perpendicular to the surface, and friction if applicable. Then write the second-law equation according to the chosen positive direction. A diagram also helps you check that the tension force appears with the correct sign in both equations.
考官反复强调，跳过清晰受力图的学生往往会因为符号分配错误而丢分。一张合格的受力图将每个物体单独画出，所有力以箭头从物体中心出发：张力 T 背离物体，重力 mg 竖直向下，法向反作用力 R 垂直于表面，如有摩擦力也一并画出。然后根据选定的正方向写出牛顿第二定律方程。受力图还能帮你检查张力是否在两方程中都以正确的符号出现。
9. Worked Example – Atwood Machine with Numerical Values | 数值例题——阿特伍德机
string. Take g = 9.8 m s⁻². Let the 5 kg mass accelerate downwards. Equation for 5 kg: 5g – T = 5a. Equation for 3 kg: T – tuting a back: T = 3g + 3a = 3×9.8 + 3×2.45 = 36.75 N (or using the other equation: T = 5g – 5a = 49 – 12.25 = 36.75 N). The tension is the same both ways — a perfect validation.
一个滑轮系统有 3 kg 和 5 kg 两个质量竖直悬挂。求加速度和绳中张力。取 g = 9.8 m s⁻²。规定 5 kg 质量向下加速。对 5 kg：5g – T = 5a。对 3 kg：T – 3g = 3a。相加：2g = 8a → a = 2g/8 = g/4 = 2.45 m s⁻²。代回 a：T = 3g + 3a = 3×9.8 + 3×2.45 = 36.75 N（或用另一方程验证：T = 5g – 5a = 49 – 12.25 = 36.75 N）。张力结果一致——完美的验证。
10. Common Pitfalls and How to Avoid Them | 常见陷阱与规避方法
One pervasive error is assuming the tension equals the weight of the hanging mass. T is only equal to weight when the system is stationary or moving at constant velocity. The moment acceleration is non-zero, tension differs from weight. Another trap is mixing up the mass that accelerates up with the one that accelerates down, leading to sign inconsistencies. Always explicitly write 'equation of motion for m₁ (↑ positive)' before setting up equations. Finally, when a particle hits the ground or the string breaks, the subsequent motion is projectile-like or under new forces; treat that as a separate stage.
一个普遍错误是假设张力等于悬挂物体的重力。只有当系统静止或匀速运动时 T 才等于重量。一旦加速度不为零，张力就与重量不同。另一个陷阱是混淆向上加速和向下加速的质量，导致符号不一致。务必在列方程前明确写出“对 m₁ 的运动方程（↑为正）”。最后，当物体撞击地面或绳子断裂时，后续运动类似抛体运动或受新力作用；要将之作为单独阶段处理。
11. Examiner Tips and Graph Interpretation | 考官建议与图像解读
In multi-part pulley questions, you may be asked to sketch a velocity–time or acceleration–time graph. Remember that while the string remains taut and the pulley smooth, acceleration is constant. Graph segments are therefore straight lines. If the string breaks, acceleration changes instantaneously. Annotate your graphs with key values, such as the common acceleration or the instant when a mass hits the ground. Doing this not only answers the question but provides a built-in checking mechanism for your numerical answers.
在多部分的滑轮问题中，你可能需要绘制速度–时间或加速度–时间图像。要记住，只要绳子绷紧且滑轮光滑，加速度就恒定不变。因此图像段是直线。若绳子断裂，加速度会瞬间改变。在图像上标注关键数值，比如共同加速度或物体着地的时刻。这样做不仅能回答问题，还能为你的数值答案提供内置检查。
12. Putting It All Together for the Edexcel Exam | Edexcel 考试综合运用
Pulleys test your ability to model systems with internal constraints, apply Newton' grams, assign a consistent acceleration direction, write F = ma for each mass, solve simultaneously, and interpret the results. Practise past paper questions that mix pulleys with momentum, energy, or variable forces to build confidence. If you master the pulley framework, a whole class of mechanics problems becomes a source of straightforward marks.
滑轮问题考察你对含内部约束系统建模、对连接体应用牛顿第二定律以及从容处理不同几何构型的能力。步骤总是相同的：陈述假设，画受力图，指定一致的加速度方向，对每个质量写 F = ma，联立求解，并解释结果。多练习将滑轮与动量、能量或变力结合的历年考题以增强信心。一旦掌握滑轮框架，一整类力学题目就会成为你拿分的稳定来源。
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