Dynamics Analysis of Pulley-Connected Objects | 滑轮连接物体的动力学分析

📚 Dynamics Analysis of Pulley-Connected Objects | 滑轮连接物体的动力学分析

Pulley systems are a classic topic in A-Level mechanics, testing your ability to apply Newton’s laws to connected particles. In this article, we will systematically analyze the dynamics of objects connected by strings passing over pulleys, from basic two-mass systems to more complex configurations.

滑轮系统是A-Level力学中的经典考点，考查你将牛顿定律应用于连接物体的能力。本文将系统分析通过绳子绕过滑轮连接的物体的动力学，从基本双物体系统到更复杂的配置。

1. Fundamental Assumptions | 基本假设

Before solving any pulley problem, we must establish the standard simplifications used in A-Level mechanics. These assumptions make the mathematics tractable while preserving physical accuracy for light, idealised systems.

在解决任何滑轮问题之前，我们必须确立A-Level力学中使用的标准简化假设。这些假设使得数学处理变得可行，同时对于轻质理想化系统保持物理准确性。

The string is inextensible, meaning its length does not change so particles connected by it move with equal acceleration.

绳子是不可伸长的，这意味着其长度不变，因此连接在其上的物体具有相同的加速度。
The string is light, so its mass is negligible and tension is constant throughout.

绳子是轻质的，因此其质量可忽略不计，张力处处相等。
The pulley is smooth, producing no friction, and the string does not slip.

滑轮是光滑的，不产生摩擦，绳子不打滑。
The string is taut at all times — slack strings transmit no tension.

绳子始终处于绷紧状态——松弛的绳子不传递张力。
Only the weight and tension act on each particle; air resistance is ignored.

每个物体仅受重力和张力作用；忽略空气阻力。

2. The Basic Two-Mass System | 基本双物体系统

Consider two particles of masses m₁ and m₂, with m₁ > m₂, connected by a light inextensible string passing over a smooth fixed pulley. The heavier mass m₁ will accelerate downward, while m₂ accelerates upward. We take the positive direction for each particle along its direction of motion.

考虑质量分别为 m₁ 和 m₂ 的两个物体（m₁ > m₂），通过一根轻质不可伸长绳子绕过光滑固定滑轮连接。较重的物体 m₁ 向下加速，而 m₂ 向上加速。我们取每个物体沿运动方向为正方向。

For m₁ moving downward, applying Newton’s second law:

对于向下运动的 m₁，应用牛顿第二定律：

m₁g − T = m₁a

For m₂ moving upward:

对于向上运动的 m₂：

T − m₂g = m₂a

Here, T is the tension in the string, g is the gravitational acceleration (9.8 m/s² on Earth), and a is the common acceleration magnitude.

其中 T 是绳子中的张力，g 是重力加速度（地球上为 9.8 m/s²），a 是共同加速度的大小。

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

Adding the two equations eliminates T, allowing us to solve directly for acceleration. The result is one of the fundamental formulas of connected-particle dynamics.

将两个方程相加可以消去 T，从而直接求解加速度。结果是连接物体动力学中最重要的公式之一。

a = (m₁ − m₂)g ÷ (m₁ + m₂)

Substituting this value back into either equation gives the tension:

将该值代回任一方程可求得张力：

T = 2m₁m₂g ÷ (m₁ + m₂)

Note that the tension is not equal to the weight of either mass; it lies between m₂g and m₁g. This is because the system is accelerating — the tension must both support m₂ and provide the net force that accelerates m₁.

注意张力不等于任何一方的重力；它介于 m₂g 和 m₁g 之间。这是因为系统在加速——张力既要支撑 m₂，又要提供加速 m₁ 的净力。

4. Physical Interpretation | 物理意义分析

Several limiting cases help build intuition. If m₁ = m₂, the acceleration is zero and the tension equals mg — the system is in equilibrium. If m₂ approaches zero, the acceleration approaches g, and the tension approaches zero, as expected for a freely falling body.

几个极限情形有助建立直观理解。若 m₁ = m₂，则加速度为零，张力等于 mg——系统处于平衡状态。若 m₂ 趋近于零，则加速度趋近于 g，张力趋近于零，这与自由落体的情况一致。

From Newton’s third law, the force exerted on the pulley by the string is 2T downward (if both sides hang vertically). This is important when determining the reaction force at the pulley support.

根据牛顿第三定律，绳子对滑轮施加的力为 2T 向下（若两侧均垂直悬挂）。这在确定滑轮支架处的反作用力时很重要。

5. Systems with Inclined Planes | 斜面系统

A common extended problem places one mass on a smooth inclined plane at angle θ to the horizontal. The string runs parallel to the plane up to the pulley, and the other mass hangs vertically. The component of weight acting down the plane is m₁g sin θ.

一个常见的扩展问题是将一个物体放置在倾角为 θ 的光滑斜面上。绳子平行于斜面延伸至滑轮，另一个物体垂直悬挂。沿斜面向下的重力分量为 m₁g sin θ。

For the mass m₁ on the plane, if it accelerates upward along the plane (i.e., toward the pulley):

对于斜面上的质量 m₁，若它沿斜面向上加速（即朝向滑轮）：

T − m₁g sin θ = m₁a

For the hanging mass m₂ moving downward:

对于向下运动的悬挂质量 m₂：

m₂g − T = m₂a

Adding these gives a = (m₂ − m₁ sin θ)g ÷ (m₁ + m₂). The tension follows by substitution. This analysis generalises naturally to rough planes by replacing m₁g sin θ with m₁g sin θ + μm₁g cos θ when friction opposes motion up the plane.

两式相加得 a = (m₂ − m₁ sin θ)g ÷ (m₁ + m₂)。通过代入可求得张力。此分析可自然推广到粗糙斜面，只需将 m₁g sin θ 替换为 m₁g sin θ + μm₁g cos θ（当摩擦力阻碍物体沿斜面向上运动时）。

6. Connected Particles Moving in Different Directions | 不同运动方向的连接物体

In some problems, both masses move along inclined planes or one moves horizontally on a table while the other hangs vertically over the edge. The method remains the same: define a positive direction for each particle, write Newton’s second law per particle, then solve the simultaneous equations.

在某些问题中，两个物体都沿斜面运动，或一个在桌面上水平运动而另一个在桌边垂直悬挂。方法保持不变：为每个物体定义正方向，对每个物体写出牛顿第二定律，然后联立求解。

Consider a mass m₁ on a smooth horizontal table connected by a string over a pulley at the edge to a hanging mass m₂. The hanging mass accelerates downward with a, and the table mass accelerates horizontally with the same a. The equations are T = m₁a and m₂g − T = m₂a, giving a = m₂g ÷ (m₁ + m₂).

考虑桌面上的质量 m₁，通过绕过桌边滑轮的绳子连接悬挂质量 m₂。悬挂物体以 a 向下加速，桌面物体以相同的 a 水平加速。方程为 T = m₁a 和 m₂g − T = m₂a，解得 a = m₂g ÷ (m₁ + m₂)。

7. Work–Energy Approach | 功能关系方法

When you are asked for velocities or distances rather than acceleration, the work–energy principle often provides a faster route. The total mechanical energy of the system (kinetic plus potential) is conserved if there is no friction. For a two-mass system, as m₁ descends a distance h, m₂ rises by h, and the common speed v satisfies:

当题目要求速度或距离而非加速度时，功能关系原理往往提供更快捷的路径。若无摩擦，系统的总机械能（动能加势能）守恒。对于双物体系统，当 m₁ 下降距离 h 时，m₂ 上升 h，共同速度 v 满足：

½(m₁ + m₂)v² = (m₁ − m₂)gh

This equation directly follows from the change in gravitational potential energy being converted into kinetic energy of both masses. This method avoids finding the tension entirely.

此方程直接源于重力势能的变化转化为两个物体的动能。这种方法完全避免了求张力。

Equivalently, from the kinematic equation v² = u² + 2as with initial speed u = 0 and acceleration a = (m₁ − m₂)g ÷ (m₁ + m₂), we obtain the same result — a good consistency check in exams.

等价地，由运动学方程 v² = u² + 2as，初速度 u = 0，加速度 a = (m₁ − m₂)g ÷ (m₁ + m₂)，我们得到相同结果——这是考试中很好的自检方法。

8. Common Pitfalls and Exam Tips | 常见错误与考试技巧

Students frequently lose marks on pulley questions due to sign errors. Always draw a clear free-body diagram and mark the direction of acceleration explicitly. For a string that changes direction over a pulley, do not try to assign a single global positive direction — use separate “positive along motion” directions per particle.

学生在滑轮题目中常因符号错误失分。务必画出清晰的受力分析图并明确标注加速度方向。对于经过滑轮改变方向的绳子，不要试图指定一个全局正方向——应分别为每个物体使用”沿运动方向为正”的方向。

Check units: masses in kg, forces in N, g = 9.8 m/s² unless told otherwise.

检查单位：质量用 kg，力的单位用 N，除非另有说明 g = 9.8 m/s²。
Distinguish mass from weight: weight = mg, always a force measured in newtons.

区分质量与重量：重量 = mg，始终是以牛为单位的力。
Identify the direction of acceleration before writing equations — the heavier side (or the side with the larger down-slope force) will move downward/forward.

写方程前先判定加速度方向——较重的一侧（或沿斜面合力较大的一侧）将向下/向前运动。
Remember tension is internal to the whole system; when using F = ma on the combined system, the net external force is the weight difference, not the sum of tensions.

注意张力是内力；对整体系统使用 F = ma 时，合外力是重量差，而非张力之和。

9. Worked Example | 例题详解

A particle A of mass 4 kg is connected by a light inextensible string over a smooth pulley to a particle B of mass 2 kg. The system is released from rest. Find the acceleration of the system and the tension in the string, taking g = 9.8 m/s².

质量 4 kg 的物体 A 通过轻质不可伸长绳子绕过光滑滑轮连接质量 2 kg 的物体 B。系统从静止释放。求系统的加速度和绳中张力，取 g = 9.8 m/s²。

Solution | 解题： For A (4 kg) descending: 4g − T = 4a. For B (2 kg) ascending: T − 2g = 2a. Adding gives 2g = 6a, so a = g ÷ 3 = 9.8 ÷ 3 = 3.27 m/s². Substituting back: T = 2g + 2a = 2 × 9.8 + 2 × 3.27 = 19.6 + 6.53 = 26.1 N.

解答： 对 A（4 kg）下降：4g − T = 4a。对 B（2 kg）上升：T − 2g = 2a。两式相加得 2g = 6a，故 a = g ÷ 3 = 9.8 ÷ 3 = 3.27 m/s²。代回：T = 2g + 2a = 2 × 9.8 + 2 × 3.27 = 19.6 + 6.53 = 26.1 N。

Check with the formula: T = 2m₁m₂g ÷ (m₁ + m₂) = 2 × 4 × 2 × 9.8 ÷ 6 = 156.8 ÷ 6 = 26.1 N ✓

用公式验证：T = 2m₁m₂g ÷ (m₁ + m₂) = 2 × 4 × 2 × 9.8 ÷ 6 = 156.8 ÷ 6 = 26.1 N ✓

A common follow-up asks for the speed after A has descended 1.5 m. Using v² = 2as: v² = 2 × 3.27 × 1.5 = 9.81, so v = 3.13 m/s. Alternatively, by energy conservation: ½(6)v² = (4 − 2)×9.8×1.5 = 29.4, giving v² = 9.8, v = 3.13 m/s. Both methods agree.

常见追问是求 A 下降 1.5 m 后的速度。利用 v² = 2as：v² = 2 × 3.27 × 1.5 = 9.81，故 v = 3.13 m/s。或者用能量守恒：½(6)v² = (4 − 2)×9.8×1.5 = 29.4，得 v² = 9.8，v = 3.13 m/s。两种方法结果一致。

10. Summary of Key Formulas | 关键公式总结

The table below consolidates the essential results for the basic pulley configurations encountered in A-Level examinations.

下表总结A-Level考试中遇到的常见滑轮配置的基本结果。

Configuration \| 配置	Acceleration a \| 加速度 a	Tension T \| 张力 T
Two hanging masses m₁ > m₂ \| 两个悬挂质量 m₁ > m₂	(m₁ − m₂)g ÷ (m₁ + m₂)	2m₁m₂g ÷ (m₁ + m₂)
m₁ on smooth slope θ, m₂ hanging \| m₁ 在光滑斜面 θ，m₂ 悬挂	(m₂ − m₁ sin θ)g ÷ (m₁ + m₂)	m₁g sin θ + m₁a
m₁ on smooth table, m₂ hanging \| m₁ 在光滑桌面，m₂ 悬挂	m₂g ÷ (m₁ + m₂)	m₁m₂g ÷ (m₁ + m₂)

Remember that these results assume the direction of motion is known consistent with the formulas. If in doubt, always return to first principles: draw the diagram, assign directions, and write Newton’s second law for each particle.

记住这些结果假设运动方向与公式中的假设一致。若有疑问，务必回到第一性原理：画图、标注方向、对每个物体写出牛顿第二定律。

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Dynamics Analysis of Pulley-Connected Objects | 滑轮连接物体的动力学分析