📚 Newton’s Laws of Motion | 牛顿运动定律
Newton’s Laws of Motion form the bedrock of mechanics in AQA A-level Mathematics. Understanding these three laws allows you to analyse forces, predict motion, and solve problems involving particles, pulleys, and inclined planes. This article breaks down each law, its mathematical formulation, and common exam applications, ensuring you are fully prepared for both straightforward and multi‑step questions.
牛顿运动定律是 AQA A‑level 数学中力学的基石。掌握这三条定律可以让你分析受力、预测运动，并解决质点、滑轮和斜面问题。本文逐一拆解每条定律、数学表达式以及常见考试应用，帮助你为直接计算题和多步骤综合题做好充分准备。
1. Force and the Newton | 力与牛顿
Before stating the laws, recall that force is a vector quantity measured in newtons (N). One newton is the force required to accelerate a mass of 1 kg at 1 m s⁻². In AQA mechanics all forces must be resolved into components, and the net force determines acceleration.
在陈述定律之前，记得力是矢量，单位是牛顿（N）。1 牛顿是将 1 kg 质量以 1 m s⁻² 加速所需的力。在 AQA 力学中，所有力都必须分解为分量，净力决定了加速度。
2. Newton’s First Law – Inertia | 牛顿第一定律——惯性
An object remains at rest or in uniform motion in a straight line unless acted upon by a resultant external force. This law introduces the concept of equilibrium: when the resultant force is zero, velocity is constant (which includes zero velocity).
任何物体都保持静止或匀速直线运动状态，除非有合外力迫使它改变。这条定律引入了平衡的概念：当合力为零时，速度恒定（包括速度为零的情况）。
In exam questions, if a particle is moving at constant speed or is stationary, you can immediately set the vector sum of all forces to zero. This is the foundation for resolving forces in static or dynamic equilibrium problems.
在考试题中，如果质点匀速运动或静止，你可以立即将所有力的矢量和设为零。这是解决静态或动态平衡问题中分解力的基础。
3. Newton’s Second Law – F = ma | 牛顿第二定律——F = ma
The acceleration of an object is directly proportional to the resultant force and inversely proportional to its mass. Mathematically, this is expressed as:
物体的加速度与合外力成正比，与其质量成反比。数学上表达为：
F = m a
where F is the resultant force in newtons, m is mass in kilograms, and a is acceleration in m s⁻². This vector equation means that acceleration is always in the direction of the resultant force.
其中 F 是合力，单位牛顿；m 是质量，单位千克；a 是加速度，单位 m s⁻²。这个矢量方程意味着加速度的方向总是与合力的方向一致。
To apply the second law, you must first identify all forces acting on a body, resolve them parallel and perpendicular to the direction of motion (or intended motion), and then use ΣF = ma in the direction of acceleration. Perpendicular to motion, ΣF = 0 if there is no acceleration in that direction.
应用第二定律时，必须首先确定作用在物体上的所有力，沿运动方向（或预期运动方向）和垂直方向分解，然后在加速度方向上使用 ΣF = ma。在垂直于运动的方向上，如果没有加速度，合力为零。
4. Newton’s Third Law – Action and Reaction | 牛顿第三定律——作用与反作用
If body A exerts a force on body B, then body B exerts an equal and opposite force on body A. These forces are of the same type, have the same magnitude, act along the same line, but on different bodies.
如果物体 A 对物体 B 施加一个力，那么物体 B 同时对物体 A 施加一个大小相等、方向相反的力。这两个力类型相同、大小相等、作用在同一直线上，但作用在不同的物体上。
This law is critical when analysing connected particles, contact forces, and tension in strings. Remember: the ‘equal and opposite’ forces do not cancel because they act on different objects; hence they never appear together in a free‑body force diagram for one particle.
在分析连接体、接触力和绳子的张力时，这条定律至关重要。记住：‘大小相等、方向相反’的力不会相消，因为它们作用在不同物体上；因此它们永远不会同时出现在一个质点的受力分析图中。
5. Free‑Body Diagrams and Resultant Forces | 受力图与合力
A free‑body diagram shows a single particle (or body) and all the forces acting on it. Only include forces that act on that body, not forces exerted by it. Label each force with its type: weight (mg), normal reaction (R), tension (T), friction (F), driving force, etc.
受力图展示单个质点（或物体）及其受到的所有力。仅包括作用在该物体上的力，而不包括它施加的力。用力的类型标明每个力：重力 (mg)、法向反作用力 (R)、张力 (T)、摩擦力 (F)、牵引力等。
From the diagram, choose a positive direction, usually the direction of acceleration, and resolve forces. This step is essential before applying ΣF = ma. For systems involving inclined planes, resolve weight into components parallel and perpendicular to the plane.
根据受力图，选择一个正方向（通常是加速度方向），然后分解力。这一步在应用 ΣF = ma 之前必不可少。对于涉及斜面的系统，需将重力分解为平行和垂直于斜面的分量。
6. Resolving Forces on an Inclined Plane | 斜面上的力分解
Consider a particle of mass m on a smooth plane inclined at angle θ to the horizontal. Weight mg acts vertically downwards. The components are:
考虑一个质量为 m 的质点放在倾角为 θ 的光滑斜面上。重力 mg 竖直向下。分量如下：
Parallel to plane: m g sin θ
Perpendicular to plane: m g cos θ
The normal reaction R balances the perpendicular component: R = mg cos θ. If the plane is smooth, the only force causing acceleration down the slope is mg sin θ, so a = g sin θ.
法向反作用力 R 平衡垂直分量：R = mg cos θ。如果斜面光滑，唯一引起沿斜面下滑加速度的力是 mg sin θ，因此 a = g sin θ。
When friction F acts up the slope opposing motion or tendency, the resultant force parallel to the plane becomes mg sin θ − F, and the second law gives mg sin θ − F = ma.
当摩擦力 F 沿斜面向上阻止运动或运动趋势时，平行于斜面的合力变为 mg sin θ − F，第二定律给出 mg sin θ − F = ma。
7. Friction and Limiting Equilibrium | 摩擦力与极限平衡
Friction opposes relative motion or the tendency to move. For two dry surfaces, the maximum (limiting) friction is given by Fmax = μR, where μ is the coefficient of friction and R is the normal reaction. This relation holds only when the object is on the point of sliding.
摩擦力阻止相对运动或运动趋势。对于两个干燥表面，最大（极限）摩擦力由 Fmax = μR 给出，其中 μ 是摩擦系数，R 是法向反作用力。这个关系仅在物体即将滑动时成立。
If an object is in equilibrium, the frictional force can be any value up to μR, not necessarily at its maximum. Always check whether the question states ‘limiting equilibrium’ or ‘on the point of moving’ before using F = μR.
如果物体处于平衡状态，摩擦力可以是任意不超过 μR 的值，不一定等于最大值。在使用 F = μR 之前，务必检查题目是否写明‘极限平衡’或‘即将运动’。
8. Connected Particles – Tension and Pulleys | 连接体——张力和滑轮
A common AQA setup involves two particles connected by a light inextensible string passing over a smooth pulley. Because the string is light (mass‑free) and inextensible, the tension is the same throughout and the accelerations of the particles have the same magnitude.
一种常见的 AQA 题型涉及两个质点，通过一根轻质且不可伸长的绳子跨过光滑滑轮相连。因为绳子轻质（无质量）且不可伸长，绳上张力处处相等，两质点的加速度大小相同。
For each particle, draw a separate free‑body diagram, apply ΣF = ma in the direction of motion, and solve the simultaneous equations. For example, for a mass m₁ descending and m₂ ascending (m₁ > m₂), the equations are:
对每个质点分别画受力图，在运动方向应用 ΣF = ma，然后解联立方程。例如，m₁ 下降，m₂ 上升 (m₁ > m₂)，方程为：
m₁ g − T = m₁ a
T − m₂ g = m₂ a
Adding eliminates T and gives a = (m₁ − m₂)g / (m₁ + m₂). Tension can then be found by substitution.
相加消去 T 得到 a = (m₁ − m₂)g / (m₁ + m₂)。张力可通过代回求得。
9. Pulleys with One Mass on a Table | 一个物体在桌面上的滑轮系统
In many problems, a particle on a rough horizontal table is connected by a string over a pulley to a hanging particle. The hanging mass provides the driving force, while friction on the table mass opposes motion.
在许多问题中，粗糙水平桌面上的质点通过绳子跨过滑轮与悬挂的质点相连。悬挂的质量提供驱动力，而桌面上质点的摩擦力阻碍运动。
Applying Newton’s second law to each mass, for the hanging mass m₂: m₂ g − T = m₂ a. For the table mass m₁: T − F = m₁ a, where F is the frictional force, which may be μR if the mass is moving or limiting. Usually R = m₁ g on a horizontal table.
对每个质量应用牛顿第二定律，对悬挂质量 m₂：m₂ g − T = m₂ a。对桌面质量 m₁：T − F = m₁ a，其中 F 是摩擦力，如果物体正在运动或处于极限状态，F = μR。在水平桌面上通常 R = m₁ g。
10. Connected Particles via a Tow Bar or Coupling | 通过拖杆或挂钩连接的粒子
Two cars or blocks may be connected by a light rigid tow bar, or a string in a straight line. In such cases both bodies have the same acceleration. The tension or thrust in the connecting bar is internal to the system, but must be considered when analysing a single body.
两辆汽车或两个物块可能由一根轻质刚性拖杆或一条直线上的绳子连接。在这种情况下，两个物体具有相同的加速度。连接杆中的张力或推力是系统的内力，但在分析单个物体时必须考虑。
When the system is treated as a whole, internal forces cancel, so you can write ΣF = mtotal a for the entire system. Then, to find the internal force, isolate one part and apply the second law.
当把系统看作一个整体时，内力相互抵消，因此可以写出 ΣF = m总 a。然后，为了求内力，隔离其中一部分并应用第二定律。
11. Common Mistakes and How to Avoid Them | 常见错误及如何避免
• Confusing mass and weight: Weight (N) = mass (kg) × g, and must be resolved when on an incline.
• Forgetting to define a positive direction: leads to sign errors in equations.
• Mixing up the bodies in Newton’s third law: the forces are on *different* bodies.
• Using F = μR when friction is not limiting: only use it for limiting equilibrium or slipping.
• Assuming tension equals the weight of a hanging mass; tension is often less while accelerating.
• 混淆质量和重量：重量 (N) = 质量 (kg) × g，在斜面上必须分解。
• 忘记定义正方向：导致方程中出现符号错误。
• 在牛顿第三定律中弄错物体：力作用在*不同*物体上。
• 摩擦力未达到极限时使用 F = μR：仅在极限平衡或滑动时使用。
• 假设张力等于悬挂物体的重量；在加速时张力通常较小。
12. Exam Strategy for Newton’s Law Questions | 牛顿定律考题的应试策略
Start by reading the question carefully and noting whether surfaces are smooth or rough, ropes light, pulleys smooth. Sketch a clear diagram for every particle, showing all forces. State a positive direction and write ΣF = ma for each particle. Keep your working systematic: substitution errors are the biggest cause of lost marks.
首先仔细审题，注意表面是光滑还是粗糙、绳子是否轻质、滑轮是否光滑。为每个质点画出清晰的受力图，展示所有力。说明正方向，并对每个质点写出 ΣF = ma。保持解题步骤井井有条：代换错误是失分的最主要原因。
Check your acceleration—does it make physical sense? In connected‑particle questions, the hanging mass should accelerate downwards only if heavier. Use the whole‑system approach to verify your acceleration value quickly.
检查你求出的加速度——它在物理上合理吗？在连接体问题中，只有当悬挂质量较重时才会向下加速。使用整体法快速验证你的加速度值。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导