Newton's Laws: Key Points for Edexcel Maths | 牛顿定律考点精讲

📚 Newton’s Laws: Key Points for Edexcel Maths | 牛顿定律考点精讲

Newton’s laws of motion form the backbone of mechanics in Edexcel Maths. Whether you are tackling connected particles, inclined planes, or pulleys, a solid grasp of these three laws allows you to construct accurate equations and solve problems systematically. This article breaks down every essential concept you need to master for the exam, from free-body diagrams to friction and equilibrium.

牛顿运动定律是Edexcel数学力学部分的基石。无论是处理连接体、斜面还是滑轮问题，牢固掌握这三条定律都能让你准确建立方程，系统性地解题。本文拆解了你需要掌握的每个核心概念，从自由体受力图到摩擦和平衡，助你备战考试。

1. Newton’s First Law: Inertia and Equilibrium | 牛顿第一定律：惯性与平衡

Newton’s first law states that an object remains at rest or moves with constant velocity unless acted upon by a resultant external force. In mathematical mechanics, this means if the net force is zero, the body is in equilibrium – either stationary or moving with constant speed in a straight line.

牛顿第一定律指出，除非受到合外力作用，物体将保持静止或匀速直线运动。在数学力学中，这意味着如果合力为零，物体处于平衡状态——要么静止，要么沿直线匀速运动。

When a particle is in equilibrium, the vector sum of all forces acting on it is zero: ΣF = 0. This gives you two scalar equations when resolving horizontally and vertically, or parallel and perpendicular to an inclined plane.

当质点处于平衡状态时，所有作用力的矢量和为零：ΣF = 0。这为你在水平和竖直方向（或沿斜面和垂直于斜面）分解力时提供了两个标量方程。

Exam tip: Always state ‘by Newton’s first law’ when setting ΣF = 0 for a body in equilibrium. It shows the examiner you understand the underlying principle.

考试技巧：当为平衡物体列出 ΣF = 0 时，务必注明“根据牛顿第一定律”。这表明你理解背后的原理。

2. Newton’s Second Law: F = ma | 牛顿第二定律：F = ma

The second law relates the resultant force acting on a particle to its acceleration: F = ma, where F is the net force (in N), m is the mass (in kg), and a is the acceleration (in m s⁻²). This vector equation is applied in the direction of motion or along a chosen axis.

第二定律将作用在质点上的合力与其加速度联系起来：F = ma，其中F为合力（牛顿），m为质量（千克），a为加速度（米每二次方秒）。这一矢量方程应用在运动方向或所选坐标轴方向上。

When solving problems, always write ‘Resultant force = ma’ and list all force components along the direction of acceleration. Remember that acceleration and resultant force act in the same direction.

解题时，务必写出“合力 = ma”，并列出沿加速度方向的所有分力。请记住加速度与合力方向相同。

In two dimensions, you often resolve forces and apply F = ma independently in perpendicular directions. For a particle on a smooth horizontal surface pulled by a string at an angle, the horizontal component gives Fcosθ = ma.

在二维问题中，你通常会分解力并在相互垂直的方向上独立应用 F = ma。对于光滑水平面上被斜向绳子拉动的质点，水平分量给出 Fcosθ = ma。

3. Newton’s Third Law: Action–Reaction Pairs | 牛顿第三定律：作用力与反作用力

Newton’s third law states: if body A exerts a force on body B, then body B exerts a force on body A of equal magnitude but opposite direction. These forces act on different bodies and are of the same type.

牛顿第三定律指出：若物体A对物体B施加一个力，则物体B同时对物体A施加一个大小相等、方向相反的力。这两个力作用在不同物体上，且属于同种类型。

Common mistake: The normal reaction force from a table on a book and the weight of the book do NOT form an action–reaction pair. The reaction pair to the weight is the gravitational pull of the book on the Earth.

常见错误：桌面对书本的法向反作用力与书本的重力并不构成作用-反作用力对。重力的反作用力是书本对地球的万有引力。

In connected-particle problems, the tension in a light inextensible string is the same at both ends, and the action–reaction principle helps you understand the forces at each connection point.

在连接体问题中，轻绳且不可伸长的条件下，绳子两端的张力大小相等，而作用-反作用原理有助于你理解每个连接点处的受力。

4. Free-Body Diagrams: Drawing the Forces | 自由体受力图：绘制受力

A clear free-body diagram is your first step to success. Isolate the particle or body and draw all forces: weight (mg, vertically down), normal reaction (perpendicular to the contact surface), tension (along the string), friction (opposite to impending motion), and any applied forces.

画出清晰的自由体受力图是成功的第一步。将质点或物体隔离，画出所有力：重力 (mg，竖直向下)、法向反作用力（垂直于接触面）、张力（沿绳子方向）、摩擦力（与预期运动方向相反）以及任何外加力。

Label each force with a symbol and direction. For inclined planes, draw the weight vector, then mark its components parallel and perpendicular to the plane: mg sinθ and mg cosθ.

用符号和方向标注每个力。对于斜面问题，画出重力矢量，然后标出平行和垂直于斜面的分量：mg sinθ 和 mg cosθ。

Always adopt a consistent sign convention, usually taking the direction of acceleration as positive. This minimizes algebraic errors.

始终采用一致的符号规则，通常以加速度方向为正。这能最大程度减少代数错误。

5. Resolving Forces: From Vectors to Equations | 力的分解：从矢量到方程

Resolving forces means breaking each force into components along perpendicular axes. For a force F at angle θ to the horizontal: horizontal component = F cosθ, vertical component = F sinθ.

力的分解是指将每个力沿垂直坐标轴分解为分量。对于与水平方向成θ角的力F：水平分量 = F cosθ，竖直分量 = F sinθ。

Use resolution to convert the vector statement ΣF = ma into scalar equations. For equilibrium problems, set ΣF_x = 0 and ΣF_y = 0. For dynamics, use ΣF_x = ma_x and ΣF_y = ma_y.

通过分解将矢量方程 ΣF = ma 转化为标量方程。对于平衡问题，设 ΣF_x = 0 和 ΣF_y = 0。对于动力学问题，使用 ΣF_x = ma_x 和 ΣF_y = ma_y。

When a particle rests on an inclined plane, it is often easiest to resolve parallel and perpendicular to the slope, rather than horizontally and vertically.

当质点静止在斜面上时，通常沿斜面和垂直于斜面分解比水平竖直分解更简便。

6. Inclined Planes: Gravity’s Components | 斜面问题：重力的分量

On an inclined plane at angle θ to the horizontal, the weight mg splits into: component parallel to the plane = mg sinθ (down the slope), and component perpendicular to the plane = mg cosθ (into the plane).

在与水平面成θ角的斜面上，重力mg分解为：平行于斜面的分量 = mg sinθ（沿斜面向下），垂直于斜面的分量 = mg cosθ（指向斜面内）。

The normal reaction R balances the perpendicular component if there is no acceleration perpendicular to the plane: R = mg cosθ. Friction, if present, acts parallel to the plane, opposing motion.

如果在垂直于斜面方向没有加速度，法向反作用力R会平衡该方向的分量：R = mg cosθ。如果存在摩擦，摩擦力则平行于斜面作用，与运动方向相反。

Applying Newton’s second law along the slope: mg sinθ − F_friction = ma (if acceleration is down the slope). Always define your positive direction at the start.

沿斜面应用牛顿第二定律：mg sinθ − 摩擦力 = ma（如果以沿斜面向下为正）。在开始解题时就应明确正方向。

7. Connected Particles: Shared Acceleration | 连接体问题：共同加速度

When two particles are connected by a light inextensible string passing over a pulley or along a surface, they share the same speed and the same magnitude of acceleration (provided the string remains taut).

当两个质点由一根跨过滑轮或沿表面延伸的轻绳连接且绳子不可伸长时，它们具有相同大小的速度和加速度（前提是绳子绷紧）。

Treat each particle separately: draw its free-body diagram, apply F = ma along its direction of motion, and write the equations. Then solve simultaneously to find acceleration and tension.

分别处理每个质点：画出各自的自由体受力图，沿其运动方向应用F = ma，列出方程。然后联立求解加速度和张力。

Massless, inextensible string ensures uniform tension T throughout. A smooth pulley changes the direction of tension without changing its magnitude.

轻绳且不可伸长保证了张力T处处相同。光滑滑轮仅改变张力的方向而不改变其大小。

8. Pulley Systems: Changing Direction | 滑轮系统：改变方向

In Edexcel Maths problems, pulleys are usually smooth and light. This means the tension on both sides is the same, and you can ignore the pulley’s mass and any resistance.

在Edexcel数学问题中，滑轮通常视为光滑且轻质。这意味着两侧张力相等，并且可以忽略滑轮质量及任何阻力。

Set up equations for each hanging mass: for mass A, weight − tension = mass × acceleration, taking the direction of motion as positive. If one mass accelerates downward, the other accelerates upward with the same magnitude.

为每个悬挂物建立方程：对质量A，重力 − 张力 = 质量 × 加速度，以运动方向为正。如果一个物体向下加速，则另一个以相同大小的加速度向上运动。

Sometimes a particle rests on a table while another hangs over the edge. The hanging weight provides the driving force, while the table may be smooth or rough. Combine the equations to find a and T.

有时一个质点放在桌面上，另一个悬挂在桌边。悬挂物的重力提供驱动力，桌面可能光滑或粗糙。联立方程求出a和T。

9. Friction: Static and Kinetic | 摩擦力：静摩擦与动摩擦

Friction is a resistive force that acts opposite to motion or impending motion. For a stationary body, static friction can vary up to a maximum value F_max = μ_s R, where μ_s is the coefficient of static friction and R is the normal reaction.

摩擦力是一种阻碍运动或运动趋势的力。对于静止物体，静摩擦力可以在最大静摩擦力 F_max = μ_s R 范围内变化，μ_s 为静摩擦系数，R 为法向反作用力。

When a body is moving or on the point of sliding, friction reaches its limiting value and is given by F = μ R, where μ is the coefficient of kinetic (or limiting) friction.

当物体正在运动或处于滑动的临界点时，摩擦力达到极限值，并可用 F = μ R 表示，其中 μ 为动（或极限）摩擦系数。

Always use R correctly: R is the normal reaction, not mg, unless the surface is horizontal and no other vertical forces act. Then R = mg. On an incline, R = mg cosθ.

务必正确使用R：R 是法向反作用力，不一定是 mg，除非表面水平且没有其他竖直力作用。此时 R = mg。在斜面上，R = mg cosθ。

Friction opposes relative motion, so its direction in your diagram must be opposite to the velocity or the direction it would slide if friction were absent.

摩擦力阻碍相对运动，因此受力图中其方向必须与运动方向相反，或者与无摩擦时将要滑动的方向相反。

10. Equilibrium and Limiting Equilibrium | 平衡与极限平衡

A body is in equilibrium when the resultant force is zero. If friction is involved, the equilibrium may be maintained even when an external force is applied, as long as F ≤ μR.

当合力为零时，物体处于平衡状态。如果涉及摩擦，即使有外力作用，只要 F ≤ μR，物体仍可保持平衡。

Limiting equilibrium is the state where the body is just about to move. At this point, the friction force is exactly μR, and any slight increase in the applied force causes motion. Set F = μR and ΣF = 0.

极限平衡是物体刚要运动的状态。此时摩擦力恰等于 μR，任何微小的外力增加都会导致运动。设 F = μR 且 ΣF = 0。

For a block on a rough slope, you can derive the angle at which sliding begins: tanθ = μ. This is a classic result that often appears in exams.

对于放在粗糙斜面上的物块，可推导开始滑动的临界角：tanθ = μ。这是考试中常见的经典结论。

11. Combining Newton’s Laws with Kinematics | 结合运动学与牛顿定律

Many exam questions link Newton’s second law with constant acceleration equations (SUVAT). After finding acceleration using F = ma, use v = u + at, s = ut + ½at², v² = u² + 2as, or s = ½(u+v)t to find displacement, time, or final velocity.

许多考题将牛顿第二定律与匀加速运动方程 (SUVAT) 联系起来。用 F = ma 求出加速度后，再用 v = u + at, s = ut + ½at², v² = u² + 2as 或 s = ½(u+v)t 求位移、时间或末速度。

When a braking force is applied, acceleration is negative relative to the direction of motion. Apply F = ma to find deceleration, then use kinematics to calculate stopping distance.

当施加制动力时，加速度相对于运动方向为负。应用 F = ma 求出减速度，再通过运动学计算制动距离。

Be careful with signs: if your positive direction is up, then gravity is −mg, and acceleration due to gravity is −9.8 m s⁻².

注意符号：若取向上为正，则重力为 −mg，重力加速度为 −9.8 米每二次方秒。

12. Common Mistakes and Exam Strategies | 常见错误与应试策略

Mistake 1: Forgetting that the normal reaction is not always equal to mg. Always resolve perpendicular to the surface to find R.

错误1：忘记法向反作用力并不总是等于 mg。务必通过垂直于表面的分解来求 R。

Mistake 2: Misidentifying the direction of friction. Friction always opposes relative motion or impending motion, not necessarily the direction of applied force.

错误2：弄错摩擦力的方向。摩擦力总是阻碍相对运动或相对运动趋势，而不一定与外力方向相反。

Mistake 3: Inconsistent sign conventions. Once you choose a positive direction, all forces and accelerations must be assigned signs accordingly.

错误3：符号规则不统一。一旦选定正方向，所有力和加速度都必须相应地赋予正负号。

Strategy: Always write down the relevant law (‘By Newton’s second law…’) and show clear resolution steps. Even if your final answer is wrong, you can earn method marks for a correct equation set‑up.

应试策略：务必写出相关定律（“由牛顿第二定律……”），并展示清晰的分解步骤。即使最终答案错误，正确的方程建立也能获得方法分。

Strategy: Practice past paper questions on connected particles, inclined planes, and limiting equilibrium. These account for a significant proportion of Mechanics marks.

应试策略：练习历年真题中的连接体、斜面和极限平衡问题。这些内容在力学部分中占相当大的分值。

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Newton’s Laws: Key Points for Edexcel Maths | 牛顿定律考点精讲