📚 Newton’s Laws of Motion: Key Concepts for IB and WJEC Physics | 牛顿定律考点精讲（IB & WJEC物理）
Newton’s laws of motion form the bedrock of classical mechanics and appear in virtually every IB and WJEC physics examination. Mastering these principles is essential for tackling problems involving forces, acceleration, equilibrium, and connected systems. This revision guide provides a structured, bilingual walkthrough of the core ideas, common pitfalls, and exam-style techniques you need to succeed.
牛顿运动定律是经典力学的基石，几乎出现在每一份 IB 和 WJEC 物理试卷中。掌握这些原理是解决力、加速度、平衡以及连接体等问题的关键。这份复习指南以结构化的双语方式，梳理核心思想、常见误区以及考试必备的解题技巧，帮助你顺利拿分。
1. First Law: Inertia and Equilibrium | 第一定律：惯性与平衡
Newton’s first law states that an object will remain at rest or in uniform motion in a straight line unless acted upon by a net external force. This property of matter is called inertia, and it explains why a passenger lurches forward when a bus brakes suddenly.
牛顿第一定律指出，除非受到净外力的作用，否则物体将保持静止或沿直线做匀速运动。物质的这一属性叫做惯性，它解释了为什么公交车突然刹车时乘客会向前倾倒。
In exam questions, the first law is often tested through equilibrium scenarios where the net force is zero. If an object moves at constant velocity or stays still, the vector sum of all forces acting on it must be zero.
在考试中，第一定律常通过净力为零的平衡情景来考查。如果一个物体以恒定速度运动或保持静止，那么作用在它上面的所有力的矢量和必须为零。
Key forces to consider include weight (mg downward), normal reaction (perpendicular to surfaces), tension, friction, and applied forces. To apply the first law, draw a free-body diagram and set up equations such as ΣFₓ = 0 and ΣFᵧ = 0.
需要考虑的关键力包括重力（mg，向下）、法向反作用力（垂直于接触面）、张力、摩擦力以及外加力。要应用第一定律，需要画出受力分析图并列出 ΣFₓ = 0 和 ΣFᵧ = 0 这样的方程。
ΣF = 0 ⇌ a = 0
2. Second Law: F = ma and Net Force | 第二定律：F = ma 与合力
The second law quantifies how the velocity of an object changes when it is subject to an unbalanced force. The net force acting on a body is equal to the product of its mass and acceleration. The direction of acceleration is always the same as the direction of the net force.
第二定律量化了物体在不平衡力作用下速度如何变化。作用在物体上的净力等于其质量与加速度的乘积。加速度的方向始终与净力的方向相同。
Mathematically, this is expressed as ΣF = ma. It is crucial to understand that ΣF is the vector sum of all forces, not just a single push or pull. When multiple forces act, you must resolve them into components and find the resultant before applying F = ma.
数学上，这表示为 ΣF = ma。理解 ΣF 是所有力的矢量和而不仅仅是某一个推力或拉力是至关重要的。当有多个力作用时，你必须先将它们分解为分量并求出合力，然后再应用 F = ma。
Fₙₑₜ = ma or ΣF = ma
Common exam mistakes involve using individual forces instead of the net force. Always check whether the object is accelerating in the direction of a particular force. For example, if a block is pulled along a rough horizontal surface, the net horizontal force is applied force minus friction, giving ma.
常见考试错误是使用单个力而不是净力。务必检查物体是否沿着某个具体力的方向加速。例如，一个木块在粗糙水平面上被拉动，水平方向上的净力是拉力减去摩擦力，从而得到 ma。
The unit of force is the newton (N). One newton is the force required to give a mass of 1 kg an acceleration of 1 m s⁻². Dimensionally, this reflects kg m s⁻².
力的单位是牛顿（N）。1 牛顿的定义是使 1 kg 的物体产生 1 m s⁻² 加速度所需的力。量纲上，这对应着 kg m s⁻²。
3. Third Law: Action-Reaction Pairs | 第三定律：作用力与反作用力
Newton’s third law states that if body A exerts a force on body B, then body B exerts an equal and opposite force on body A. These forces are always of the same type, act along the same line, and never cancel each other out because they act on different bodies.
牛顿第三定律指出，如果物体 A 对物体 B 施加一个力，那么物体 B 也同时对物体 A 施加一个大小相等、方向相反的力。这一对力总是同种性质的力，沿同一直线作用，永远不会相互抵消，因为它们作用在不同的物体上。
A classic example is a book resting on a table. The Earth pulls the book downward (weight), and the book pulls the Earth upward with an equal force. Simultaneously, the book pushes down on the table, and the table pushes up on the book (normal force). Note that weight and normal force are not an action-reaction pair; they act on the same object.
典型例子是放在桌上的书。地球向下拉书（重力），书同时以同样大小的力向上拉地球。与此同时，书向下压桌子，桌子向上推书（法向力）。注意重力和法向力并非一对作用力与反作用力，因为它们作用在同一个物体上。
In connected body problems, the third law helps to identify tension forces: if a rope pulls a block, the block pulls back on the rope with the same magnitude. Understanding action-reaction pairs prevents confusion when drawing free-body diagrams.
在连接体问题中，第三定律有助于确定张力：如果绳子拉一个物块，物块也会用相同大小的力向后拉绳子。理解作用力与反作用力对可以避免在画受力分析图时产生混淆。
4. Free-Body Diagrams | 受力分析图
A free-body diagram is a simplified sketch showing all the forces acting on a single object. Mastering this tool is half the battle in mechanics. The diagram should clearly indicate the direction of each force and label it appropriately, e.g. W, N, T, f.
受力分析图是一个简化的示意图，用来表示作用在单个物体上的所有力。掌握这个工具是解决力学问题成功的一半。图中应清楚标明每个力的方向并用 W、N、T、f 等符号适当标记。
Start by isolating the object of interest. Draw it as a point or a simple shape. Then add vectors: weight always acts vertically downward from the centre of mass; normal reaction acts perpendicular to the contact surface; tension acts along the rope away from the object; friction opposes relative motion or the tendency of motion.
首先隔离感兴趣的物体，将其画成一个点或简单的形状。然后添加矢量：重力总是从质心竖直向下；法向反作用力垂直于接触面；张力沿着绳子方向远离物体；摩擦力与相对运动或运动趋势的方向相反。
If the object is on an inclined plane, weight should be resolved into components parallel and perpendicular to the plane. Never draw the components on the diagram without also showing the original weight vector unless you are asked to use a rotated coordinate system.
如果物体在斜面上，重力应分解为平行和垂直于斜面的分量。除非题目要求使用旋转坐标系，否则不要在图上只画出分量而不画出原来的重力矢量。
5. Resolving Forces and Components | 力的分解与合成
Forces are vectors, so they can be resolved into perpendicular components. This technique is indispensable when forces do not act along the same line. In most IB and WJEC problems, you will use a Cartesian coordinate system oriented conveniently, e.g. with one axis along an incline.
力是矢量，因此可以分解为相互垂直的分量。当力不在同一直线上时，这一技巧不可或缺。在大多数 IB 和 WJEC 问题中，你会使用一个方便定向的笛卡尔坐标系，例如将一个坐标轴沿着斜面方向。
To resolve a force F at angle θ to the horizontal, the horizontal component is F cos θ and the vertical component is F sin θ. Always check whether the angle is measured from the horizontal or vertical, as this changes the trigonometric function.
要将与水平方向成 θ 角的力 F 分解，水平分量为 F cos θ，竖直分量为 F sin θ。务必检查角度是从水平方向还是竖直方向量起的，因为这会改变三角函数的使用。
When dealing with inclined planes of angle θ, the weight mg is broken into mg sin θ down the plane and mg cos θ into the plane. These form the basis for writing Newton’s second law equations along and perpendicular to the slope.
处理倾角为 θ 的斜面时，重力 mg 被分解为沿斜面向下的 mg sin θ 和垂直于斜面的 mg cos θ。这些分量构成了沿斜面方向和垂直斜面方向列写牛顿第二定律方程的基础。
Fₓ = F cos θ, Fᵧ = F sin θ
6. Friction: Static and Kinetic | 摩擦力：静摩擦与动摩擦
Frictional forces arise when two surfaces are in contact and there is relative motion or a tendency for motion. Static friction fₛ acts when objects are not moving relative to each other, up to a maximum value given by fₛ(max) = μₛ N, where μₛ is the coefficient of static friction and N is the normal reaction.
当两个表面接触并且存在相对运动或运动趋势时，就会产生摩擦力。静摩擦力 fₛ 在物体相对静止时作用，其最大值由 fₛ(max) = μₛ N 给出，其中 μₛ 是静摩擦系数，N 是法向反作用力。
Kinetic friction fₖ takes over once sliding begins. It is usually slightly less than the maximum static friction and is calculated as fₖ = μₖ N, where μₖ is the coefficient of kinetic friction. Both frictional forces act parallel to the contact surface and oppose motion.
一旦开始滑动，动摩擦力 fₖ 就会取而代之。它通常略小于最大静摩擦力，计算公式为 fₖ = μₖ N，其中 μₖ 是动摩擦系数。两种摩擦力都平行于接触面，并且与运动方向相反。
In equilibrium problems with static friction, the actual friction force is whatever is required to maintain equilibrium, up to the limit μₛ N. If the required friction exceeds this limit, the object will start to slide.
在涉及静摩擦的平衡问题中，实际的摩擦力是维持平衡所需的任何值，最大不超过极限值 μₛ N。如果所需摩擦力超过这一极限，物体就会开始滑动。
fₖ = μₖ N, fₛ ≤ μₛ N
7. Inclined Planes | 斜面问题
An object on a smooth inclined plane accelerates down the slope with a = g sin θ, because the component of weight along the plane is mg sin θ. If friction is present, the net force becomes mg sin θ – f, and the acceleration is a = g(sin θ – μ cos θ) for kinetic friction.
位于光滑斜面上的物体会以 a = g sin θ 的加速度沿斜面下滑，因为重力沿斜面的分量是 mg sin θ。如果存在摩擦，净力变为 mg sin θ – f，对于动摩擦，加速度为 a = g(sin θ – μ cos θ)。
To find the normal reaction, apply Newton’s first or second law perpendicular to the plane: since there is no acceleration perpendicularly (unless the plane itself is accelerating), N = mg cos θ. This normal force then determines the frictional force.
要求法向反作用力，需在垂直于斜面方向上应用牛顿第一或第二定律：由于垂直方向上没有加速度（除非斜面本身在加速），N = mg cos θ。这个法向力随后决定了摩擦力。
Many WJEC questions ask for the angle of repose, i.e. the angle at which an object just begins to slide. This occurs when mg sin θ = μₛ mg cos θ, giving tan θ = μₛ. This relationship is extremely useful and appears regularly in multiple-choice items.
许多 WJEC 试题会问及静止角，即物体刚好开始滑动时的角度。这发生在 mg sin θ = μₛ mg cos θ 时，得出 tan θ = μₛ。这一关系极为实用，经常出现在选择题中。
8. Connected Bodies and Tension | 连接体与张力
Problems involving two or more masses connected by a light inextensible string are common in both IB and WJEC specifications. The key is to treat the entire system as one to find acceleration, then isolate individual masses to find tension.
涉及两个或多个由轻质不可伸长的绳子连接的物体的题目在 IB 和 WJEC 大纲中都很常见。关键是将整个系统视为一个整体来求加速度，再隔离单个物体来求张力。
If two masses m₁ and m₂ hang over a frictionless pulley with m₁ > m₂, the net force on the system is (m₁ – m₂)g. The total mass being accelerated is m₁ + m₂, so the acceleration is a = (m₁ – m₂)g/(m₁ + m₂). This is a standard result that you should be able to derive quickly.
如果两个质量 m₁ 和 m₂ 悬挂在一个无摩擦滑轮上且 m₁ > m₂，系统所受净力为 (m₁ – m₂)g。被加速的总质量为 m₁ + m₂，因此加速度 a = (m₁ – m₂)g/(m₁ + m₂)。这是一个标准结论，你应该能快速推导出来。
Tension T is less than the weight of the heavier mass because that mass is accelerating downward. Applying Newton’s second law to m₁: m₁ g – T = m₁ a. After solving for a, substitute back to find T = 2 m₁ m₂ g/(m₁ + m₂).
张力 T 小于较重物体的重量，因为该物体正在向下加速。对 m₁ 应用牛顿第二定律：m₁ g – T = m₁ a。解出 a 后，代回即可得到 T = 2 m₁ m₂ g/(m₁ + m₂)。
When masses are on a horizontal surface with one hanging off the edge, the same system approach applies: net force is the weight of the hanging mass, and total mass includes both. Friction may be added to the horizontal surface to increase complexity.
当一个物体在水平桌面上，另一个悬挂在桌边时，同样的整体法适用：净力是悬挂物的重力，总质量包括两者。还可以在水平面加入摩擦以增加题目难度。
9. Pulleys and Atwood Machines | 滑轮与阿特伍德机
The Atwood machine, a classic device consisting of two unequal masses connected over a pulley, is a favourite for examining Newton’s second law. The analysis requires careful assignment of a positive direction – usually the direction of the heavier mass’s motion.
阿特伍德机是由两个不等的质量通过滑轮连接构成的经典装置，深受考查牛顿第二定律的青睐。分析时需要仔细设定正方向——通常是较重物体运动的方向。
If the pulley has mass or friction, the problem extends to rotational dynamics, but for IB Standard and most WJEC problems, you assume a light, frictionless pulley. This ensures the tension is uniform on both sides and the string just transmits force.
如果滑轮具有质量或存在摩擦，问题就会拓展到转动动力学，但在 IB 标准和大多数 WJEC 题目中，常假设滑轮轻质且无摩擦。这样可以保证两侧张力均匀，绳子仅传递力。
Always begin by drawing separate free-body diagrams for each mass. For mass m₁ (descending): m₁ g – T = m₁ a. For mass m₂ (ascending): T – m₂ g = m₂ a. Adding these eliminates T and yields the system acceleration.
始终先为每个物体单独绘制受力分析图。对于 m₁（下降）：m₁ g – T = m₁ a。对于 m₂（上升）：T – m₂ g = m₂ a。将两式相加消去 T 即可得出系统加速度。
Check your work by testing extreme cases: if masses are equal, acceleration should be zero; if one mass is vastly larger, acceleration should approach g. Such physical plausibility checks can catch algebraic errors.
通过检验极端情况来检查你的解答：如果两质量相等，加速度为零；如果一个质量远大于另一个，加速度应趋近于 g。这种物理合理性检查可以揪出代数错误。
10. Circular Motion and Centripetal Force | 圆周运动与向心力
Although circular motion is sometimes treated as a separate topic, it is a direct application of Newton’s second law. Any object moving in a circle at constant speed has a centripetal acceleration directed toward the centre, given by a = v²/r = ω²r.
尽管圆周运动有时被视作一个独立课题，但它本质上是牛顿第二定律的直接应用。任何以恒定速率做圆周运动的物体都具有指向圆心的向心加速度，表达式为 a = v²/r = ω²r。
The net force causing this acceleration is the centripetal force, F = mv²/r. This is not a new type of force, but the resultant of real forces such as tension, gravity, friction, or the normal reaction. A common mistake is to add ‘centripetal force’ to a free-body diagram as an extra force when it should emerge from summing other forces.
引起这一加速度的净力就是向心力，F = mv²/r。向心力并不是一种新的力，而是真实力（如张力、重力、摩擦力或法向反作用力）的合力。一个常见错误是在受力分析图中把“向心力”当作一个额外的力添加上去，实际上它应从其他力的总和中得出。
For a car rounding a banked curve without friction, the horizontal component of the normal force provides the centripetal force. Equating N sin θ to mv²/r and N cos θ to mg leads to the design speed v = √(rg tan θ). This is a standard derivation in both IB and WJEC.
对于汽车在没有摩擦的倾斜弯道上转弯的情况，法向力的水平分量提供了向心力。令 N sin θ = mv²/r，N cos θ = mg，即可得到设计速度 v = √(rg tan θ)。这是 IB 和 WJEC 中都要求掌握的标准推导。
In vertical circular motion, speed changes along the path, so the net force varies. At the top of a loop, minimum speed is achieved when the tension or normal force just reaches zero, giving mg = mv²/r. Such boundary conditions are typical of higher-tier exam questions.
在竖直面内的圆周运动中，速度沿路径变化，因此净力也相应变化。在圆环顶端，当张力或法向力恰好为零时达到最小速度，此时 mg = mv²/r。这类临界条件是高难度试题的典型特征。
a_c = v²/r, F_c = mv²/r
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