📚 Newton's Laws: IB Edexcel Mathematics Key Points | IB Edexcel 数学：牛顿定律 考点精讲
Newton's laws of motion form the foundation of classical mechanics and are a central topic in both IB Mathematics (Analysis & Approaches HL or Applications & Interpretation HL) and Edexcel A Level Mathematics (Mechanics). Mastering these principles is essential for solving problems involving forces, motion, and connected particles.
牛顿运动定律是经典力学的基石，也是 IB 数学（分析与方法 HL 或应用与解释 HL）以及 Edexcel A Level 数学（力学）的核心考点。掌握这些原理是解决力、运动和连接体问题的关键。
1. Newton's First Law: The Law of Inertia | 牛顿第一定律：惯性定律
A body remains at rest or moves with constant velocity unless acted upon by a net external force. In vector terms, if ΣF = 0 then acceleration a = 0.
物体保持静止或匀速直线运动状态，除非受到净外力作用。用矢量表示：若 ΣF = 0，则加速度 a = 0。
This law defines an inertial frame of reference and implies that objects naturally resist changes to their state of motion.
该定律定义了惯性参考系，意味着物体自然抗拒运动状态的改变。
2. Newton's Second Law: F = ma | 牛顿第二定律：F = ma
The resultant force acting on a particle is equal to the product of its mass and acceleration: F = ma. Force and acceleration are always in the same direction.
作用在质点上的合外力等于其质量与加速度的乘积：F = ma。力与加速度的方向始终相同。
F = ma
When multiple forces act, use the vector sum: ΣF = ma. In component form, ΣFₓ = maₓ and ΣFy = ma_y.
当有多个力作用时，使用矢量和：ΣF = ma。在分量形式下，ΣFₓ = maₓ 且 ΣFy = ma_y。
ways resolve forces or choose a direction of positive acceleration before applying F = ma.
该方程是连接运动学与动力学的关键。在应用 F = ma 之前，务必分解力或选定加速度正方向。
3. Newton's Third Law: Action–Reaction Pairs | 牛顿第三定律：作用力与反作用力
If body A exerts a force on body B, then body B exerts an equal and opposite force on body A: FAB = –FBA. These forces act on different bodies and are of the same type.
若物体 A 对物体 B 施加一个力，则物体 B 对 A 施加一个大小相等、方向相反的力：FAB = –FBA。这两个力作用在不同物体上且性质相同。
This law is critical when analysing normal reaction forces, tension, and contact forces in connected systems. Never include action–reaction pairs on the same free-body diagram.
分析连接系统中的法向反力、张力和接触力时，该定律至关重要。切勿将一对作用力与反作用力画在同一受力图上。
4. Resolving Forces into Components | 力的分解
For a force F at an angle θ to the horizontal, its horizontal component is F cos θ and its vertical component is F sin θ. Always define a clear coordinate system.
对于与水平方向成 θ 角的力 F，其水平分量为 F cos θ，竖直分量为 F sin θ。务必明确坐标轴方向。
Fₓ = F cos θ, Fy = F sin θ
In IB and Edexcel problems, you often need to resolve weight (mg) on an inclined plane: parallel to the slope = mg sin θ, perpendicular to the slope = mg cos θ.
在 IB 和 Edexcel 试题中，常需分解斜面上的重力：平行于斜面的分力为 mg sin θ，垂直于斜面的分力为 mg cos θ。
5. Friction: Static and Kinetic | 摩擦力：静摩擦力与动摩擦力
Static friction: F ≤ μₛ R, where μₛ is the coefficient of static friction and R is the normal reaction. It acts to prevent relative motion.
静摩擦力：F ≤ μₛ R，其中 μₛ 为静摩擦系数，R 为法向反力。它阻碍相对运动的发生。
Kinetic (dynamic) friction: F = μₖ R, acting opposite to the direction of motion. In most exam questions, μₖ is simply denoted as μ.
动摩擦力：F = μₖ R，方向与运动方向相反。在多数考题中，动摩擦系数简记为 μ。
Remember: friction always opposes relative motion or the tendency of motion. On the point of sliding, the limiting friction is Fmax = μR.
注意：摩擦力始终阻碍相对运动或运动趋势。在将要滑动时，极限摩擦力为 Fmax = μR。
6. Motion on an Inclined Plane | 斜面上的运动
For a particle of mass m on a smooth incline of angle θ, the acceleration down the slope is a = g sin θ (no friction).
对于质量为 m 的质点，在倾角为 θ 的光滑斜面上，沿斜面下滑的加速度为 a = g sin θ（无摩擦）。
When friction is present, apply Newton's second law along the incline: mg sin θ – Ffriction = ma. The normal reaction is R = mg cos θ.
有摩擦时，沿斜面应用牛顿第二定律：mg sin θ – F摩擦 = ma。法向反力为 R = mg cos θ。
| Smooth incline (no friction) | a = g sin θ |
| Rough incline (kinetic friction) | a = g(sin θ – μ cos θ) |
7. Tension in Strings and Pulleys | 绳中张力与滑轮
Tension is the force transmitted through a light inextensible string. In such a string, the tension is the same on both sides of a smooth pulley.
张力是通过轻质不可伸长的绳子传递的力。对于光滑滑轮，绳子两端的张力大小相等。
When particles are connected by a string passing over a pulley, treat each particle separately. Write an equation of motion for each and solve simultaneously.
当质点通过跨过滑轮的绳子连接时，应对各质点单独分析，分别列出运动方程并联立求解。
T – m₁g = m₁a, m₂g – T = m₂a (for a vertical pulley system)
In IB and Edexcel, common setups include two masses over a pulley, or one mass on a table connected to a hanging mass.
在 IB 和 Edexcel 中，常见模型包括滑轮两侧各挂一重物，或桌面上的物体通过绳子与悬挂重物相连。
8. Connected Particles on a Horizontal Surface | 水平面上的连接体
For two blocks connected by a light string on a smooth horizontal surface, pull force F accelerates the whole system: F = (m₁ + m₂)a.
对于光滑水平面上用轻绳连接的两个木块，拉力 F 使系统整体加速：F = (m₁ + m₂)a。
To find the tension T between the blocks, isolate the leading block or the trailing block. For example, for the trailing block: T = m₂ a.
为求两木块间的张力 T，单独分析其中一个木块。例如，对其中的后一个木块：T = m₂ a。
If friction is present, include the frictional force on each mass before solving. This tests your ability to handle multiple force equations.
若有摩擦，应在求解前计入每个物体所受的摩擦力。这考查处理多个力平衡方程的能力。
9. Equilibrium: ΣF = 0 | 平衡状态：ΣF = 0
A particle is in equilibrium when the vector sum of all forces acting on it is zero. This yields two scalar equations: ΣFₓ = 0 and ΣFy = 0.
当作用在一个质点上的所有力的矢量和为零时，质点处于平衡状态。由此得到两个标量方程：ΣFₓ = 0 和 ΣFy = 0。
These equations are used to find unknown forces, tensions, or angles in static systems. Lami's theorem may also be applied for three-force equilibrium.
这些方程用于求解静态系统中的未知力、张力或角度。对于三力平衡问题，也可应用拉密定理。
Lami's theorem: F₁ / sin α = F₂ / sin β = F₃ / sin γ
10. Non‑Equilibrium and Acceleration | 非平衡状态与加速度
resultant force. The steps are: draw a clear force diagram, resolve forces, apply ΣF = ma.
当ΣF ≠ 0 时，质点沿合外力方向加速。解题步骤为：绘制清晰的受力图，分解力，应用 ΣF = ma。
In problems involving lifts (elevators), the normal reaction R changes: R – mg = ma (upward acceleration) or mg – R = ma (downward).
在电梯问题中，法向反力 R 会变化：R – mg = ma（向上加速）或 mg – R = ma（向下加速）。
Always define the positive direction clearly to avoid sign errors. Consistent sign conventions are the key to success.
务必明确正方向，避免符号错误。保持一致的符号约定是解题成功的关键。
11. Momentum and Impulse (Extension) | 动量与冲量（拓展）
Newton's second law can be expressed in terms of momentum: F = dp/dt. For constant mass, this reduces to F = ma. Impulse J = FΔt = Δp = mv – mu.
牛顿第二定律可用动量表示为：F = dp/dt。当质量恒定时，简化成 F = ma。冲量 J = FΔt = Δp = mv – mu。
In IB and Edexcel, impulse–momentum problems often appear in collision or particle motion contexts. Use vector subtraction for change in momentum.
在 IB 和 Edexcel 中，冲量‑动量问题常出现在碰撞或质点运动背景中。动量变化需用矢量减法计算。
12. Exam Strategy and Common Pitfalls | 应试策略与常见错误
Always start with a clear, large free-body diagram. Label all forces, including weight, normal reaction, tension, friction, and applied forces.
始终从清晰且足够大的受力图开始。标出所有力，包括重力、法向反力、张力、摩擦力和外加力。
Resolve forces before writing equations. Choose axes cleverly — often parallel and perpendicular to the direction of motion. Double‑check the direction of acceleration.
在列方程前先分解力。巧妙选择坐标轴——通常沿运动方向及垂直方向。再三检查加速度的方向。
.8 m s⁻² unless otherwise specified. In connected particle questions, write separate equations and solve simultaneously.
注意单位：质量用 kg，力用 N，加速度用 m s⁻²。除非题目另有说明，取 g = 9.8 m s⁻²。在连接体问题中，应分写方程并联立求解。
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