📚 Edexcel A Level Mechanics Part 2: Forces, Kinematics and Moments | 爱德思A Level力学第2部分：力、运动学与力矩
Mechanics in Edexcel A Level Mathematics is assessed in Paper 3, alongside Statistics. It tests your ability to model real-life motion and forces using clear mathematical language. This expanded article covers the most important mechanics topics in the second mechanics strand: quantities and units, SUVAT kinematics, Newton's laws, connected particles, friction, moments, equilibrium, calculus-based kinematics, and exam technique. Each section includes worked-style reminders and common errors to avoid.
在爱德思 A Level 数学中，力学与统计共同在 Paper 3 中考查。力学考查你使用清晰的数学语言对现实中的运动与力进行建模的能力。这篇扩写文章涵盖力学第二部分最重要的主题：力学量与单位、SUVAT 运动学、牛顿定律、连接体、摩擦、力矩、平衡、微积分运动学以及考试技巧。每一节都包含解题提示与常见错误提醒。
1. Quantities, Units and Modelling Assumptions | 力学量、单位与建模假设
Before applying any formula, identify the fundamental quantities: displacement s (m), initial velocity u (m s⁻¹), final velocity v (m s⁻¹), acceleration a (m s⁻²), time t (s), mass m (kg) and force F (N). Always use base SI units in every calculation, and convert km h⁻¹ to m s⁻¹ by dividing by 3.6 or multiplying by 1000/3600. Write down the quantity, symbol and unit before substituting numbers, because unit errors lose method marks even when the algebra is correct.
在套用任何公式之前，先要识别基本力学量：位移 s（m）、初速度 u（m s⁻¹）、末速度 v（m s⁻¹）、加速度 a（m s⁻²）、时间 t（s）、质量 m（kg）和力 F（N）。所有计算都应使用基本 SI 单位；将 km h⁻¹ 换算为 m s⁻¹ 时，除以 3.6，或乘以 1000/3600。在代入数字之前先写出量、符号和单位，因为单位错误即使代数正确也会失分。
| Quantity | Symbol | SI Unit |
|---|---|---|
| Displacement | s | metre (m) |
| Initial velocity | u | m s⁻¹ |
| Final velocity | v | m s⁻¹ |
| Acceleration | a | m s⁻² |
| Time | t | second (s) |
| Mass | m | kilogram (kg) |
| Force | F | newton (N) |
| Weight | W | newton (N) |
| Friction | Ff | newton (N) |
Modelling assumptions are part of the mark scheme. State them clearly: the object is a particle, the string is light and inextensible, the pulley is smooth, air resistance is negligible, or the rod is uniform. These assumptions justify using simplified equations and ignoring rotation or elasticity unless the question says otherwise.
建模假设本身就是评分标准的一部分。要清楚说明：物体视为质点、绳子轻且不可伸长、滑轮光滑、空气阻力可忽略，或杆是均匀的。这些假设使我们可以使用简化方程，并在题目未说明时忽略转动或弹性。
2. SUVAT Equations for Constant Acceleration | 匀加速运动 SUVAT 方程
quation that contains the three known quantities and the one unknown. For vertical motion, take g = 9.8 m s⁻² unless told otherwise, and decide a positive direction at the start.
五个 SUVAT 方程只在加速度恒定时适用。它们是 v = u + at、s = ut + ½at²、v² = u² + 2as、s = ½(u + v)t 和 s = vt − ½at²。选择包含三个已知量和一个未知量的方程。对于竖直运动，除非另有说明，取 g = 9.8 m s⁻²，并在开始时规定正方向。
v = u + at s = ut + ½at² v² = u² + 2as
s positive, then a = +9.8 m s⁻². Never mix directions in one equation. For a particle thrown upwards, the velocity at maximum height is zero, and the time to maximum height is t = u/g.
符号约定至关重要。若取向上为正，则抛体运动 a = −9.8 m s⁻²，起点以上的位移为正。若取向下为正，则 a = +9.8 m s⁻²。不要在同一个方程中混用方向。对于上抛质点，最高点速度为 0，到达最高点的时间为 t = u/g。
3. Vertical Motion Under Gravity and Projectiles | 竖直运动与抛体运动
wards from a height h. Use s = ut + ½at² with a = −g, and solve the quadratic for t. Reject negative times unless the context allows them. The total distance travelled is not the same as displacement when the object reverses direction.
在重力作用下的竖直运动中，唯一的作用力是重力，因此加速度为向下的 g。常见题型是球从高度 h 竖直上抛。使用 s = ut + ½at²，取 a = −g，并解二次方程求 t。除非情境允许，否则舍去负时间。当物体反向运动时，总路程与位移不同。
, resolve the initial velocity into horizontal and vertical components: uₓ = u cos θ and uy from the vertical motion, and the horizontal range is R = u² sin 2θ / g for level ground.
对于抛体运动，将初速度分解为水平与竖直分量：uₓ = u cos θ，uy = u sin θ。水平加速度为 0，因此 x = u cos θ × t。竖直加速度为 −g，因此 y = u sin θ × t − ½gt²。飞行时间由竖直运动求出，水平射程在水平地面为 R = u² sin 2θ / g。
A frequent error is to use the same time for both directions without justification. The two directions are linked only by time, so first find t from the vertical motion, then substitute into the horizontal equation. At the highest point, the vertical velocity is zero, but the horizontal velocity is unchanged.
常见错误是未加说明就对两个方向使用相同时间。两个方向仅通过时间联系，因此先从竖直运动求出 t，再代入水平方程。在最高点，竖直速度为 0，但水平速度不变。
4. Newton's Laws and Free-Body Diagrams | 牛顿定律与受力分析
Newton' mains at rest or moves with constant velocity unless acted on by a resultant force. Newton's second law gives Fnet = ma, where Fnet is the vector sum of all forces. Newton' a force on B, then B exerts an equal and opposite force on A, acting on different bodies.
牛顿第一定律指出，除非受到合力作用，质点保持静止或匀速直线运动。牛顿第二定律给出 Fnet = ma，其中 Fnet 是所有力的矢量和。牛顿第三定律指出，若 A 对 B 施加力，则 B 对 A 施加大小相等、方向相反的力，且作用在不同物体上。
ds, normal reaction R perpendicular to the surface, tension T along the string, friction Ff ontal surface, vertical equilibrium gives R = mg unless there is a vertical applied force.
写方程前一定要画受力图。标出向下的重力 W = mg、垂直于表面的法向反力 R、沿绳的张力 T、阻碍运动的摩擦力 Ff 以及任何外力。将力沿运动方向和平行方向分解。对于水平面上的物体，若没有竖直外力，竖直方向平衡给出 R = mg。
If the resultant force is zero, the object is in equilibrium and acceleration is zero. If the resultant force is constant, acceleration is constant and SUVAT can be used. If the force varies, you may need calculus or energy methods. Check that every force in your diagram appears in the equation.
若合力为 0，物体处于平衡，加速度为 0。若合力恒定，加速度恒定，可以使用 SUVAT。若力变化，可能需要微积分或能量方法。检查受力图中的每一个力是否都出现在方程中。
5. Connected Particles, Pulleys and Lift Problems | 连接体、滑轮与电梯问题
ted by a light inextensible string over a smooth pulley, with m₁ > m₂, the acceleration is a = (m₁ − m₂)g / (m₁ + m₂). The tension is T = m₁(g − a) or T = m₂(g + a).
对于连接体，先整体求加速度，再隔离一个物体求张力。对于两个质量 m₁ 和 m₂ 用轻质不可伸长绳跨过光滑滑轮连接，且 m₁ > m₂，加速度为 a = (m₁ − m₂)g / (m₁ + m₂)。张力为 T = m₁(g − a) 或 T = m₂(g + a)。
If it moves at constant velocity or is at rest, R = mg. A common error is to use the weight instead of the normal reaction for apparent weight, or to forget that the acceleration of the lift changes the normal reaction.
在电梯问题中，视重是法向反力 R。若电梯向上加速，R = m(g + a)。若向下加速，R = m(g − a)。若匀速运动或静止，R = mg。常见错误是用重力代替法向反力求视重，或忘记电梯加速度会改变法向反力。
able connected to a hanging mass, resolve vertically for the hanging mass and horizontally for the table mass. Friction acts on the table mass, so include Ff e equation. If the string is slack, tension is zero, and the particles move independently.
对于粗糙水平桌面上连接悬挂质量的质点，对悬挂质量竖直分解，对桌面质量水平分解。摩擦力作用在桌面质量上，因此方程中要加入 Ff = μR。若绳松弛，张力为 0，两个物体独立运动。
6. Friction and Inclined Planes | 摩擦与斜面
Friction opposes relative motion or the tendency to move. The limiting friction is Fmax s F ≤ μR; only at the point of sliding does F = μR. Always check whether the question says "limiting equilibrium" or "about to move".
摩擦阻碍相对运动或运动趋势。极限摩擦力为 Fmax = μR，其中 μ 是摩擦系数，R 是法向反力。在静力平衡中，摩擦满足 F ≤ μR；只有在即将滑动的临界状态才有 F = μR。一定要检查题目说的是 "极限平衡" 还是 "即将运动"。
e. The normal reaction is R = mg cos θ if there is no other perpendicular force. The acceleration down a smooth plane is a = g sin θ. On a rough plane, the acceleration down the slope is a = g(sin θ − μ cos θ) when sliding.
在倾角为 θ 的斜面上，将重力分解为沿斜面的 mg sin θ 和垂直斜面的 mg cos θ。若没有其他垂直力，法向反力为 R = mg cos θ。光滑斜面上的下滑加速度为 a = g sin θ。粗糙斜面上滑动时，下滑加速度为 a = g(sin θ − μ cos θ)。
ined plane. That is only true when the surface is horizontal. Another error is to resolve weight incorrectly: the component perpendicular to the plane is mg cos θ, not mg sin θ. Draw the angle carefully and check which component is adjacent to θ.
常见错误是在斜面上使用 R = mg。这只有在水平面时才成立。另一个错误是分解重力出错：垂直斜面的分量是 mg cos θ，不是 mg sin θ。仔细画角，检查哪个分量与 θ 相邻。
7. Moments and Rigid Bodies | 力矩与刚体
wton metre (N m). A rigid body is in equilibrium when the resultant force is zero and the resultant moment about any point is zero.
力关于某点的力矩是力与从该点到力的作用线的垂直距离的乘积：M = F × d。单位是牛顿米（N m）。刚体处于平衡的条件是合力为 0，且关于任意点的合力矩为 0。
M = F × d ΣM_clockwise = ΣM_anticlockwise
When solving a moments problem, choose a point that eliminates an unknown force. For example, take moments about a support to eliminate the reaction there. For a uniform rod, weight acts at the midpoint. For a non-uniform rod, the centre of mass is not at the midpoint, so let it be x from one end and use moments to find x.
解力矩问题时，选择能消去未知力的点。例如，关于某个支点取矩以消去该处的反力。对于均匀杆，重力作用在中点。对于非均匀杆，重心不在中点，因此设其距一端为 x，用取矩求 x。
Common errors include using the distance along the rod instead of the perpendicular distance, forgetting the weight of the rod, and mixing clockwise and anticlockwise directions. If a force acts at an angle, use F sin θ or F cos θ to find the perpendicular component before calculating the moment.
常见错误包括使用沿杆的距离而不是垂直距离、忘记杆的重力、混淆顺时针与逆时针方向。若力斜向作用，先求垂直分量 F sin θ 或 F cos θ 再计算力矩。
8. Equilibrium, Resultants and Resolving Forces | 平衡、合力与力的分解
to the horizontal, its components are F cos θ horizontally and F sin θ vertically. Set ΣFx = 0 and ΣFy equilibrium, their lines of action are concurrent, and a triangle of forces can be drawn.
对于在共面力作用下平衡的质点，分别沿水平和竖直方向分解。若力 F 与水平方向成 θ 角，则分量为水平 F cos θ 和竖直 F sin θ。令 ΣFx = 0，ΣFy = 0。对于三个非平行力平衡，它们的作用线共点，可以画出力三角形。
Lami' / sin γ, where α, β, γ are the angles opposite the respective forces. However, resolving forces is usually safer and easier to mark. Always state the direction of the resultant, for example "resultant is 12 N at 30° above the horizontal".
拉密定理适用于三个力平衡：F₁ / sin α = F₂ / sin β = F₃ / sin γ，其中 α、β、γ 是各力对面的角。不过，分解力通常更稳妥也更容易得分。一定要说明合力的方向，例如 "合力为 12 N，与水平方向成 30° 向上"。
A common error is to subtract forces that are not in the same direction. Forces at an angle must be resolved first. Another error is to forget that tension acts away from the particle along the string, while thrust acts towards the particle along the rod.
常见错误是把方向不同的力直接相减。斜向力必须先分解。另一个错误是忘记张力沿绳远离质点，而推力沿杆指向质点。
9. Kinematics Graphs and Calculus Methods | 运动学图像与微积分方法
On a displacement-time graph, the gradient is velocity. On a velocity-time graph, the gradient is acceleration, and the area under the graph is displacement. Areas below the time axis count as negative displacement. On an acceleration-time graph, the area is change in velocity. Sketching the graph can reveal key times such as when velocity is zero or when displacement returns to zero.
在位移-时间图像中，斜率是速度。在速度-时间图像中，斜率是加速度，图像与时间轴围成的面积是位移。时间轴下方的面积算作负位移。在加速度-时间图像中，面积是速度变化量。画出图像可以帮助发现关键时间，如速度为 0 或位移回到 0 的时刻。
y: a = dv/dt = d²s/dt². Conversely, displacement is the integral of velocity: s = ∫ v dt, and velocity is the integral of acceleration: v = ∫ a dt. Use initial conditions to find the constant of integration.
当加速度不恒定时，使用微积分。速度是位移的导数：v = ds/dt。加速度是速度的导数：a = dv/dt = d²s/dt²。反过来，位移是速度的积分：s = ∫ v dt，速度是加速度的积分：v = ∫ a dt。用初始条件求积分常数。
t³ − 2t² + 2t + D, and the initial displacement determines D. Always include units and check whether the question asks for distance travelled or displacement.
例如，若 a = 6t − 4，则 v = 3t² − 4t + C。若 t = 0 时 v = 2，则 C = 2。然后 s = t³ − 2t² + 2t + D，初始位移决定 D。一定要带单位，并检查题目问的是路程还是位移。
10. Moments in Context: Beams, Ladders and Non-Uniform Rods | 力矩应用：梁、梯子与非均匀杆
Beam problems usually involve a uniform beam resting on two supports, with a load or a person standing on it. Take moments about one support to find the reaction at the other. Then resolve vertically to find the remaining reaction. If the beam is about to tilt, one reaction becomes zero, and that condition gives the critical position of the load.
梁问题通常涉及放置在两个支点上的均匀梁，上面有负载或人站立。关于一个支点取矩求另一个支点的反力，再竖直方向分解求剩下的反力。若梁即将倾斜，一个反力变为 0，该条件给出负载的临界位置。
Ladder problems combine moments, friction and equilibrium. A uniform ladder of length L and mass m rests against a smooth vertical wall and a rough horizontal floor. The wall reaction is horizontal, the floor reaction has normal and friction components, and weight acts at L/2. Take moments about the foot of the ladder to find the wall reaction, then resolve to find friction and the normal reaction.
梯子问题综合了力矩、摩擦与平衡。长度为 L、质量为 m 的均匀梯子斜靠在光滑竖直墙和粗糙水平地面上。墙的反力是水平的，地面的反力有法向和摩擦分量，重力作用在 L/2 处。关于梯脚取矩求墙的反力，再分解求摩擦力和法向反力。
For a non-uniform rod, the centre of mass is at an unknown distance from one end. Use the fact that the rod balances at a given point to find this distance. If the rod is suspended by two strings, take moments about one string to find the tension in the other. Always draw a clear diagram with all distances marked
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