📚 Forces: Vectors and Moments | 力：矢量与力矩
Forces govern all interactions in mechanics, from the gentle pull of a spring to the immense thrust of a rocket. At A-Level, mastering forces means moving beyond simple pushes and pulls into a world of vector addition, resolution, and the turning effects that decide whether a bridge stands or a ladder slips. This article unpacks the core concepts of vectors and moments, laying down clear definitions, step-by-step methods, and worked examples that align with the CIE Physics syllabus. Whether you are drawing free-body diagrams or balancing a beam, understanding these principles will give you the confidence to tackle any mechanics problem.
力支配着力学中的一切相互作用，从弹簧的轻柔拉伸到火箭的巨大推力。在A-Level阶段，掌握力意味着超越简单的推拉，进入矢量加法、分解以及决定桥梁是否屹立、梯子是否滑动的转动效应的世界。本文将剖析矢量与力矩的核心概念，给出清晰的定义、逐步的方法以及与CIE物理大纲一致的典型例题。无论你是在画受力图还是使横梁平衡，理解这些原理都将让你有信心解决任何力学问题。
1. Scalar and Vector Quantities | 标量与矢量
In physics, measurable quantities are classified as either scalars or vectors. A scalar quantity is fully described by its magnitude (size) and a unit; for example, a mass of 5 kg or a temperature of 20 °C. Scalars add and subtract using ordinary arithmetic and do not involve direction.
在物理学中，可测量的物理量被分为标量或矢量。标量仅由其大小（量值）和单位完全描述；例如，5 kg的质量或20 °C的温度。标量使用普通的算术进行加减，不涉及方向。
A vector quantity, however, possesses both magnitude and direction. Displacement (5 m due north), velocity (30 m s⁻¹ east), force (10 N downwards) and acceleration are all vectors. Vectors cannot be combined by simple arithmetic; their direction must be accounted for using geometric or component methods.
然而，矢量同时具有大小和方向。位移（向北5 m）、速度（向东30 m s⁻¹）、力（向下10 N）和加速度都是矢量。矢量不能通过简单的算术来合成；必须使用几何法或分量法来考虑它们的方向。
2. Vector Addition and Resolution | 矢量加法与分解
Adding vectors is most easily visualised by the tip-to-tail method: draw the first vector, then place the tail of the second vector at the tip of the first. The resultant vector goes from the start of the first vector to the end of the last. For perpendicular vectors, the magnitude of the resultant can be found using Pythagoras' theorem, and its direction by trigonometry.
矢量加法最直观的方法是三角形法则：画出第一个矢量，然后将第二个矢量的尾部放在第一个矢量的尖端。合矢量从第一个矢量的起点指向最后一个矢量的末端。对于相互垂直的矢量，合力的大小可以用勾股定理求出，方向则通过三角函数确定。
R = √F₁² + F₂² ; θ = tan⁻¹(F₂/F₁)
When a vector is not aligned with the axes, resolving it into perpendicular components makes calculation far easier. A force F at an angle θ to the horizontal has a horizontal component F cos θ and a vertical component F sin θ. Choosing mutually perpendicular directions (often horizontal and vertical) allows us to convert a two-dimensional vector problem into two independent one-dimensional problems.
当矢量不与坐标轴对齐时，将其分解为相互垂直的分量会使计算简单得多。一个与水平方向成θ角的力F，水平分量为F cos θ，垂直分量为F sin θ。选择相互垂直的方向（通常是水平和垂直）可以将一个二维矢量问题转化为两个独立的一维问题。
3. Free-body Diagrams | 受力图
A free-body diagram is a simplified sketch that shows all the forces acting on a single body. Each force is drawn as an arrow pointing in the direction it acts, with its length roughly proportional to its magnitude. Common forces include weight (mg, acting downwards from the centre of gravity), normal contact force (perpendicular to the surface), tension (along a string or cable), friction (parallel to the interface, opposing motion) and applied pushes or pulls.
受力图是一个简化的示意图，它表示作用在单个物体上的所有力。每个力都画成一个指向作用方向的箭头，其长度大致与力的大小成正比。常见的力包括重力（mg，从重心向下作用）、法向接触力（垂直于表面）、张力（沿绳子或缆绳方向）、摩擦力（平行于接触面，与运动方向相反）以及外加的推力或拉力。
Drawing a clear free-body diagram is the essential first step before applying equilibrium conditions or Newton’s second law. Always isolate the body, label each force with its symbol and value if known, and note the direction by an adjacent angle or broken line.
在应用平衡条件或牛顿第二定律之前，画出清晰的受力图是必不可少的首要步骤。一定要隔离物体，标注每个力的符号与已知数值，并借助旁边的角度或虚线注明方向。
4. Moments: The Turning Effect of a Force | 力矩：力的转动效应
A moment is the turning effect produced by a force about a pivot or axis. Its magnitude depends on two factors: the size of the force and the perpendicular distance from the pivot to the line of action of the force. In symbols:
力矩是力绕支点或轴产生的转动效应。其大小取决于两个因素：力的大小以及从支点到力作用线的垂直距离。符号表示为：
M = F × d⊥
where F is the force (in newtons) and d⊥ endicular distance (in metres). The SI unit of moment is the newton-metre (N m). To find d⊥ y the distance along the lever by the sine of the angle between the force and the lever: d⊥ force whose line of action passes directly through the pivot produces zero moment, since d⊥ = 0.
式中F是力（单位为牛顿），d⊥是垂直距离（单位为米）。力矩的国际单位是牛顿·米（N m）。为了求出d⊥，通常需将沿杠杆的距离乘以力与杠杆之间夹角的正弦：d⊥ = d sin θ。作用线直接通过支点的力产生的力矩为零，因为此时d⊥ = 0。
5. Calculating Moments and Perpendicular Distance | 计算力矩与垂直距离
a nut: if you pull at an angle, only the component F sin θ creates a turning effect; the component F cos θ pulls along the spanner and gives no moment. Therefore, M = F d sin θ.
垂直距离是从支点量到力的作用线，而不一定是力的作用点。想象用扳手拧螺母：如果你以某个角度拉，只有分量F sin θ产生转动效应；分量F cos θ沿着扳手方向拉，不产生力矩。因此，M = F d sin θ。
When multiple forces act on a body, each force contributes its own moment. By convention, moments that tend to cause clockwise rotation are often taken as positive, while anticlockwise moments are negative, or vice versa. Consistency in sign convention is crucial when summing moments.
当多个力作用在一个物体上时，每个力都产生各自的力矩。按照惯例，倾向于引起顺时针转动的力矩常取为正，而逆时针力矩为负，或者反过来。在对力矩进行求和时，符号约定的一致性至关重要。
6. Principle of Moments | 力矩原理
For a body in rotational equilibrium, the total clockwise moment about any pivot must equal the total anticlockwise moment about that same pivot. This is the principle of moments:
对于处于转动平衡的物体，绕任意支点的总顺时针力矩必须等于绕同一支点的总逆时针力矩。这就是力矩原理：
Σ M_clockwise = Σ M_anticlockwise
This principle allows us to calculate unknown forces or distances in balanced systems such as seesaws, bridges and crane arms. The pivot can be chosen arbitrarily; selecting a point through which an unknown force passes simplifies the calculation, because that force then contributes zero moment.
利用这一原理，我们可以计算跷跷板、桥梁和起重机臂等平衡系统中未知的力或距离。支点可以任意选取；选择一个未知力通过的点可以简化计算，因为该力对该点的力矩为零。
7. Centre of Gravity and Stability | 重心与稳定性
The centre of gravity (C.G.) of a body is the point where the entire weight of the object may be considered to act. For a uniform symmetrical object, the C.G. lies at its geometric centre. For irregular shapes, it can be found experimentally by suspending the object from two different points and drawing vertical lines: their intersection locates the C.G.
物体的重心是指可以认为物体全部重量都集中作用于该点。对于均匀对称的物体，重心位于其几何中心。对于不规则形状，可以通过从两个不同点悬挂物体并画出铅垂线来实验确定重心：两条线的交点即重心所在。
Stability is closely linked to the position of the centre of gravity. An object is stable if its C.G. is low and its base area is wide. If the line of action of the weight falls outside the base, the object will topple. This is why racing cars are low and wide, and why high-loaded lorries are more likely to overturn.
稳定性与重心的位置密切相关。如果一个物体的重心低且支撑面积宽，它就是稳定的。如果重力的作用线超出支撑面以外，物体就会倾覆。这就是赛车又低又宽、以及高载重货车更容易侧翻的原因。
8. Couples and Torque | 力偶与转矩
A couple consists of two equal and opposite parallel forces whose lines of action do not coincide. A couple produces a pure rotational effect without any resultant linear force. The moment of a couple (often called torque) is given by:
力偶由两个大小相等、方向相反且作用线不重合的平行力组成。力偶产生纯粹的转动效应，而不会有任何净的平动合力。力偶的力矩（常称为转矩）由下式给出：
Torque = F × d
where F is the magnitude of one of the forces and d is the perpendicular distance between the two lines of action (the arm of the couple). Unlike a single moment that depends on a pivot, the torque of a couple is the same about any point. Steering wheels, screwdrivers and electric motors all rely on couples to produce rotation.
式中F是其中一个力的大小，d是两条作用线之间的垂直距离（力偶臂）。与依赖于某个支点的单一力矩不同，力偶的转矩绕任意点都相同。方向盘、螺丝刀和电动机都是依靠力偶来产生转动的。
9. Equilibrium Conditions for Rigid Bodies | 刚体的平衡条件
For a rigid body to be in complete equilibrium, two conditions must be satisfied simultaneously:
一个刚体要处于完全平衡状态，必须同时满足两个条件：
1. Translational equilibrium: The vector sum of all forces acting on the body must be zero. In component form: Σ Fx = 0 and Σ Fy = 0.
1. 平动平衡：作用在物体上的所有力的矢量和必须为零。用分量形式表示为：Σ Fx = 0 且 Σ Fy = 0。
2. Rotational equilibrium: The algebraic sum of all moments about any point must be zero: Σ M = 0.
2. 转动平衡：绕任意点的所有力矩的代数和必须为零：Σ M = 0。
These conditions are the bedrock of statics problems. Start by drawing a free-body diagram, resolve forces into convenient perpendicular directions, apply Σ Fx = 0 and Σ Fy e a pivot to take moments. Solving the equations simultaneously yields the unknown forces.
这些条件是静力学问题的基础。首先画受力图，将力分解到方便的垂直方向上，应用Σ Fx = 0和Σ Fy = 0，然后选取一个支点计算力矩。联立求解这些方程就可以得到未知力的大小。
10. Worked Examples and Common Pitfalls | 典型例题与常见错误
Example: A uniform beam of length 5.0 m and weight 200 N rests on two supports at its ends. A load of 600 N is placed 2.0 m from the left end. Find the reaction forces at the supports. (Take g = 10 N kg⁻¹)
例题：一根长5.0 m、重200 N的均匀横梁，两端搁在两个支座上。在距离左端2.0 m处放有一个600 N的负载。求支座的反作用力。（取g = 10 N kg⁻¹）
t eliminates R₁. The weight of the beam acts at its centre, 2.5 m from left, giving an anticlockwise moment of 200 N × 2.5 m = 500 N m. The load gives 600 N × 2.0 m = 1200 N m, also anticlockwise. The right support produces a clockwise moment R₂ × 5.0 m. Applying the principle of moments: 500 + 1200 = 5.0 R₂, so R₂ = 340 N. Then R₁ = 800 − 340 = 460 N.
设左支座反力为R₁，右支座反力为R₂。由竖直方向平衡：R₁ + R₂ = 200 N + 600 N = 800 N。绕左支座取矩可消去R₁。横梁的重力作用在其中心，距左端2.5 m，产生逆时针力矩200 N × 2.5 m = 500 N m。负载产生600 N × 2.0 m = 1200 N m，也是逆时针。右支座产生顺时针力矩R₂ × 5.0 m。应用力矩原理：500 + 1200 = 5.0 R₂，得R₂ = 340 N。因此R₁ = 800 − 340 = 460 N。
Common pitfalls: forgetting to use the perpendicular distance to the line of action, incorrectly assuming the weight acts at the point of support, mixing up clockwise and anticlockwise signs, and neglecting to check that both force and moment equilibrium hold. Always test your answer by taking moments about a second point to ensure consistency.
常见错误：忘记使用到作用线的垂直距离，错误地认为重力作用在支撑点上，混淆顺时针和逆时针的符号，以及忽略同时检查力和力矩平衡。始终要绕第二个点取矩来验证答案的一致性。
Published by TutorHao | Physics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图