📚 A-Level WJEC Maths: Moments and Equilibrium Key Points | A-Level WJEC 数学：力矩与平衡 考点精讲
In A-Level WJEC Mathematics, moments and equilibrium form a cornerstone of the Mechanics component. Understanding how forces cause rotation and the conditions that keep a rigid body stationary is essential for solving real-world problems, from balancing a seesaw to analysing the forces on a ladder leaning against a wall. This article breaks down every crucial concept, provides clear worked-style explanations, and highlights common pitfalls to help you master the topic.
在 A-Level WJEC 数学中，力矩与平衡是力学部分的核心。理解力如何引起转动，以及保持刚体静止的条件，对于解决从跷跷板平衡到分析靠墙梯子受力等现实问题至关重要。本文将逐个拆解关键概念，提供清晰的解释，并指出常见错误，帮助你掌握这一主题。
1. What is a Moment? | 什么是力矩？
A moment is the turning effect of a force about a pivot (or point). It depends on two things: the size of the force and the perpendicular distance from the pivot to the line of action of the force. The larger the force or the greater the distance, the bigger the turning effect.
力矩是力对支点（或某一点）产生的转动效应。它取决于两个因素：力的大小以及支点到力作用线的垂直距离。力越大或距离越大，转动效应就越强。
The moment is calculated using the formula: Moment = Force × Perpendicular distance. In symbols, M = F × d, where d is measured in metres (m) and F in newtons (N). The SI unit of a moment is the newton-metre (Nm). Note that work also has units of Nm but is measured in joules; moments are not a form of energy, so we strictly use Nm.
力矩计算公式为：力矩 = 力 × 垂直距离。用符号表示为 M = F × d，其中 d 以米（m）为单位，F 以牛（N）为单位。力矩的国际单位是牛·米（Nm）。注意，虽然功的单位也是 Nm，但用焦耳表示；力矩不是能量形式，因此必须使用 Nm。
Always take the perpendicular distance. If a force is applied at an angle, you must either resolve the force into a component perpendicular to the pivot–force line, or use M = F d sin θ, where θ is the angle between the force and the line from pivot to point of application.
务必使用垂直距离。如果力以某个角度施加，你必须将力分解为垂直于支点与力作用点连线的分量，或者使用公式 M = F d sin θ，其中 θ 是力与从支点到作用点连线之间的夹角。
2. Sign Convention: Clockwise vs Anticlockwise | 正负号规定：顺时针与逆时针
To solve equilibrium problems, you need a consistent sign convention. Usually, anticlockwise moments are taken as positive, and clockwise moments as negative. The choice is up to you, but you must apply it consistently throughout a problem.
解决平衡问题时，需要采用统一的正负号规定。通常规定逆时针力矩为正，顺时针力矩为负。具体选择可自定，但必须在整个问题中始终如一。
When writing the principle of moments, the sum of clockwise moments equals the sum of anticlockwise moments if you use magnitudes only. But if you use signed moments, you set the algebraic sum of all moments about any point to zero.
在书写力矩原理时，如果只考虑大小，则顺时针力矩之和等于逆时针力矩之和。如果使用带符号的力矩，则对任一点的所有力矩代数和为零。
Exam tip: always state which direction you are taking as positive at the start. For example: ‘Take clockwise as positive.’ Then write an equation summing all moments.
应考提示：一开始就要说明你规定哪个方向为正。例如：“取顺时针方向为正。”然后列出所有力矩的代数和方程。
3. The Principle of Moments | 力矩原理
The principle of moments states that for a system in rotational equilibrium, the total clockwise moment about any pivot equals the total anticlockwise moment about that same pivot. Equivalently, the net moment is zero.
力矩原理指出，对于处于转动平衡的系统，绕任一转动轴的顺时针力矩总和等于逆时针力矩总和。等价地说，合外力矩为零。
This principle is used to find unknown forces or distances in balanced systems, such as a uniform beam resting on supports, or a rod carrying weights. You must choose a pivot point wisely—often you pick a point where an unknown force acts, so its moment is zero and does not appear in the equation.
该原理用于求解平衡系统中的未知力或距离，例如放置在支架上的均匀梁，或挂有重物的杆。你需要巧妙地选择支点——通常选择有未知力作用的那一点，这样该力的力矩为零，不会出现在方程中。
Strictly, the principle of moments is a necessary condition for rotational equilibrium. Together with the condition that the resultant force is zero (translational equilibrium), we get the full equilibrium conditions for a rigid body.
严格来说，力矩原理是转动平衡的必要条件。加上合外力为零（平动平衡），我们就得到了刚体的完整平衡条件。
4. Equilibrium of a Rigid Body | 刚体的平衡
A rigid body is in equilibrium when it is either at rest or moving with constant velocity (no acceleration). This requires two conditions to be met simultaneously.
当刚体处于静止或匀速运动（无加速度）状态时，就说它处于平衡状态。这需要同时满足两个条件。
Condition 1: Resultant force = 0 in all directions. Usually this means resolving forces horizontally and vertically. The sum of components left equals sum of components right; up equals down.
条件一：各方向的合外力为零。通常这意味着水平和竖直方向分解力。向左分量之和等于向右分量之和；向上的等于向下的。
Condition 2: Resultant moment = 0 about any point. You can take moments about any point you like, but the sum of moments must be zero. This condition rules out rotational acceleration.
条件二：对任意点的合外力矩为零。你可以选择任意点求力矩，但力矩总和必须为零。这一条件排除了转动加速度。
Many WJEC exam questions ask you to use both conditions together. For a uniform beam, first take moments about a support to find a reaction, then resolve vertically to find the other reaction.
许多 WJEC 考题要求同时使用这两个条件。对于均匀梁，通常先对某一支点求力矩以求出支反力，然后竖直方向分解力求另一个支反力。
| Translational equilibrium | Sum of forces = 0 (e.g. ΣF↑ = ΣF↓) |
| Rotational equilibrium | Sum of moments about any point = 0 |
5. Taking Moments About Different Points | 对不同点求力矩
A powerful technique is that the net moment is zero about any point in equilibrium. This means you can choose the most convenient pivot to simplify the algebra. Typically, you pick a point through which the line of action of an unknown force passes; that force then contributes no moment.
一个强有力的技巧是，在平衡状态下，对任意点的合外力矩都为零。这意味着你可以选择最方便的支点来简化计算。通常，选择某个未知力的作用线所经过的点，这样该力就不产生力矩。
For example, if a ladder leans against a smooth wall, the wall reaction is perpendicular to the wall. You might take moments about the contact point with the ground to eliminate the ground friction and normal reaction from the moment equation. Or take moments about the top to eliminate the wall reaction.
例如，一把梯子靠在光滑墙壁上，墙的反力垂直于墙面。你可以对梯脚与地面的接触点求力矩，从而将地面摩擦力和法向反力从力矩方程中消去。或者对梯顶求力矩，消去墙的反力。
Always label your diagram clearly with all forces and pivot. Use the perpendicular distance from the pivot to each force’s line of action. If a force does not act at a single point, such as the weight of the rod, treat it as acting at the centre of mass.
始终在图上清晰标出所有力和支点。使用从支点到各力作用线的垂直距离。如果某个力不是集中作用，比如杆的重力，则将其视为作用在质心处。
6. Uniform and Non-Uniform Rods | 均匀与非均匀杆
A uniform rod has its weight evenly distributed; its centre of mass is at its geometric centre. The weight can be treated as a single force acting at the midpoint. For a uniform rod of length L, the weight acts at L/2 from either end.
均匀杆的重量均匀分布，其质心位于几何中心。重力可视为作用在中点的一个力。对于长度为 L 的均匀杆，重力作用在距离两端各 L/2 处。
A non‑uniform rod has a centre of mass that is not necessarily in the middle. The problem may give you the position of the centre of mass, or ask you to find it. You will need to include the weight acting at that unknown distance from a reference point.
非均匀杆的质心不一定在中间。题目可能给出质心位置，或要求你求出质心。你需要将重力力作用在距离参考点未知的位置上。
Typical question: ‘A non‑uniform rod AB of length 5 m and weight 60 N rests horizontally on two supports… Find the distance of the centre of mass from A.’ You set up two equations (moments and vertical forces) and solve for the unknown distance.
典型问题：“一根长 5 m、重 60 N 的非均匀杆 AB 水平搁在两个支架上……求质心到 A 的距离。”你建立两个方程（力矩和竖直方向受力），求解未知距离。
7. Centre of Mass and Its Role in Moments | 质心及其在力矩中的作用
The weight of an extended object acts through its centre of mass. Therefore, when calculating the moment of the weight about a point, you use the perpendicular distance from that point to the vertical line through the centre of mass.
广延物体的重力通过其质心作用。因此，在计算重力对某点的力矩时，需使用该点到通过质心的竖直线之间的垂直距离。
For composite bodies, such as a rod with several particles attached, the centre of mass can be found using moments: sum of (mass × distance from a reference point) divided by total mass. This idea is often tested in WJEC papers.
对于组合体，如附有几个质点的杆，可用力矩求出质心：各（质量 × 到参考点距离）之和除以总质量。这一思想在 WJEC 试卷中经常考查。
If a particle is placed on a beam, its weight creates a moment. You treat it as a point mass, so the perpendicular distance is simply the distance along the beam from the pivot.
如果梁上放置了一个质点，其重力会产生力矩。你将其视作一个质点，因此垂直距离就是从支点沿梁到该点的距离。
8. Ladder Problems | 梯子问题
Ladder problems are classic equilibrium questions involving friction. A ladder rests against a rough or smooth wall and ground. The forces include weight, normal reactions, and friction. You need to use both horizontal/vertical force balance and the moment equation.
梯子问题是涉及摩擦力的经典平衡题。梯子靠在粗糙或光滑的墙和地面上。作用力包括重力、法向反力和摩擦力。你需要同时使用水平和竖直方向的力平衡以及力矩方程。
If the wall is smooth, there is no vertical friction at the wall, only a horizontal reaction. The ground may be rough, providing both normal reaction and a horizontal friction force. Always draw a free‑body diagram and mark all forces clearly.
如果墙是光滑的，墙处没有竖直摩擦力，只有水平反力。地面可能是粗糙的，提供法向反力和水平摩擦力。务必画出隔离体图，清晰标出所有力。
Take moments about a point that eliminates as many unknowns as possible. Usually, the moment about the foot of the ladder eliminates both the normal reaction and friction at the ground. Then resolve forces.
选择一个能消去尽可能多未知数的点求力矩。通常，对梯脚求力矩可同时消去地面的法向反力和摩擦力。然后再分解力。
Remember, when the ladder is on the point of slipping, friction reaches its limiting value, F = μR, where μ is the coefficient of friction. This extra equation helps solve for μ or the angle of the ladder.
记住，当梯子即将滑动时，摩擦力达到极限值 F = μR，其中 μ 是摩擦系数。这个额外方程有助于求解 μ 或梯子角度。
9. Hinges and Fixed Points | 铰链与固定点
When a body is attached to a hinge, the hinge exerts a reaction force that can have both horizontal and vertical components. Unlike a simple pivot, a hinge can provide a reaction in any direction. You must treat the hinge reaction as two perpendicular components, or as a single force at an unknown angle.
物体与铰链连接时，铰链施加的支反力可以有水平分量和竖直分量。与简单支点不同，铰链能在任意方向提供反力。你必须把铰链反力处理为两个互相垂直的分量，或者一个大小方向未知的力。
A typical problem: a uniform door of weight W is held horizontally by two hinges. You take moments about one hinge to find the vertical component at the other, then resolve. Always check if the hinges are well‑oiled (smooth) or can exert thrust in any direction.
典型问题：一扇重 W 的均质门水平地由两个铰链支撑。你对一个铰链求力矩以求出另一个铰链的竖直分力，然后再分解力。务必检查铰链是否润滑（光滑），或者能否提供任意方向的推力。
When taking moments about a hinge, the reaction at that hinge has no moment because its line of action passes through the pivot. This technique is extremely useful.
对铰链求力矩时，该铰链处的反力力矩为零，因为其作用线通过支点。这种技巧极其有用。
10. Tilting and Toppling | 倾斜与倾倒
A rigid body on a support will tilt (or topple) when the line of action of its weight falls outside its base. In WJEC mechanics, questions often ask: ‘Find the maximum weight that can be hung from a beam without it tilting.’
当刚体重力的作用线落在支撑面之外时，它将倾斜（或倾倒）。在 WJEC 力学中，常问：“求出能够悬挂在梁上而不使梁倾斜的最大重量。”
At the point of tilting, one of the support reactions becomes zero. Suppose a beam rests on two supports A and B. As a weight moves towards A, the reaction at B decreases. The beam is about to tilt when the reaction at B becomes exactly zero. At that instant, the beam is just in contact with B but exerts no force on it.
在即将倾斜的瞬间，某一支反力变为零。假设一根梁搁在 A、B 两个支点上。当重物向 A 移动时，B 点的支反力逐渐减小。当 B 点支反力恰好为零时，梁即将倾斜。此时梁与 B 点刚接触但不施加力。
To solve, set the reaction at the support that is about to lift to zero, then take moments about the other support. This gives the limiting position or the maximum load.
解题时，将即将离地的那个支点的支反力设为零，然后对另一支点求力矩。由此得出极限位置或最大荷载。
11. Typical WJEC Exam Questions and Approach | 典型 WJEC 考题与解题思路
WJEC A-Level mechanics papers tend to test moments and equilibrium in scaffolded parts. For example: (a) mark forces on the diagram, (b) find a reaction by taking moments, (c) find the other reaction by resolving, (d) use limiting friction or tilting condition. Always read the whole question before starting, because earlier parts are designed to help the later ones.
WJEC A-Level 力学试卷往往以递进式问题考查力矩与平衡。例如：（a）在图上标出力，（b）通过求力矩计算支反力，（c）分解力求另一支反力，（d）利用极限摩擦力或倾斜条件。解题前务必通读全题，因为前面的小问往往为后面的小问做铺垫。
Common trick: if a string is attached to a beam and exerts a tension at an angle, resolve that tension into horizontal and vertical components or use the perpendicular distance directly from the pivot to the line of tension.
常见技巧：如果绳子系在梁上并以一个角度施加拉力，可将该拉力分解为水平与竖直分量，或者直接使用从支点到拉力作用线的垂直距离。
Units and accuracy: give final answers to 2 or 3 significant figures unless exact. Include units (N, Nm, etc.). If you get a negative reaction, it may indicate the direction you assumed is opposite—interpret with the physical situation.
单位和精确度：除非给出精确值，最终答案取 2 或 3 位有效数字。带上单位（N、Nm 等）。如果得出负值反力，可能表明假设方向与实际相反——结合物理情景进行解释。
12. Summary and Key Points for Revision | 总结与复习要点
To succeed with WJEC moments and equilibrium questions, remember: (1) Draw a clear, labelled diagram. (2) State your sign convention and chosen pivot. (3) Apply the two equilibrium conditions: ΣF = 0 (in two perpendicular directions) and ΣM = 0 about any convenient point. (4) Choose the pivot to eliminate unknowns. (5) Use the limiting friction relation F = μR when slipping is about to occur. (6) For tilting, set the rising reaction to zero.
想在 WJEC 力矩与平衡题中取得好成绩，请记住：（1）画出清晰、有标注的示意图。（2）声明正负规定和所选支点。（3）应用两个平衡条件：ΣF = 0（分别沿两个垂直方向）以及对任意合适点的 ΣM = 0。（4）选择支点以消去未知量。（5）即将滑动时使用极限摩擦关系 F = μR。（6）对于倾斜问题，将上升的那一支反力设为零。
Practice with past papers, particularly those with ladders, hinges, and non‑uniform beams. Reinforce the idea that moments are independent of translational effects—they are a separate condition that must be satisfied. A solid understanding of perpendicular distances is the foundation of every solution.
通过历年真题进行练习，尤其是涉及梯子、铰链和非均匀梁的题目。强化一个观念：力矩与平动效应相互独立——它们是需要分别满足的独立条件。对垂直距离的扎实理解是每一道题解答的基础。
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