📚 Moments and Equilibrium: Key Points for IB and WJEC Mathematics | 力矩与平衡：IB与WJEC数学考点精讲
Moments and equilibrium form a vital part of mechanics within IB and WJEC mathematics. Understanding how forces cause rotation and how to balance moments is essential for solving statics problems involving beams, ladders, and levers. This article systematically covers key concepts, formulas, and exam techniques to help you master the topic.
力矩与平衡是IB及WJEC数学力学部分的重要内容。理解力如何引起转动以及如何使力矩平衡，是解决横梁、梯子和杠杆等静力学问题的关键。本文系统地讲解核心概念、公式与考试技巧，帮助你掌握该专题。
1. Definition of Moment | 力矩的定义
The moment of a force about a point is the measure of its turning effect. It depends on both the magnitude of the force and the perpendicular distance from the point to the line of action of the force.
力对某点的力矩是衡量力对该点转动效应的物理量，它取决于力的大小以及从该点到力作用线的垂直距离。
If a force tends to rotate the object clockwise, the moment is often taken as negative; if anticlockwise, positive. This sign convention is arbitrary but must be used consistently.
如果力使物体有顺时针转动趋势，力矩通常取负值；若为逆时针，则取正值。这种符号约定是任意的，但必须统一使用。
2. Moment Formula and Perpendicular Distance | 力矩公式与垂直距离
The fundamental formula for the magnitude of a moment is:
力矩大小的基本公式为：
M = F × d
Here F is the magnitude of the force and d is the perpendicular distance from the pivot to the line of action of the force. The distance must always be measured at 90° to the force vector.
式中F为力的大小，d为从支点到力作用线的垂直距离。该距离必须始终与力矢量成90°测量。
If the force is not perpendicular to the lever arm, you must either resolve the force into perpendicular components or find the perpendicular distance directly using trigonometry.
若力不垂直于作用臂，则必须将力分解为垂直分量，或利用三角函数直接求垂直距离。
3. Units and Direction of Moment | 力矩的单位与方向
The SI unit of moment is the newton-metre (N m). Note that although this unit has the same dimensions as energy (joules), moments are not energy and the unit should not be written as joules.
力矩的国际单位是牛顿·米（N m）。注意虽然该单位与能量（焦耳）的量纲相同，但力矩并非能量，不能写作焦耳。
Direction is indicated by plus or minus signs. A common convention is to take anticlockwise moments as positive and clockwise moments as negative. In equilibrium problems, the sum of clockwise moments equals the sum of anticlockwise moments, so you can also choose to equate magnitudes.
方向用正负号表示。常见的约定是将逆时针力矩取正，顺时针取负。在平衡问题中，顺时针力矩之和等于逆时针力矩之和，因此也可以选择直接等同大小。
4. Principle of Moments | 力矩原理
For a system to be in rotational equilibrium, the sum of all clockwise moments about any pivot must equal the sum of all anticlockwise moments about that same pivot.
系统若处于转动平衡，则对任意支点，所有顺时针力矩之和必须等于所有逆时针力矩之和。
Σ Mclockwise = Σ Manticlockwise
This principle is used to find unknown forces or distances when an object is stationary and not rotating.
该原理用于当物体静止且不转动时求解未知力或距离。
5. Conditions for Equilibrium | 平衡条件
Complete static equilibrium requires two conditions to be satisfied simultaneously:
完全的静力平衡需要同时满足两个条件：
- Translational equilibrium: The vector sum of all forces acting on the object is zero. Σ F = 0
- Rotational equilibrium: The sum of the moments about any point is zero. Σ M = 0
- 平动平衡：作用在物体上的所有力的矢量和为零。Σ F = 0
- 转动平衡：对任意点的合力矩为零。Σ M = 0
In WJEC and IB mathematics exams, you typically apply these conditions to rigid bodies such as rods, beams, and ladders resting on supports.
在WJEC和IB数学考试中，通常会将这两个条件应用于刚体，如放在支座上的杆、横梁和梯子。
6. Reaction Forces and Supports | 支座反力与支撑
When a beam is supported, reaction forces arise at the supports. These forces are perpendicular to the surface if the support is smooth or a roller, but can have components if hinged or rough.
当横梁被支撑时，支座处会产生反力。如果支撑面光滑或为滚轴，反力垂直于表面；若为铰接或粗糙支座，则可能有分量。
To solve equilibrium problems, always draw a clear diagram showing all forces, including weight, applied loads, and reaction forces. Label distances from a chosen pivot.
求解平衡问题时，务必画出清晰的示意图，标明所有力，包括重力、外加载荷和支座反力。同时标注选定的支点到各力的距离。
A common exam step is to take moments about a point where an unknown force acts, eliminating that unknown from the moment equation.
常见考试技巧是取未知力作用点作为支点来求力矩，从而在力矩方程中消去该未知量。
7. Solving Uniform Beams with Point Loads | 均质横梁与集中载荷的求解
Consider a uniform beam of length L and weight W, supported at two points and carrying an additional point load. The weight acts at the centre of the beam because it is uniform.
考虑一根长度为L、自重为W的均质横梁，由两个支点支撑并承受额外的集中载荷。由于材质均匀，自重作用于梁的中点。
Steps to solve:
求解步骤：
- Identify all forces and mark their positions.
- Choose a pivot (usually one of the supports).
- Write the moment equation using the principle of moments.
- Solve for one unknown, then use Σ F = 0 to find the remaining reaction force.
- 识别所有力并标出其位置。
- 选取支点（通常为某一支座）。
- 利用力矩原理列出力矩方程。
- 求解一个未知量，再利用Σ F = 0求出剩余的支座反力。
Always check that both equilibrium conditions are satisfied as a verification step.
始终验证两个平衡条件是否均满足，作为检验步骤。
8. Distributed Loads and Their Moments | 分布载荷及其力矩
tributed length. For a UDL of w N/m over a length a, the total load is w × a and it acts at the centre of that segment.
作用在梁上的均布载荷（UDL）可等效为一个集中力，其大小等于总载荷，作用点位于分布长度的中点。若载荷强度为w N/m，作用长度为a，则总载荷为w × a，作用在该段的中心。
When taking moments, treat this resultant force as a single point load. The distance to the chosen pivot must be measured to the midpoint of the distributed load.
计算力矩时，将这一等效合力当作单个集中载荷处理。到选定支点的距离应量至分布载荷的中点。
9. Inclined Forces and Resolving | 倾斜力与力的分解
If a force acts at an angle θ to the beam, you must either resolve it into components perpendicular and parallel to the beam, or use the perpendicular distance method.
若力与横梁成θ角，则必须将其分解为垂直于梁和平行于梁的分量，或采用垂直距离法。
F⟂ = F sin θ
Only the perpendicular component contributes to the moment about a point on the beam. The parallel component passes through the pivot and creates no moment.
只有垂直分量才会对横梁上某点产生力矩。平行分量通过支点，不产生力矩。
= x sin θ, where x is the distance along the beam. Using M = F × d gives the same result.
另一种方法是延长力作用线并求出垂直距离d = x sin θ，其中x为沿梁的距离。采用M = F × d可得到相同结果。
10. Tilting and Toppling Problems | 倾斜与翻倒问题
Tilting occurs when the line of action of the weight of an object falls outside its base of support. In WJEC exam questions, you often determine the maximum load that can be placed before a beam tilts.
当物体重力的作用线超出支撑基底时，会发生倾斜。在WJEC考试题中，常需确定在横梁倾斜前所能施加的最大载荷。
At the point of tilting, the reaction force at one support becomes zero. The beam is still in equilibrium, but only just. Set the reaction at the lifting support to zero and take moments about the other support to find the critical load or distance.
在即将倾斜的临界点，某一支座的反力变为零。横梁仍处于平衡状态，但处于临界。设抬起端支座反力为零，对另一支座取力矩，即可求出临界载荷或临界距离。
11. Common Pitfalls and Exam Tips | 常见错误与考试技巧
- Forgetting the perpendicular distance: Always check whether the given distance is perpendicular to the force. If not, use trigonometry.
- Sign errors: Be consistent with the sign convention for clockwise and anticlockwise moments throughout the entire solution.
- Missing forces: Include all forces, especially the weight of the beam itself, which acts at its centre.
- Incorrect equivalent load for UDL: The resultant acts at the midpoint of the distributed segment, not at the support.
- Ignoring equilibrium of forces: After solving moments, apply Σ F = 0 to find reaction forces and verify consistency.
- 忘记垂直距离：务必检查给定距离是否与力垂直。若不垂直，使用三角函数换算。
- 符号错误：整个解题过程中保持一致的正负号约定。
- 遗漏力：将所有力纳入，尤其是横梁自重，其作用在梁的中心。
- 均布载荷等效错误：等效合力作用在分布段的中点，而非支座处。
- 忽略力的平衡：在利用力矩求解后，应用Σ F = 0求出反力并验证一致性。
In the exam, always draw a large, clear diagram and annotate all known values. It’s easy to gain method marks even if the final arithmetic is wrong.
考试时，务必画出大而清晰的示意图，并标注所有已知量。即便最后计算有误，也容易获得过程分。
12. Key Equations Summary | 关键公式总结
The following table summarises the essential equations for moments and equilibrium in IB and WJEC mathematics.
下表总结了IB与WJEC数学中力矩与平衡的基本公式。
| Quantity 量 | Formula 公式 | Notes 备注 |
|---|---|---|
| Moment of a force 力矩 | M = F × d | d is perpendicular distance d为垂直距离 |
| Principle of moments 力矩原理 | Σ Mcw = Σ Macw | Choose a pivot 任选支点 |
| Force equilibrium 力平衡 | Σ F = 0 | Horizontal & vertical 水平与竖直 |
| Moment of inclined force 倾斜力力矩 | M = F sin θ × r | r is distance along beam r为沿梁距离 |
| Equivalent UDL force 均布载荷等效力 | Feq = w × L | Acts at L/2 from start 作用于起始端L/2处 |
Master these concepts and you will be well prepared to tackle any moments and equilibrium problem confidently.
掌握了这些概念，你将能够自信地应对任何力矩与平衡问题。
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