📚 Moment and Equilibrium | 力矩与平衡
In A‑Level Mechanics, the concept of a moment measures the turning effect of a force about a point. Understanding moments and the conditions for equilibrium is essential for analysing rigid bodies, beams, ladders, and other structures. This article summarises the key principles, definitions, and typical problem‑solving strategies for the CIE Mathematics syllabus.
在 A‑Level 力学中，力矩衡量力对某一点的转动效应。理解力矩与平衡条件是分析刚体、梁、梯子及其他结构的基础。本文总结了CIE数学大纲中的核心原理、定义和典型解题策略。
1. What is a Moment? | 什么是力矩？
A moment is the turning effect produced when a force is applied away from a pivot. The size of a moment depends on two factors: the magnitude of the force and the perpendicular distance from the pivot to the line of action of the force. The greater the force or the distance, the larger the turning effect.
力矩是力在离支点一定距离处产生的转动效应。力矩的大小取决于两个因素：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，转动效应越显著。
2. Definition and Formula | 定义与公式
For a single force, the moment M about a point is given by:
对于单个力，关于某一点的力矩 M 由下式给出：
M = F × d
where 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 SI unit of moment is the newton‑metre (N m). Moments are vector quantities: by convention, anticlockwise moments are taken as positive and clockwise moments as negative, or vice versa, as long as the sign convention is consistent.
其中 F 是力的大小，d 是支点到力作用线的垂直距离。力矩的 SI 单位是牛顿·米（N m）。力矩是矢量：通常规定逆时针力矩为正，顺时针为负，反之亦可，但必须保持符号约定一致。
3. Principle of Moments | 力矩原理
When a rigid body is in rotational equilibrium, the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about that same point. This is known as the principle of moments. It allows us to write an equation that does not involve the reaction force at that pivot if we take moments about a point where an unknown force acts.
当刚体处于转动平衡时，关于 任意 一点的顺时针力矩之和等于关于同一点的逆时针力矩之和。这就是力矩原理。若我们取矩点选在未知力作用点上，可以在方程中消去该未知的反作用力。
For a light rod in equilibrium under several parallel forces, we often take moments about one of the supports to find the other support reaction.
对于受多个平行力作用的轻杆，通常对其中一个支点取矩以求另一个支点的反力。
4. Equilibrium Conditions for Rigid Bodies | 刚体的平衡条件
A rigid body is in static equilibrium when two independent conditions are satisfied simultaneously:
刚体处于静力平衡时必须同时满足两个独立条件：
1. Translational equilibrium: The vector sum of all forces acting on the body is zero.
2. Rotational equilibrium: The sum of moments about any point is zero.
1. 平动平衡：作用在刚体上所有力的矢量和为零。
2. 转动平衡：关于 任意 一点的合力矩为零。
In two dimensions, this gives up to three independent equations: sum of horizontal components = 0, sum of vertical components = 0, and sum of moments about a chosen point = 0. These equations are sufficient to solve for unknown forces and distances.
在二维问题中，最多可以列出三个独立方程：水平分量之和为零、竖直分量之和为零、对选定点的力矩之和为零。这些方程足以求解未知的力与距离。
5. Couples and Torque | 力偶与扭矩
A couple consists of two equal, parallel but opposite forces whose lines of action do not coincide. A couple produces rotation without any resultant linear force. The moment of a couple (also called torque) is independent of the point about which moments are taken and is given by:
力偶由一对大小相等、方向相反且作用线不共线的平行力组成。力偶只产生转动，不产生净力。力偶矩（又称扭矩）与取矩点无关，可由下式给出：
Moment of couple = F × d
where d is the perpendicular distance between the two parallel lines of action. Couples are very useful when analysing systems such as steering wheels or rotating machinery.
其中 d 是两条平行作用线之间的垂直距离。在分析方向盘或旋转机械等系统时，力偶非常有用。
6. Centre of Gravity and Centre of Mass | 重心与质心
The centre of gravity (C.G.) of a body is the point through which the entire weight of the body may be considered to act, regardless of its orientation. For a uniform rod, the C.G. is at its midpoint. For a uniform lamina, the C.G. can be found using symmetry or by taking moments of individual masses. In uniform gravitational fields, the centre of gravity coincides with the centre of mass.
物体的重心是无论物体处于何种方位，其全部重量可视为集中作用的点。对于均匀杆，重心在其中点。对于均匀薄板，可通过对称性或各部分质量取矩求得重心。在均匀重力场中，重心与质心重合。
Problems often involve a non‑uniform rod, where the centre of mass is not at the geometric centre. The position of the centre of mass may be given or must be found by balancing moments.
题目常涉及非均匀杆，其质心不在几何中心。质心位置可能直接给出，或需通过力矩平衡求出。
7. Reactions at Supports and Pivots | 支点与转轴的反作用力
When a beam rests on supports or is hinged at a pivot, the supports exert normal reaction forces perpendicular to the beam (if smooth). A hinge can provide both a horizontal and a vertical component of reaction. When taking moments, it is often strategic to choose a point through which an unknown reaction passes, so its moment is zero and it disappears from the equation.
当梁搁在支座上或铰接于转轴时，光滑支座施加垂直于梁的法向反力。铰链可提供水平和竖直两个分量的反力。取矩时，策略上常选择未知反力作用点作为矩心，使其力臂为零，从而在方程中消去该未知量。
8. Tilting and Limiting Equilibrium | 倾倒与极限平衡
A body placed on a rough or smooth surface will tilt (or topple) when the line of action of its weight falls outside the base of support. The limiting case occurs when the reaction at one support becomes zero. To find the greatest load that can be applied before tilting, set the reaction at the pivot about which tilting occurs to zero and apply the principle of moments about the other support.
放置在粗糙或光滑表面的物体，当其重力作用线超出支撑基底时会倾倒。极限情况发生在某一支点的反力为零时。求倾倒前可施加的最大载荷时，可将倾倒支点的反力设为零，并对另一支点运用力矩原理。
9. Common Problem‑Solving Strategy | 常见解题策略
1. Draw a clear, labelled diagram showing all forces, distances and the chosen pivot.
2. Resolve forces vertically and horizontally if needed.
3. Choose a point to take moments – pick a point that eliminates as many unknown forces as possible.
4. Write an equation using the principle of moments (anticlockwise moments = clockwise moments) or set total moment = 0.
5. Solve simultaneous equations to find unknown reactions, forces or distances.
6. Check that your answers are physically reasonable (e.g., positive reactions).
1. 绘制清晰标注的示意图，标出所有力、距离和所选矩心。
2. 如有需要，将力沿水平和竖直方向分解。
3. 选取矩心——选择能消去尽可能多未知力的点。
4. 利用力矩原理（逆时针力矩=顺时针力矩）建立方程，或令总力矩为零。
5. 解联立方程求出未知反力、力或距离。
6. 检查答案的物理合理性（如反力应为正值）。
10. Sample Problem with Worked Solution | 典型例题详解
Problem: A uniform beam AB of length 6 m and weight 200 N rests horizontally on two supports at C and D, where AC = 1 m and DB = 1 m. A load of 300 N is placed at the point 1.5 m from A. Find the reactions at C and D.
问题： 一根长 6 m、重 200 N 的均匀梁 AB 水平放置在 C 和 D 两个支点上，AC = 1 m，DB = 1 m。在距 A 端 1.5 m 处放置 300 N 的载荷。求 C、D 处的反力。
Solution: Let the reactions at C and D be RC and RD. The weight 200 N acts at the midpoint, 3 m from A. Distances from A: C is at 1 m, D is at 5 m (since DB = 1 m), load at 1.5 m.
解： 设C和D的反力为 RC 和 RD。重力 200 N 作用在梁中点，距A端3 m。各力距A端距离：C在1 m，D在5 m（因DB=1 m），载荷在1.5 m。
Resolving vertically:
竖直方向合力为零：
RC + RD = 200 + 300 = 500 N
Take moments about C (so RC has zero moment):
对 C 取矩（使 RC 力矩为零）：
Anticlockwise moment of RD about C = RD × (5 – 1) = 4RD.
Clockwise moments about C: weight: 200 × (3 – 1) = 400 N m; load: 300 × (1.5 – 1) = 150 N m.
RD 对 C 的逆时针力矩 = RD × (5 – 1) = 4RD。
对 C 的顺时针力矩：重力：200 × (3 – 1) = 400 N m；载荷：300 × (1.5 – 1) = 150 N m。
Equating clockwise and anticlockwise moments:
令顺时针力矩等于逆时针力矩：
4RD = 400 + 150 = 550
RD = 137.5 N
Then RC = 500 – 137.5 = 362.5 N.
代入得 RC = 500 – 137.5 = 362.5 N。
Thus the reactions are 362.5 N at C and 137.5 N at D.
因此，C处的反力为 362.5 N，D处的反力为 137.5 N。
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