📚 Moments and Equilibrium: Key Points Explained | 力矩与平衡考点精讲
In both IB and OCR Mathematics (Mechanics), the topic of moments and equilibrium is fundamental to understanding how forces cause rotation and how objects remain in balance. This revision guide covers essential concepts, common problem types, and exam techniques to help you master moments and equilibrium.
在 IB 和 OCR 数学（力学）中，力矩与平衡是理解力如何引起转动以及物体如何保持平衡的基础。本复习指南涵盖了基本概念、常见题型与考试技巧，助你掌握力矩与平衡。
1. What is a Moment? | 什么是力矩？
The moment of a force about a point is a measure of its turning effect. It is defined as the product of the force and the perpendicular distance from the point to the line of action of the force.
力对一点的力矩是衡量其转动效应的量。它定义为力与从该点到力作用线的垂直距离的乘积。
Moment = F × d
where d is the perpendicular distance from the pivot to the line of action of the force. The SI unit is newton-metre (N m).
其中 d 是从支点到力作用线的垂直距离。国际单位是牛顿·米 (N m)。
When the force is not perpendicular to the distance, you must use the perpendicular component: Moment = F d sinθ, where θ is the angle between the force and the lever arm.
当力不垂直于距离时，必须使用垂直分量：力矩 = F d sinθ，其中 θ 是力与力臂之间的夹角。
By convention, anticlockwise moments are taken as positive, and clockwise moments as negative. This sign convention is crucial when summing moments.
通常规定，逆时针力矩取正，顺时针力矩取负。在求合力矩时，这一符号规定至关重要。
2. Principle of Moments | 力矩原理
For an object in rotational equilibrium, the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about that same point.
对于处于转动平衡的物体，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。
Mathematically, Σ clockwise moments = Σ anticlockwise moments, or Σ M = 0 when using a sign convention.
数学上，顺时针力矩总和 = 逆时针力矩总和，或当使用符号规定时，合力矩为零 Σ M = 0。
This principle is often used to find unknown forces or distances in static problems. Always choose a point that eliminates one unknown force to simplify the calculation.
该原理常用于求解静力学问题中的未知力或距离。应始终选择能够消除一个未知力的点来简化计算。
3. Conditions for Equilibrium | 平衡条件
A rigid body is in static equilibrium if it satisfies two conditions: (1) the resultant force in any direction is zero (translational equilibrium), and (2) the resultant moment about any point is zero (rotational equilibrium).
刚体静力平衡需满足两个条件：(1) 任意方向上的合力为零（平动平衡），(2) 对任意一点的合力矩为零（转动平衡）。
These can be expressed as ΣF = 0 (both horizontally and vertically) and ΣM = 0. In many exam questions, you will resolve forces horizontally/vertically and take moments simultaneously.
可表示为 ΣF = 0（水平方向与竖直方向合力均为零）以及 ΣM = 0。在许多考题中，需要同时分解水平和竖直方向的力并取力矩。
Example: A uniform rod of length 2 m and weight 30 N is pivoted at one end and held horizontally by a vertical string at the other end. Find the tension in the string and the reaction at the pivot.
例子：一根长 2 m、重 30 N 的均匀杆，一端铰支，另一端由竖直绳子拉住保持水平。求绳子张力及铰支座反力。
Resolving vertically: R + T = 30. Taking moments about pivot: T × 2 = 30 × 1 (since weight acts at centre). Solving gives T = 15 N, R = 15 N.
竖直方向分解：R + T = 30。对铰支座取矩：T × 2 = 30 × 1（重力作用于中点）。解得 T = 15 N，R = 15 N。
4. Taking Moments About a Point | 对点取矩
When taking moments, always specify the point clearly. The perpendicular distance must be measured from that point. If a force passes through the point, its moment is zero, which is a useful trick to eliminate unknowns.
取力矩时，务必清晰指定取矩点。垂直距离须从该点量起。若力通过该点，其力矩为零，这是消除未知量的常用技巧。
In OCR exam problems, you often take moments about a support or a pivot where there is an unknown reaction force, so that reaction does not appear in the moment equation.
在 OCR 考题中，常对存在未知反力的支座或铰点取矩，从而使该反力不出现在力矩方程中。
Remember: distance is the perpendicular distance. If a force is inclined, use sin of the angle or resolve the force into components and take moments of each component.
记住：距离是指垂直距离。若力是倾斜的，需使用角度的正弦值，或将力分解为分量后对各分量取矩。
5. Uniform and Non-Uniform Rods | 均匀杆与非均匀杆
For a uniform rod, the weight acts at its centre (midpoint). For a non-uniform rod, the centre of mass location is either given or must be determined using moments. You can treat the weight as a concentrated force at the centre of mass.
对于均匀杆，重力作用于其中心（中点）。对于非均匀杆，重心位置要么给出，要么需用力矩求得。可将重力视为作用在质心处的集中力。
In problems with a non-uniform beam, you will need to use both the principle of moments and force balance to find the centre of mass or the reaction forces.
在涉及非均匀梁的问题中，需同时运用力矩原理与力的平衡来求质心或反力。
Example: A non-uniform rod of weight 40 N is supported at distances 0.8 m and 1.2 m from one end, and the reactions are 24 N and 16 N respectively. Find the distance of the centre of mass from the left end.
例子：一根重 40 N 的非均匀杆在距一端 0.8 m 和 1.2 m 处被支撑，反力分别为 24 N 和 16 N。求质心离左端的距离。
Taking moments about left end: 40x = 24×0.8 + 16×1.2 = 38.4, so x = 0.96 m.
对左端取矩：40x = 24×0.8 + 16×1.2 = 38.4，得 x = 0.96 m。
6. Centre of Mass and Moments | 质心与力矩
The centre of mass is the point where the entire weight of the object appears to act. For composite bodies or systems of particles, you can find the centre of mass by taking moments of the weights about a reference point.
质心是物体全部重力似乎作用的点。对于组合体或质点系，可通过将各重量对参考点取矩来求质心。
In OCR and IB, you may be asked to find the centre of mass of a system of particles placed along a light rod, or of a lamina. The method: sum of (mass × position) divided by total mass.
在 OCR 和 IB 中，可能要求计算放置在轻杆上的一系列质点组成的系统的质心，或薄板的质心。方法：(质量 × 位置) 的总和除以总质量。
x̄ = (Σ mᵢ xᵢ) / Σ mᵢ
This is an application of the moment concept where weight is proportional to mass.
这运用了力矩概念，其中重量与质量成正比。
For symmetrical objects, the centre of mass lies on axes of symmetry.
对于对称物体，质心位于对称轴上。
7. Resolving Forces and Moments Together | 力与力矩的综合求解
Many equilibrium problems require you to set up three equations: horizontal force balance, vertical force balance, and moment balance (usually taken about a point where an unknown acts). Solve these simultaneously.
许多平衡问题需要建立三个方程：水平力平衡、竖直力平衡以及力矩平衡（通常对未知力作用点取矩）。联立求解。
If a force is at an angle, resolve it into its horizontal and vertical components. The moment of an inclined force can be calculated by F × perpendicular distance, or by summing the moments of its components.
若力有角度，则将其分解为水平和竖直分量。倾斜力的力矩可通过 F × 垂直距离计算，或通过对其分量取力矩求和计算。
A typical problem: a ladder leaning against a smooth wall, with friction at the ground. Resolve horizontally and vertically, take moments about the point of contact with ground to find the normal reaction at the wall.
典型问题：梯子靠在光滑墙上，地面有摩擦。分解水平与竖直方向，对与地面接触点取矩，求出墙上的法向反力。
Always include all forces: weight, normal reactions, friction, tensions.
务必包含所有力：重力、法向反力、摩擦力、张力。
8. Hinged and Supported Beams | 铰支梁与支撑梁
A hinge or pivot can provide a reaction force in any direction unless otherwise specified. For a smooth hinge, there may be horizontal and vertical components. A roller support provides only a reaction perpendicular to the surface.
铰链或枢轴若无特别说明，可在任意方向提供反力。对于光滑铰链，可能存在水平与竖向分量。滚轴支座只提供垂直于表面的反力。
When a beam is supported at several points, you can take moments about one support to find another reaction. For beams with a uniformly distributed load (UDL), replace it with a single resultant force at the centre of the load.
当梁有多个支点时，可对某一支座取矩求另一支座反力。对于均布荷载 (UDL)，可用作用在荷载中心的单个合力替代。
In OCR problems, a uniformly distributed load of w N/m over a length L results in a
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