Moments and Equilibrium: Key Concepts for IB & CIE Mathematics | 力矩与平衡考点精讲

📚 Moments and Equilibrium: Key Concepts for IB & CIE Mathematics | 力矩与平衡考点精讲

In the Mechanics units of IB and CIE Mathematics, the study of moments and equilibrium lays the foundation for analysing rigid bodies under the action of forces. A solid understanding of these concepts allows you to solve a wide range of practical problems – from balancing a uniform beam to determining unknown reaction forces in structures. This article presents a structured, exam-focused review of the key principles, problem-solving techniques, and common pitfalls associated with moments and equilibrium.

在IB和CIE数学的力学模块中，力矩与平衡是分析刚体受力状态的基础。扎实掌握这些概念，你就能解决从平衡均匀梁到确定结构支座反力等各类实际问题。本文围绕考试要求，系统梳理了力矩与平衡的核心原理、解题方法和常见易错点，帮助你高效备考。

1. What Is a Moment? | 什么是力矩？

A moment measures the turning effect of a force about a given point or pivot. It depends on two factors: the magnitude of the force and the perpendicular distance from the pivot to the line of action of the force.

力矩衡量的是力对某一点（或转轴）产生的转动效应。它取决于两个因素：力的大小以及转轴到力作用线的垂直距离。

Moment = Force × Perpendicular Distance

力矩 = 力 × 垂直距离

In mathematical notation we often write M = Fd, where d is the perpendicular distance. The SI unit is the newton-metre (N m). The direction of a moment is either clockwise or anticlockwise, and in calculations we usually assign a positive sign to one direction and a negative sign to the other.

数学中常用 M = Fd 表示，其中 d 为垂直距离。国际单位是牛·米（N m）。力矩有顺时针和逆时针两种转向，计算时通常规定某一转向为正、另一转向为负。

2. Calculating the Moment of a Force | 力的力矩计算

To find the moment, always identify the perpendicular distance from the pivot to the line of action. If the force is not perpendicular, resolve it into components and use the component perpendicular to the line joining the pivot to the point of application.

计算力矩时，务必找出从转轴到力作用线的垂直距离。若力不垂直，可将力分解，然后用垂直于转轴与作用点连线方向的分量来计算。

For a force F applied at an angle θ to a lever arm of length r, the moment about the pivot is either F r sin θ or simply F times the lever arm’s perpendicular projection. Always draw a clear diagram and label the right-angled triangle used to find the perpendicular distance.

对于与杠杆臂长 r 成角度 θ 的力 F，绕转轴的力矩为 F r sin θ，或者说 F 乘以力臂在垂直方向的投影长度。解题时务必画出清晰示意图，并标出用于求垂直距离的直角三角形。

Common mistake: using the component along the lever arm rather than the perpendicular one. Remind yourself that a force whose line of action passes through the pivot produces zero moment.

常见错误：用了沿杠杆臂方向的分量而非垂直分量。请牢记：力作用线通过转轴时力矩为零。

3. Principle of Moments | 力矩原理

When a rigid body is in rotational equilibrium, the sum of clockwise moments about any point equals the sum of anticlockwise moments about that same point. This is the principle of moments.

当刚体处于转动平衡状态时，对任一点取矩，顺时针力矩之和等于逆时针力矩之和，这便是力矩原理。

Σ clockwise moments = Σ anticlockwise moments

Σ 顺时针力矩 = Σ 逆时针力矩

This principle is extremely powerful. In many problems you can take moments about a carefully chosen point – often the point where an unknown force acts – to eliminate that unknown from the equation.

这一原理非常有用。许多问题中，你可以巧妙选择取矩点——通常选在某个未知力作用点——从而在方程中消去该未知力。

4. Conditions for Equilibrium | 平衡条件

A rigid body is in static equilibrium when two conditions are satisfied simultaneously: (1) the vector sum of all external forces is zero (translational equilibrium), and (2) the sum of all moments about any point is zero (rotational equilibrium).

刚体处于静力平衡时，需同时满足两个条件：(1) 所有外力的矢量和为零（平动平衡）；(2) 对任意点的合力矩为零（转动平衡）。

ΣF↑ = ΣF↓ and ΣF→ = ΣF←, ΣM_{any point} = 0

向上合力 = 向下合力，向右合力 = 向左合力，对任意点 ΣM = 0

In two-dimensional problems, resolve forces horizontally and vertically, then use the principle of moments. You typically need three independent equations to find up to three unknowns.

在二维问题中，分别列出水平方向、竖直方向的力平衡方程，再结合力矩原理。通常需要三个独立方程来求解最多三个未知量。

5. Solving Equilibrium Problems | 求解平衡问题

A systematic approach is essential. Start by drawing a large, clear free-body diagram showing all forces, including weight, normal reactions, tensions and friction if present. Mark all known distances and angles.

系统性的步骤至关重要。先画一幅大而清晰受力图，标出所有力，包括重力、支持力、张力和可能存在的摩擦力。标注所有已知距离和角度。

Choose a convenient point for taking moments – usually one through which an unknown force passes. Write a moment equation, then write force-resolving equations. Solve the system of equations step by step. Always check whether a reaction force has a non-vertical component that you must include.

选择一个方便的取矩点——通常选在一个未知力通过的点。列出力矩方程，再列出力的分解方程。逐步解方程组。务必检查支反力是否有非竖直分量需要包含在内。

6. Uniform and Non-Uniform Rods | 均匀与非均匀杆

For a uniform rod, the weight acts at its midpoint. This symmetry simplifies moment calculations. For a non-uniform rod, the centre of mass is not at the geometric centre, and its position may need to be determined experimentally or algebraically using equilibrium conditions.

对于均匀杆，重力作用于杆的中点，这种对称性简化了力矩计算。对于非均匀杆，质心并不在几何中心，其位置可能需要通过实验或平衡条件来确定。

Typical exam question: a non-uniform beam is balanced horizontally on a support. Given forces and distances, find the position of its centre of gravity. Let the unknown distance be x, take moments about the support, and solve.

典型考题：一根非均匀梁在一支点上水平平衡，已知力和距离，求其重心的位置。设未知距离为 x，对支点取矩并求解即可。

7. Centre of Gravity and Its Application | 重心及其应用

The centre of gravity is the point through which the entire weight of a body appears to act. For regular shapes and composite bodies, you can find it using moments. The weight can be considered as a single force acting at this point, greatly simplifying problems involving extended objects.

重心是物体全部重力看起来所作用的点。对于规则形状和组合体，可利用力矩求出重心。重力可视为作用于该点的一个单一力，这极大地简化了涉及广延物体的平衡问题。

When a body is suspended freely, it comes to rest with its centre of gravity vertically below the point of suspension. This fact is often used in experimental determination of the centre of gravity and appears in deduction-style questions.

将物体自由悬挂时，它静止后重心位于悬挂点正下方。这一事实常用于实验测定重心，也会出现在推理性考题中。

8. Couples and Torque | 力偶与扭矩

A couple consists of two equal, parallel but oppositely directed forces separated by a perpendicular distance d. The resultant force of a couple is zero, but it produces a pure turning effect. The moment of a couple (torque) is given by the product of one force and the perpendicular distance between their lines of action.

力偶由大小相等、方向相反且作用线平行但不共线的两个力组成，它们相隔垂直距离 d。力偶的合力为零，但产生纯转动效应。力偶矩（扭矩）等于其中一个力的大小乘以两力作用线间的垂直距离。

Moment of a couple = F × d

力偶矩 = F × d

In IB and CIE exams, couples often appear in the context of wrenches, steering wheels, or systems where two equal and opposite forces act to produce rotation without translation. Recognising a couple helps you directly write the torque term without taking separate moments for each force.

在IB和CIE考试中，力偶常出现在扳手、方向盘等情景，或两个等大反向力同时作用而产生转动而没有平动的系统中。识别出力偶能让你直接写出扭矩项，而不必对两个力分别取矩。

9. Equilibrium of Bodies on Inclined Planes | 斜面上物体的平衡

When a body rests on a rough inclined plane, equilibrium involves resolving the weight parallel and perpendicular to the plane, and using friction to balance the parallel component. The moment condition often comes into play when the body is about to topple or when it is held by additional forces.

当物体静止在粗糙斜面上时，平衡分析需要将重力沿斜面和平行斜面分解，并利用摩擦力平衡平行分量。当物体即将倾倒或受到额外力作用时，往往需要引入力矩条件。

For a block on the point of toppling, the normal reaction shifts to the edge of the base. Taking moments about that edge, with the weight providing the overturning moment, allows you to find the critical angle or applied force.

对于即将倾倒的物体，支持力会移动到底面边缘。对该边缘取矩，重力产生倾覆力矩，由此可求出临界角度或外加力。

10. Common Pitfalls and Exam Tips | 常见易错点与考试技巧

• Confusing the line of action with the point of application when finding the perpendicular distance. Always drop a perpendicular from the pivot to the force line.

  求垂直距离时混淆力作用线与作用点。务必从转轴向力的作用线作垂线。

• Forgetting that the weight of a uniform rod acts at its centre, not at its end. Label the weight accurately in your diagram.

  忘记均匀杆的重力作用于中心而非端点。在图中准确标出重力位置。

• Omitting the moment contribution of a reaction force because its line of action passes through the pivot, which might not be true if the force is not perpendicular to the lever.

  忽略了实际上有力的矩贡献，只因错误地认为力线通过转轴——当力不垂直时需重新检验。

• Sign errors: choose a consistent sign convention and stick to it. Write “anticlockwise positive” on your diagram.

  符号错误：规定统一的符号约定并坚持使用。在图上标注“逆时针为正”。

• Skipping units: forces in newtons, distances in metres, moments in N m. Convert cm to m if needed.

  忽略单位：力用牛顿，距离用米，力矩用牛·米。需要时将厘米化为米。

11. Worked Example: Beam Supported at Ends | 例题：简支梁

A uniform beam of length 6 m and weight 200 N rests on two supports at its ends. A 300 N weight is placed 2 m from the left end. Find the reactions at the supports.

一根长 6 m、重 200 N 的均匀梁两端简支。一个 300 N 的重物置于距左端 2 m 处，求两支座反力。

Let the left reaction be R_A and the right reaction be R_B. The weight of the beam (200 N) acts at the centre, 3 m from either end. Taking moments about A eliminates R_A:

设左支座反力为 R_A，右支座反力为 R_B。梁重 200 N 作用于中点，距两端 3 m。对 A 点取矩可消去 R_A：

ΣM_A = 0: (300 N × 2 m) + (200 N × 3 m) − (R_B × 6 m) = 0

600 + 600 − 6R_B = 0 → R_B = 200 N

Vertically, ΣF↑ = ΣF↓: R_A + 200 = 300 + 200 → R_A = 300 N. Always check by taking moments about another point.

竖直方向合力为零：R_A + 200 = 300 + 200 → R_A = 300 N。可对另一点取矩验证结果。

12. Further Practice – Taking Moments about Different Points | 进阶练习：选择不同取矩点

In more complex problems, the choice of pivot can dramatically simplify your algebra. You may also need to consider hinged objects, ladders leaning against walls, or rods held by strings. In such cases, resolve forces into components before taking moments, and be meticulous about perpendicular distances.

在更复杂的问题中，取矩点的选择能大大简化代数运算。你可能还会遇到铰接物体、靠墙的梯子或被绳子拉住的杆。此时需先将力分解再取矩，并细心处理垂直距离。

Practise a variety of past paper questions. Start with simple beam problems, progress to non-uniform rods, then move on to hinged systems and slope-stability analysis. Always draw a clear diagram and write equations systematically.

练习各种往年真题。从简单梁问题开始，过渡到非均匀杆，再到铰接系统和斜坡稳定性分析。始终画清晰的图并系统地列出方程。

Remember: in IB and CIE Mathematics, the examiner expects you to demonstrate a logical process, not just the answer. Show your choice of pivot, your moment equation, and any resolution of forces. Correct methodology earns partial credit even if the final numerical answer is wrong.

记住：在IB和CIE数学中，考官期望你展示逻辑过程，而不仅仅是答案。标出取矩点，写明力矩方程和力的分解。即使最终数值有误，方法正确也能获得步骤分。

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