Moments and Equilibrium | 力矩与平衡

📚 Moments and Equilibrium | 力矩与平衡

In AQA A-Level Mathematics (Mechanics), moments and equilibrium are central to statics. Understanding how forces produce rotation and the conditions that keep a body at rest is vital for solving problems involving rods, beams, ladders and tilting objects. This revision guide unpacks every key concept, formula, and strategy you need to master the topic.

在 AQA A-Level 数学（力学）中，力矩与平衡是静力学的核心。理解力如何产生转动效应以及物体保持静止的条件，对于解决涉及杆、梁、梯子和倾倒物体的问题至关重要。本文精讲将梳理每个关键概念、公式和解题策略，助你全面掌握该考点。

1. What is a Moment? | 什么是力矩？

A moment is the turning effect of a force about a given point, often called the pivot or fulcrum. It measures the tendency of the force to rotate an object around that point.

力矩是力对某一点（通常称为支点或转轴）产生的转动效应。它衡量该力使物体绕该点转动的趋势。

The magnitude of a moment is defined as the product of the force and the perpendicular distance from the pivot to the line of action of the force.

力矩的大小定义为力与从支点到力作用线的垂直距离的乘积。

M = F × d

力矩 = 力 × 垂直距离

In SI units, moment is measured in newton metres (N m). The moment of a force can be either clockwise or anticlockwise, and this directional sense must be treated consistently when applying equilibrium conditions.

在国际单位制中，力矩的单位是牛顿米（N m）。力矩的方向可以是顺时针或逆时针，在应用平衡条件时必须一致地处理这一方向感。

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

To calculate a moment, first identify the pivot. Draw a straight line representing the force’s line of action, then drop a perpendicular from the pivot to this line – that perpendicular distance is d.

计算力矩时，首先要确定支点。画出代表力作用线的直线，然后从支点向该线作垂线，此垂线长度即为 d。

If the force is already perpendicular to the lever arm, the moment is simply force × distance along the arm. If the force acts at an angle, you can either resolve the force into perpendicular components or use the general expression involving the angle.

如果力已经与杆臂垂直，力矩就是力 × 沿杆臂的距离。如果力以某个角度作用，可以将力分解为垂直分量，或使用含角度的通用表达式。

For a force F applied at a distance r from the pivot, with an angle θ between the force vector and the line joining the pivot to the point of application, the moment is given by:

对于作用在距支点 r 处的力 F，若力矢量与支点到作用点连线之间的夹角为 θ，力矩为：

M = F × r sin θ

The distance r sin θ is the perpendicular distance from the pivot to the force’s line of action. Always ensure you are using the perpendicular component of the distance, not the slant distance.

其中 r sin θ 就是从支点到力作用线的垂直距离。务必确保使用的是垂直距离分量，而不是斜距。

3. The Principle of Moments | 力矩原理

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

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该点的逆时针力矩之和。

Σ clockwise moments = Σ anticlockwise moments

顺时针力矩总和 = 逆时针力矩总和

This principle is incredibly powerful because it allows you to take moments about a point that eliminates unknown forces, simplifying the algebra significantly.

这一原理非常有用，因为它允许你选取某个能消去未知力的点来取矩，从而大大简化代数运算。

In practice, you choose a pivot through which as many unknown forces as possible act, so their moments become zero. You then set up an equation linking the known and unknown forces that do create moments.

实际操作中，需选取一个尽可能多的未知力作用线通过的点作为支点，这样这些力的力矩为零。然后建立方程联系产生力矩的已知力和未知力。

4. Conditions for Equilibrium | 平衡条件

A body is in static equilibrium when it remains at rest. This requires two sets of conditions to be satisfied simultaneously: translational equilibrium and rotational equilibrium.

物体处于静力平衡时，将保持静止。这需要同时满足两组条件：平动平衡和转动平衡。

For translational equilibrium, the resultant force in any direction must be zero. Typically, this is broken into horizontal and vertical components:

对于平动平衡，任一方向上的合力必须为零。通常将其分解为水平和垂直分量：

ΣF_x = 0 and ΣF_y = 0

ΣFₓ = 0 且 ΣF_y = 0

For rotational equilibrium, the resultant moment about any point must be zero. This is often written as ΣM = 0, but you must choose a convenient pivot and consider all moments that have a non-zero perpendicular distance.

对于转动平衡，关于任意点的合力矩必须为零。通常记为 ΣM = 0，但需要选取一个方便的支点，并考虑所有具有非零垂直距离的力矩。

In most AQA problems, you will resolve forces horizontally and vertically, and then take moments about one point to get a third independent equation. This system of equations is then solved for the unknown forces or distances.

在大多数 AQA 题目中，你会水平和垂直分解力，然后对某一点取矩以获得第三个独立方程。然后求解此方程组，得出未知力或距离。

5. Uniform Rods and Beams | 均匀杆与梁

A uniform rod or beam has its weight distributed evenly along its length. The entire weight W can be considered to act at a single point – the geometric centre of the rod.

均匀杆或梁的重量沿长度均匀分布。其全部重量 W 可视为集中作用于一点——杆的几何中心。

For a uniform rod of length L, the centre of mass is at a distance L/2 from either end. When you take moments, you place the weight W at the midpoint.

对于长度为 L 的均匀杆，质心距两端均为 L/2。取矩时，将重量 W 置于中点。

A common exam problem involves a uniform beam resting on two supports, with extra loads placed along it. You are typically asked to find the reaction forces at the supports.

常见考题涉及一根均匀梁在两个支点上，中间放置额外负载。通常要求求出支点处的反作用力。

Approach: draw a clear diagram showing the beam, pivot points, weight at centre, applied loads, and reaction forces. Then take moments about one support to eliminate the reaction at that point, and use ΣF_y = 0 to find the other.

解题方法：画出示意图，标明梁、支点、中心重量、施加的负载以及反作用力。然后对其中一个支点取矩以消去该点反力，再用 ΣF_y = 0 求另一个反力。

6. Non-Uniform Rods and the Centre of Mass | 非均匀杆与质心

If a rod is non-uniform, its weight does not necessarily act at the geometric centre. The position of the centre of mass is often an unknown that you will need to find, or it may be given so you can determine other unknowns.

如果杆是非均匀的，其重量不一定作用在几何中心。质心位置通常是需要求解的未知量，或者题目给出质心位置以便求解其他未知量。

The principle of moments is used to locate the centre of mass. By supporting the rod at a point and adjusting until it balances horizontally, the pivot lies directly under the centre of mass. In calculations, you can take moments about a chosen point and set the sum equal to zero, treating the unknown distance to the centre of mass as a variable.

利用力矩原理可以确定质心位置。通过在某点支撑杆并调整至水平平衡，支点恰好位于质心正下方。在计算中，可对选定点取矩并令其为零，将未知的质心距离作为变量处理。

For example, a non-uniform rod of length L is held horizontally by two vertical strings at its ends. If the tensions in the strings are known, you can take moments about one end to locate the centre of mass.

例如，一根长 L 的非均匀杆两端用竖直绳子悬挂并保持水平。若已知绳中张力，可对一端取矩来求出质心位置。

Remember: the weight vector always acts through the centre of mass, and this fact remains true for any orientation of the rod.

切记：重力矢量总是通过质心，且这一事实对于杆的任何取向都成立。

7. Tilting and Toppling | 倾斜与倾倒

Tilting occurs when a body is on the verge of rotating about one edge or support. At the point of tilting, the normal reaction at any other support drops to zero.

当物体即将绕某一边缘或支点转动时，即发生倾斜。在倾斜的临界点，其他支点处的法向反作用力降为零。

This condition is key to solving problems about stability and maximum loads. You set the reaction at the pivot about which tilting occurs to zero, then apply the principle of moments to find the limiting load or distance.

这一条件是解决稳定性与最大负载问题的关键。将倾斜所绕支点处的反作用力设为零，然后应用力矩原理求出极限负载或距离。

For instance, consider a uniform plank overhanging a cliff. A person walks out onto the plank. The plank will tilt when the person’s weight creates a clockwise moment about the cliff edge that exceeds the counter-clockwise moment due to the plank’s own weight. At the point of tilting, the reaction from the ground underneath the inner section becomes zero.

例如，一块均匀木板伸出悬崖，一个人向外走去。当人的重量对悬崖边缘产生的顺时针力矩超过木板自身重量产生的逆时针力矩时，木板将开始倾斜。在倾倒临界点，内侧地面支持力为零。

Always identify the pivot edge and equate the moments about that edge just before tilting begins. The load that causes the net moment to reverse direction is the maximum safe load.

务必确定倾倒所绕的支点边缘，并在倾斜即将发生时对该边缘列力矩平衡方程。使净力矩反向的负载即为最大安全负载。

8. Reactions at Supports and Hinges | 支撑与铰链的反作用力

Supports and hinges exert forces on a body to maintain equilibrium. A smooth support or a roller provides a reaction force perpendicular to the contact surface. A rough surface provides both a normal reaction and a frictional force.

支撑和铰链对物体施加力以维持平衡。光滑支撑或滚轴提供垂直于接触面的反作用力。粗糙表面则提供法向反力和摩擦力。

A hinge or a pin joint can exert a force in any direction. It is usually resolved into two perpendicular components: a horizontal component and a vertical component, both unknown.

铰链或销钉连接可以在任意方向上施加力。通常将其分解为两个相互垂直的分量：水平分力和竖直分力，均为未知数。

When taking moments, it is often convenient to choose the hinge or support as the pivot, because the reaction there will have zero moment, simplifying the equation. If the hinge has two unknown components, you still eliminate both simultaneously.

取矩时，常选择铰链或支撑作为支点，因为该处反力的力矩为零，从而简化方程。若铰链有两个未知分量，用此支点可同时消去它们。

In beam problems, you might have a light rod attached to a wall by a hinge and held by a string. The hinge reaction is found by resolving forces after using moments to find the string tension.

在梁问题中，可能会有一根轻杆通过铰链连接在墙上并由绳子拉住。先用取矩求出绳中张力，再分解力求铰链反力。

9. Couples and Torque | 力偶与扭矩

A couple consists of two equal, parallel, opposite forces whose lines of action are separated by a perpendicular distance d. A couple produces pure rotation without any net translational force.

力偶由两个大小相等、平行且反向的力组成，它们的作用线相距垂直距离 d。力偶产生纯转动，无净平动力。

The moment of a couple, also called torque, is independent of the choice of pivot. It is calculated as the magnitude of one force multiplied by the perpendicular distance between the forces.

力偶的力矩也称为扭矩，与支点的选择无关。其大小等于其中一个力乘以两力作用线之间的垂直距离。

Couple moment M = F × d

力偶矩 M = F × d

Couples are important when analysing steering wheels, wrenches, or any system where two forces create a turning effect without moving the centre of mass. In equilibrium problems, the net moment from all couples plus all individual force moments must sum to zero.

分析方向盘、扳手或任何由两个力产生转动而不移动质心的系统时，力偶很重要。在平衡问题中，所有力偶矩加上所有单个力的力矩，净总和必须为零。

10. Ladder Problems (Limiting Equilibrium) | 梯子问题（极限平衡）

A classic application of moments and friction is the ladder leaning against a smooth wall. The wall is usually assumed to be smooth, so it exerts only a horizontal reaction R on the ladder. The ground is rough, providing a normal reaction N and a friction force F.

力矩与摩擦的经典应用是倚在光滑墙上的梯子。通常假设墙光滑，因此墙只对梯子施加水平反力 R。地面粗糙，提供法向反力 N 和摩擦力 F。

The ladder of length L has weight W acting at its midpoint. It makes an angle θ with the horizontal floor. The friction prevents the foot from sliding.

梯子长 L，重量 W 作用在中点，与水平地面的夹角为 θ。摩擦力阻止梯脚滑动。

For equilibrium, resolve horizontally and vertically: horizontally, R = F; vertically, N = W. Then take moments about the point of contact with the floor to eliminate N and F. The moment equation is typically:

对于平衡，水平和竖直分解：水平方向，R = F；竖直方向，N = W。然后对梯子与地面的接触点取矩以消去 N 和 F。力矩方程通常为：

R × L sin θ = W × (L/2) cos θ

This allows you to find R, and hence the friction F. If the ladder is on the point of slipping (limiting equilibrium), F = μN, where μ is the coefficient of static friction. This condition can be used to find the minimum angle for stability or the required friction coefficient.

由此可解出 R，从而得到摩擦力 F。若梯子处于即将滑动的极限平衡状态，F = μN，其中 μ 为静摩擦系数。该条件可用于求稳定的最小角度或所需摩擦系数。

11. Strategic Problem Solving in Moments | 力矩问题的解题策略

Approach every equilibrium problem in a structured way. Begin by drawing a clear, labelled free-body diagram showing all forces and their lines of action, including weight, reactions, tensions and friction.

以结构化的方式处理每个平衡问题。首先绘制清晰、带标注的隔离体图，显示所有力及其作用线，包括重量、反力、张力和摩擦力。

Choose a convenient pivot for moments – ideally one through which the greatest number of unknown forces pass. This instantly reduces the number of terms in your moment equation.

选择一个方便的取矩支点——理想情况下，让尽可能多的未知力通过该点。这能立即减少力矩方程中的项数。

Decide on a positive sense for moments (e.g. clockwise positive) and stick to it consistently. Apply ΣM = 0, then resolve forces in perpendicular directions. Solve the resulting simultaneous equations for the unknowns.

确定力矩的正方向（例如顺时针为正）并始终一致地使用。应用 ΣM = 0，再在正交方向上分解力。解出联立方程组，求出未知量。

Always check the physical realism of your answers. A negative reaction normally indicates the direction you assumed was opposite to the actual direction, but magnitudes must be positive. If a frictional force exceeds the limiting value, the object will slip.

务必检查答案的物理合理性。负的反力通常意味着你假设的方向与实际相反，但大小应为正值。若摩擦力超过极限值，物体将滑动。

12. Common Mistakes and Exam Tips | 常见错误与应试技巧

One of the most frequent errors is confusing the perpendicular distance with the slant distance. Always ensure the distance used in M = Fd is the shortest distance from the pivot to the line of action, never the length along the rod unless the force is perpendicular to the rod.

最常见的错误之一是将垂直距离与斜距混淆。切记 M = Fd 中的距离是从支点到力作用线的最短距离，除非力与杆垂直，否则绝不能直接使用杆的长度。

Another mistake is not treating the weight of a uniform beam as acting at the exact midpoint. In non-uniform beams, reading the centre of mass position incorrectly or misplacing it leads to a wrong moment arm.

另一个错误是未将均匀梁的重量准确放在中点。对于非均匀梁，若误读或错放置心位置，会导致力臂错误。

In ladder problems, forgetting that the wall is smooth and thus has only a horizontal reaction is a common slip. Also, be careful to use the correct trigonometric component when converting the ladder length into horizontal and vertical distances for moments.

在梯子问题中，常见的失误是忘记墙是光滑的因而只有水平反力。此外，在将梯长转换为力矩方程所需的水平和竖直距离时，要正确使用三角函数分量。

When a body is on the point of tilting, explicitly state that the reaction at the other support is zero, and write that down. Examiners look for that statement and its correct application in the moment equation.

当物体即将倾倒时，要明确写出另一个支撑处的反力为零，并将其代入方程。考官期望看到这个陈述及其在力矩方程中的正确应用。

Finally, always include your units (N, N m, m) and give answers to an appropriate degree of accuracy. Practising past AQA questions under

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Moments and Equilibrium | 力矩与平衡