📚 Moments and Equilibrium | 力矩与平衡
In A-Level Mechanics, moments and equilibrium form the foundation for analysing rigid bodies that do not move, or are about to rotate. A moment measures the turning effect of a force about a pivot. When a body is in equilibrium, both the resultant force and the resultant moment acting on it are zero. This article covers key concepts, typical problem‑solving strategies, and common pitfalls for Edexcel A‑Level Mathematics.
在A-Level力学中，力矩与平衡是分析静止或即将转动的刚体的基础。力矩衡量力对某一点的转动效应。当物体处于平衡状态时，作用在其上的合力和合力矩都为零。本文涵盖爱德思A-Level数学中的核心概念、典型解题策略和常见易错点。
1. Definition of a Moment | 力矩的定义
The moment of a force about a point is the product of the force and the perpendicular distance from the point to the line of action of the force.
力对某一点的力矩等于力的大小乘以该点到力作用线的垂直距离。
Moment = F × d
where F is the magnitude of the force and d is the perpendicular distance. The unit is newton‑metre (N m).
其中 F 是力的大小，d 是垂直距离。单位是牛顿·米（N m）。
2. Sense of a Moment | 力矩的方向
Moments can be clockwise or anticlockwise. It is essential to choose a consistent sign convention, for example, taking anticlockwise moments as positive.
力矩可以是顺时针或逆时针的。必须选择一致的符号约定，例如取逆时针力矩为正。
When summing moments about a pivot, label each moment as clockwise (CW) or anticlockwise (ACW). In equilibrium, the sum of clockwise moments equals the sum of anticlockwise moments about any point.
在求对支点的合力矩时，将每个力矩标记为顺时针（CW）或逆时针（ACW）。平衡时，对任意点的顺时针力矩之和等于逆时针力矩之和。
3. Principle of Moments | 力矩原理
For a body in rotational equilibrium, the total clockwise moment about any point is equal to the total anticlockwise moment about that same point. This is the principle of moments.
对于处于转动平衡的物体，对任意点的顺时针力矩之和等于对该点的逆时针力矩之和。这就是力矩原理。
Algebraically, Σ MCW = Σ MACW. This is used together with the condition for translational equilibrium, Σ F = 0, to solve for unknown forces or distances.
代数上，Σ MCW = Σ MACW。此式与平动平衡条件 Σ F = 0 联立，可求解未知力或距离。
4. Equilibrium of a Rigid Body | 刚体的平衡
A rigid body is in equilibrium when both the resultant force and the resultant moment are zero. This gives two vector conditions: Σ F = 0 and Σ M = 0 about any point.
刚体平衡时，合力和合力矩均为零。这给出两个矢量条件：对任意点 Σ F = 0 且 Σ M = 0。
In two dimensions, we usually resolve forces in two perpendicular directions (e.g. horizontal and vertical) and take moments about a convenient point to form three independent equations.
在二维问题中，通常沿两个垂直方向（如水平和竖直）分解力，并对合适的点取矩，以建立三个独立方程。
5. Uniform Rods and Centre of Mass | 均匀杆与质心
A uniform rod has its weight acting at its centre. This simplifies moment calculations because the entire weight can be considered to act at the midpoint of the rod.
均匀杆的重量作用在其中心。这简化了力矩计算，因为全部重量可视为作用在杆的中点。
For a rod of length L, the centre of mass is at distance L/2 from either end. When the rod is non‑uniform, the centre of mass must be given or calculated.
对于长度为 L 的杆，质心距离两端均为 L/2。若杆不均匀，则质心位置需给定或通过计算得出。
6. Taking Moments About a Hinge or Pivot | 对铰链或支点取矩
When a rod is hinged at a point, the reaction force at the hinge has both horizontal and vertical components. Taking moments about the hinge eliminates these unknown reactions from the moment equation.
当杆铰接于某点时，铰链处的反作用力有水平和竖直分量。对该铰链取矩可从力矩方程中消去这些未知反力。
Strategy: always choose the point where the most unknown forces act to take moments, so they contribute zero moment.
策略：始终选择有最多未知力作用的点取矩，这样这些力的力矩为零。
7. Inclined Rods and Perpendicular Distance | 倾斜杆与垂直距离
When a rod is not horizontal, the perpendicular distance from the pivot to the line of action of a force must be found. This often involves trigonometry: d = x sin θ, where x is the distance along the rod and θ is the angle between the rod and the force direction.
当杆不水平时，需找到支点到力作用线的垂直距离。这通常涉及三角学：d = x sin θ，其中 x 是沿杆的距离，θ 是杆与力方向之间的夹角。
If the force is vertical, the perpendicular distance is the horizontal distance from the pivot to the line of action. If the force is perpendicular to the rod, d is simply the distance along the rod.
若力竖直，垂直距离是支点到力作用线的水平距离。若力垂直于杆，d 就是沿杆的距离。
8. Solving Equilibrium Problems Step by Step | 逐步求解平衡问题
A systematic approach is essential: (1) Draw a clear diagram showing all forces, including weight, reactions, and applied forces. (2) Resolve forces in convenient directions. (3) Take moments about a suitable point. (4) Write equilibrium equations and solve.
系统性的方法至关重要：(1) 画出清楚显示所有力的示意图，包括重量、反作用力和外力。(2) 沿合适方向分解力。(3) 对合适的点取矩。(4) 列出平衡方程并求解。
Always check that the number of unknowns matches the number of independent equations. Common errors include missing the weight of a rod, incorrect perpendicular distances, or not considering the direction of a moment.
务必检查未知数个数与独立方程个数是否一致。常见错误包括遗漏杆的重量、垂直距离不正确或未考虑力矩方向。
9. Example: Rod with Hinge and String | 示例：带铰链和绳子的杆
A uniform rod AB of length 2 m and mass 5 kg is hinged at A to a vertical wall. It is held horizontally by a light string attached at B, making an angle of 30° with the rod. Find the tension in the string and the magnitude of the reaction at the hinge.
一根长2 m、质量5 kg 的均匀杆 AB 在 A 端铰接于竖直墙上。杆由系于 B 端且与杆成30°角的轻绳水平拉住。求绳中张力及铰链反力的大小。
Solution: Weight = 5g = 49 N (taking g = 9.8) acts at the midpoint, 1 m from A. Let tension T. Take moments about A: ACW moment = (T sin30°) × 2 = 2T × 0.5 = T. CW moment = 49 × 1 = 49. Equilibrium: T = 49 N. Resolve vertically: Ry + T sin30° = 49 → Ry = 49 − 24.5 = 24.5 N. Horizontally: Rx = T cos30° = 49 × (√3/2) ≈ 42.4 N. Reaction R = √(Rx² + Ry²) = √(42.4² + 24.5²) ≈ 49 N.
解： 重力 = 5g = 49 N（取 g = 9.8）作用在中点，距 A 1 m。设张力为 T。对 A 取矩：逆时针力矩 = (T sin30°) × 2 = 2T × 0.5 = T。顺时针力矩 = 49 × 1 = 49。平衡得 T = 49 N。竖直分解：Ry + T sin30° = 49 → Ry = 24.5 N。水平：Rx = T cos30° ≈ 42.4 N。反力 R = √(Rx² + Ry²) ≈ 49 N。
10. Non‑Uniform Rods and Tipping | 非均匀杆与倾倒
When a rod is non‑uniform, the centre of mass is not at the geometric centre. The position must be used correctly in moment calculations. Tipping occurs when the normal reaction at one support becomes zero, indicating that the rod is about to lose contact.
当杆不均匀时，质心不在几何中心。在力矩计算中必须正确使用该位置。当某一支撑处的法向反力变为零时，发生倾倒，表明杆即将脱离接触。
For a rod resting on two supports, take moments about one support and set the reaction at the other support to zero to find the critical position of a moving load.
对于搁在两个支撑上的杆，对其中一个支撑取矩，并设另一个支撑的反力为零，可求得移动载荷的临界位置。
11. Couples and Torque | 力偶与转矩
A couple consists of two equal, parallel but opposite forces. The moment of a couple is the product of one force and the perpendicular distance between the forces, independent of the point about which moments are taken.
力偶由两个大小相等、平行且方向相反的力组成。力偶矩等于其中一个力与两力间垂直距离的乘积，与取矩点无关。
Torque is often used to describe the turning effect of a couple. In equilibrium, if a couple acts on a body, it must be balanced by an equal and opposite couple.
转矩常用来描述力偶的转动效应。若有力偶作用于物体，在平衡时必须由等大反向的力偶来平衡。
12. Common Exam Pitfalls | 常见考试误区
Students often forget to include the weight of the rod, misuse the perpendicular distance (especially on inclined rods), or misinterpret the direction of the force at a hinge. Always draw the forces clearly and label the perpendicular distances.
学生常常忘记包含杆的重量、误用垂直距离（尤其在倾斜杆上）或误解铰链处力的方向。务必清晰画出力并标出垂直距离。
Another mistake is to take moments about a point that does not simplify the problem effectively. Practice choosing the pivot wisely and check that all distances are perpendicular to the forces.
另一种错误是对不能有效简化问题的点取矩。练习明智地选择支点，并检查所有距离均垂直于力。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导