📚 AQA Maths: Moments & Equilibrium Revision | AQA 数学：力矩与平衡 考点精讲
Moments are fundamental in AQA A-level mechanics, describing the turning effect of forces. Mastering equilibrium problems unlocks the ability to analyse everything from simple levers to complex ladders. This revision guide walks you through every key idea, with paired English and Chinese explanations, typical problem types, and the precise exam technique you need.
力矩是 AQA A-level 力学中的核心概念，它描述力的转动效应。掌握平衡问题之后，你就可以分析从简单杠杆到复杂梯子等各种题型。本文带你梳理每一个关键知识点，采用中英双语对照讲解，覆盖常见题型与考试技巧，帮助你在试卷中稳拿高分。
1. What is a Moment? | 什么是力矩？
A moment measures the turning effect of a force about a point or pivot. It depends on two things: the size of the force and the perpendicular distance from the pivot to the line of action of the force. If the distance is zero, there is no moment, because the force acts directly through the pivot.
力矩衡量一个力绕某点（转轴）产生的转动效果。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。如果垂直距离为零，说明力的作用线通过转轴，不产生力矩。
Moment = Force × perpendicular distance M = F × d
ce is already at 90°. When the force is applied at an angle θ to the lever, the perpendicular distance is d × sin θ where d is the distance from pivot to the point of application.
力矩的单位是 牛顿·米（N m）。一定要用垂直距离，而不是沿杆的斜距，除非力恰好垂直于杆。当力与杠杆成 θ 角时，垂直距离为 d × sin θ，其中 d 是转轴到施力点的距离。
2. Calculating a Moment in Practice | 力矩的实际计算
To find the moment of a force about A, draw a line from A perpendicular to the force's line of action. If the force is given as components (horizontal and vertical), calculate the moment of each component and add them, paying close attention to clockwise or anticlockwise sense. Trigonometry is often needed to resolve distances.
计算力对 A 点的力矩时，先画出从 A 到力作用线的垂线。如果力以水平和竖直分量的形式给出，则分别计算每个分量的力矩，再依据顺、逆时针方向求和。常需借助三角函数分解距离。
Example: of action is 2 × sin 30° = 1 m. Moment = 10 × 1 = 10 N m anticlockwise. If the pivot is not at the end, adapt the geometry accordingly.
示例：10 N 的力以与杆成 30° 的方向拉一根长 2 m 的杆，求对另一端的力矩。端点到力作用线的垂直距离为 2 × sin 30° = 1 m，力矩 = 10 × 1 = 10 N m 逆时针。若转轴不在端点，几何关系需相应调整。
3. The Principle of Moments | 力矩原理
When a 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 principle lets you form an equation linking forces and distances, often making it possible to find an unknown force or reaction.
当物体处于转动平衡时，绕任意点的顺时针力矩之和等于逆时针力矩之和。这一原理让你能够建立力与距离的方程，通常可由此解出未知力或支反力。
Σ clockwise moments = Σ anticlockwise moments
Always pick a pivot that makes the algebra easiest – often a point where an unknown force acts, so that force has zero moment and disappears from the equation. Sign conventions must be clear: you may label clockwise positive and anticlockwise negative, or simply equate magnitudes after identifying direction.
一定要选择一个让代数计算最简单的转轴——通常选某个未知力的作用点，这样该力的力矩为零，从方程中消失。正负号约定必须清晰：可以将顺时针定为正、逆时针为负，或者在分清方向后直接让大小相等。
4. Conditions for Equilibrium | 平衡条件
For a rigid body to be in complete equilibrium, two vector conditions must be met: the resultant force in any direction is zero, and the resultant moment about any point is zero. In 2D problems, this gives up to three independent equations: two force equations (usually horizontal and vertical) and one moment equation.
刚体完全平衡必须满足两个矢量条件：任意方向的合力为零，且绕任意点的合力矩为零。在二维问题中，这最多得出三个独立方程：两个力平衡方程（通常取水平和竖直方向）和一个力矩平衡方程。
AQA exam questions often ask for the magnitude and direction of a reaction at a hinge or the tension in a string. Always start by drawing a free-body diagram showing all forces, their directions, and their points of application. Then write down the equations of equilibrium methodically.
AQA 的考题经常要求计算铰链支座反力的大小与方向，或绳的张力。务必从受力示意图开始，标出所有力、方向和作用点。然后条理清晰地列出平衡方程。
5. Resolving Forces for Equilibrium | 分解力求解平衡
When forces are not parallel, resolve them into perpendicular components. The upward components must balance weight and any other vertical forces, while horizontal components balance each other. Together with the principle of moments, you can handle an enormous range of structures, from brackets to cranes.
当作用力不平行时，将其分解为互相垂直的分量。向上的分力必须与重力及其他竖直力平衡，水平分力则相互抵消。结合力矩原理，你就能处理从支架到起重机的各种结构。
Always remember: resolving vertically gives Σ Fy = 0, and horizontally Σ Fx bout any point, but it is wise to choose a point that eliminates as many unknowns as possible, particularly unknown contact forces.
永远记住：竖直方向合力为零 Σ Fy = 0，水平方向 Σ Fx = 0。力矩方程可以绕任意点取矩，但明智的选择是消去尽可能多的未知力，尤其是未知的接触力。
6. Uniform Rods and Beams | 匀质杆与横梁
If a rod is 'uniform', its weight acts at its centre. This simplification is used heavily. When a uniform rod rests on supports or is hinged, you immediately know the position of the weight vector – at the midpoint. Combine this with reaction forces and any applied loads to set up moment and force balances.
如果一根杆是“匀质”的，其重力作用点在杆的中心。这一简化被大量使用。当匀质杆搁在支座上或铰接时，你立刻就知道重力矢量的位置——在杆的中点。将其与支反力及外加荷载结合，建立力矩与力平衡方程。
A typical problem: a uniform beam of length 4 m and weight 100 N rests on two supports at its ends. An additional weight of 50 N is placed 1 m from one end. Find the reactions. Taking moments about one support, the weight of the beam (100 N at 2 m) and the added weight contribute clockwise moments; the reaction at the far support provides the balancing anticlockwise moment.
典型问题：一根长 4 m 重 100 N 的匀质横梁，两端支撑，在距一端 1 m 处加放 50 N 的重物，求两支反力。对其中一个支座取矩，梁自重（100 N 作用在 2 m 处）和附加重物产生顺时针力矩，较远端的支座反力提供逆时针力矩平衡。
7. Couples and Torque | 力偶与扭矩
about any point and equals force × perpendicular distance between the forces. Couples produce pure rotation without net force.
力偶由两个大小相等、方向相反、作用线平行且相隔一定垂直距离的力组成。力偶的力矩——即扭矩——绕任意点都相同，等于力的大小 × 两力之间的垂直距离。力偶产生纯转动效果，合力为零。
Torque of couple = F × d (where d is the perpendicular distance between the forces)
In equilibrium problems, a couple may be applied to a body and must be balanced by another couple from the supports. Recognising a couple can dramatically simplify a moment equation, because you don't need to choose a pivot – its moment is invariant.
在平衡问题中，物体上可能作用有力偶，它必须由支座产生的另一力偶来平衡。识别出力偶能极大地简化力矩方程，因为你无需选择特定转轴——它的力矩是恒定不变的。
8. Tilting and the Point of Toppling | 倾覆与翻倒点
When a body is on the point of tilting about one edge, the normal reaction at other supports becomes zero. This condition gives the limiting value of an applied force or position of a load. You write the moment equation about the tilting edge, setting the reaction at the other support to zero.
当物体处于即将绕某一棱边倾覆的临界状态时，其他支点的法向反力变为零。利用这一条件可求外加力的极限值或荷载的极限位置。此时对倾覆棱边取矩，并令另一边支座反力为零来列方程。
Consider a uniform block on a horizontal surface being pushed at increasing height. Tilting occurs when the moment of the pushing force about the front edge overcomes the restoring moment of the block's weight. This principle also explains why tall loads on a truck are more likely to topple.
设想一个匀质方块放在水平面上，推力作用点逐渐升高。当推力对前棱边的力矩超过方块自身重量的恢复力矩时，倾覆就会发生。这一原理也解释了为何卡车上的高耸货物更容易翻倒。
9. Ladder Problems Involving Friction | 梯子问题与摩擦力
A classic AQA mechanics question is the ladder leaning against a smooth wall on a rough floor. The wall is smooth, so its reaction is perpendicular to the wall (horizontal). The floor exerts both a normal reaction (upwards) and a friction force (horizontal) that stops the ladder from slipping. The ladder's weight acts at its centre.
AQA 力学中的经典题目是梯子斜靠在光滑竖直墙壁、底部立于粗糙地面上。墙壁光滑，所以墙的反力垂直于墙面（水平方向）。地面提供法向反力（向上）和阻止滑动的摩擦力（水平方向）。梯子自重作用在中心。
contact point. The moment equation involves the weight and the wall reaction, using the ladder length L and angle θ with the ground. The limiting equilibrium condition requires F ≤ μR, where μ is the coefficient of friction.
建立三个方程：竖直力平衡（R = 重量），水平力平衡（F = S，S 为墙反力），以及对梯子底端接触点的力矩方程。力矩方程包含重力和墙反力，利用梯长 L 和梯与地面的夹角 θ。极限平衡条件需满足 F ≤ μR，其中 μ 为摩擦系数。
check F ≤ μR for the required μ or find the minimum angle for which the ladder stays put.
典型的是对底端取矩：S × L sin θ = 重力 × (L/2) cos θ。解出 S，再用 F = S 求出所需摩擦力。最后用 F ≤ μR 检验所需摩擦系数，或求梯子保持静止的最小角度。
10. Common Mistakes and Exam Tips | 常见错误与考试技巧
Many students lose marks by using the wrong distance for a moment – always check it is perpendicular. Another pitfall is forgetting to include the weight of a uniform rod when it is not explicitly drawn as a force. When a beam is on the point of tilting, remember that a reaction force can vanish.
很多学生因取错力矩距离而失分——一定要确认是垂直距离。另一个陷阱是：当匀质杆的重力没有被明确画出时，忘记将其计入。当横梁处于即将倾覆状态时，切记某个支座反力可以变为零。
In ladder problems, double-check which surfaces are smooth. If the wall is rough, there is a vertical component of reaction at the wall, adding an additional unknown. Always isolate the body with a clear diagram and mark every force with its direction and symbol. Time management is key: practise past paper questions where you can form the three equations quickly.
在梯子问题中，务必确认哪个面是光滑的。若墙面粗糙，墙反力会有竖直分量，多出一个未知数。始终画出清晰的隔离体图，标出每个力的方向与符号。时间管理很关键：多练历年真题，训练自己快速列出三个平衡方程。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图