📚 A-Level Mathematics: Analysis of Rigid Body Equilibrium Conditions | A-Level数学：刚体平衡条件分析
In A-Level mechanics, a rigid body is an object that does not deform under the action of forces. For such a body to be in equilibrium, two independent conditions must be satisfied simultaneously: the resultant force must be zero, and the resultant moment about any point must be zero. This article provides a systematic analysis of rigid body equilibrium, including key definitions, problem-solving strategies, and common exam-style applications.
在A-Level力学中，刚体是指在力的作用下不发生形变的物体。要使刚体处于平衡状态，必须同时满足两个独立条件：合力为零，且对任意一点的合力矩为零。本文将系统地分析刚体平衡问题，包括关键定义、解题策略以及常见的考试题型应用。
1. Conditions for Equilibrium | 平衡条件
A rigid body is in equilibrium when it is both translationally and rotationally at rest. The first condition is translational equilibrium: the vector sum of all external forces acting on the body is zero. This ensures that the centre of mass has no linear acceleration.
刚体处于平衡状态意味着它既无平动加速度也无转动加速度。第一个条件是平动平衡：作用在刚体上所有外力的矢量和为零。这保证质心没有线性加速度。
The second condition is rotational equilibrium: the sum of all moments (torques) about any point is zero. This ensures that the body has no angular acceleration. Moments are calculated as force × perpendicular distance from the line of action to the pivot point.
第二个条件是转动平衡：对任意一点的合力矩为零。这保证刚体没有角加速度。力矩的计算公式为力 × 力的作用线到转轴的垂直距离。
ΣF = 0 and ΣM = 0 (about any point)
In two dimensions, the equilibrium conditions give three scalar equations: ΣFₓ = 0, ΣFᵧ = 0, and ΣM = 0. These equations are sufficient to solve for up to three unknown quantities.
在二维问题中，平衡条件给出三个标量方程：ΣFₓ = 0，ΣFᵧ = 0，ΣM = 0。这三个方程足以求解最多三个未知量。
2. Moment of a Force | 力的力矩
The moment of a force about a point O is defined as the product of the magnitude of the force and the perpendicular distance from O to the line of action of the force. The unit is newton-metre (N·m).
力对点O的力矩定义为力的大小与O到力的作用线垂直距离的乘积，单位为牛顿·米（N·m）。
If a force F acts at a point whose perpendicular distance from O is d, then the moment M = F × d. The direction (clockwise or anticlockwise) must be stated when summing moments.
若力F作用于距O的垂直距离为d的点，则力矩M = F × d。在求力矩和时必须标明方向（顺时针或逆时针）。
For a force applied at an angle, the perpendicular distance must be found using trigonometry. For example, if a force F makes an angle θ with a rod of length L, the moment about the end is F × L × sin θ.
对于成角度施加的力，必须通过三角函数求垂直距离。例如，若力F与长度为L的杆成θ角，则关于杆端的力矩为F × L × sin θ。
3. Resultant Force and Free-Body Diagrams | 合力与受力分析图
Drawing an accurate free-body diagram (FBD) is the first step in any rigid body equilibrium problem. Every external force acting on the body must be shown, including weights, normal reactions, friction, and applied forces.
绘制准确的受力分析图（FBD）是解决任何刚体平衡问题的第一步。必须标出作用在刚体上的所有外力，包括重力、法向反力、摩擦力和施加力。
For a body in equilibrium, the vector sum of forces in any direction is zero. Therefore, resolving horizontally and vertically gives two independent equations:
对于平衡的刚体，任意方向上的合力为零。因此，水平和竖直方向的分力方程给出两个独立方程：
ΣFₓ = 0 → sum of horizontal components = 0
ΣFᵧ = 0 → sum of vertical components = 0
If forces are not perpendicular, resolve each force into x- and y-components using cos and sin. Always define a positive direction (e.g., right and up as positive) and be consistent.
如果力不互相垂直，则用cos和sin将每个力分解为x和y方向的分量。务必定义正方向（例如右和上为正）并保持一致。
4. Choosing a Pivot Point | 选择转轴（矩心）
The moment equation ΣM = 0 is valid about any point, but a wise choice of pivot can simplify calculations dramatically. The best pivot is often the point where an unknown force acts, because that force then has zero moment about the pivot.
力矩方程ΣM = 0对任意点都成立，但明智地选择转轴可以大大简化计算。最佳转轴通常是某个未知力作用点，因为该力对此转轴的力矩为零。
For example, when a ladder leans against a smooth wall and rough ground, taking moments about the foot of the ladder eliminates both the normal reaction and the friction at the ground, leaving only the weight and the wall reaction as moment contributors.
例如，当梯子靠在光滑墙壁和粗糙地面上时，以梯子底部为转轴取矩，可以消去地面的法向反力和摩擦力，只剩下重力与墙壁反力产生力矩。
In problems with multiple unknown supports, choosing the pivot at one support yields an equation that directly solves for the force at the other support.
在具有多个未知支座的问题中，选择其中一个支座为转轴，可以直接求出另一支座处的力。
5. Types of Supports and Reactions | 支撑类型与反力
Different supports exert different types of reaction forces. A smooth contact exerts a reaction perpendicular to the surface. A rough contact exerts both a normal reaction and a frictional force parallel to the surface.
不同类型的支撑产生不同性质的反力。光滑接触面产生垂直于表面的法向反力。粗糙接触面同时产生法向反力和平行于表面的摩擦力。
A hinge or pin joint can exert a reaction force in any direction, usually represented by two perpendicular components Rₓ and Rᵧ. A roller support exerts a reaction perpendicular to the surface on which it rolls.
铰链或销钉连接可产生任意方向的反力，通常用两个互相垂直的分量Rₓ和Rᵧ表示。滚动支座产生垂直于支撑面的反力。
| Support Type | Reaction Components |
| Smooth surface | Normal only (perpendicular to surface) |
| Rough surface | Normal + Friction (parallel to surface) |
| Hinge / Pin | Two perpendicular components Rₓ, Rᵧ |
| Roller | Normal to surface only |
Identifying the correct reaction directions is essential for writing correct equilibrium equations.
正确识别反力方向对于列出正确的平衡方程至关重要。
6. Friction and Limiting Equilibrium | 摩擦力与极限平衡
Friction is a tangential force that opposes relative motion. For a body in equilibrium, friction is static friction, which can vary from zero up to a maximum value Fₘₐₓ = μR, where μ is the coefficient of friction and R is the normal reaction.
摩擦力是阻碍相对运动的切向力。对于平衡的刚体，摩擦力为静摩擦力，其大小可在零到最大值Fₘₐₓ = μR之间变化，其中μ为摩擦系数，R为法向反力。
When a body is on the point of slipping, it is said to be in limiting equilibrium. In this state, the friction force equals μR, and its direction opposes the impending motion.
当物体即将滑动时，称其处于极限平衡状态。在此状态下，摩擦力等于μR，其方向与即将发生的运动方向相反。
In many exam problems, the condition “on the point of slipping” is used to relate friction and normal reaction, providing an additional equation that completes the solution.
在许多考试题中，“即将滑动”这一条件用于建立摩擦力与法向反力的关系，为解题提供额外的方程。
7. Centre of Mass and Weight | 质心与重力
The weight of a rigid body acts through its centre of mass. For a uniform rod, the centre of mass is at its midpoint. For a uniform lamina, the centre of mass is at the geometric centre. When taking moments, the distance used is from the pivot to the vertical line through the centre of mass.
刚体的重力作用线通过其质心。对于均匀杆，质心在中点；对于均匀薄板，质心在几何中心。取矩时，所用的距离是转轴到通过质心的竖直线的水平距离。
If a body is composed of several parts, the centre of mass can be found by taking moments about a convenient axis: the total moment of the weight equals the sum of moments of the individual weights.
如果刚体由多个部分组成，可通过关于某方便轴取矩来求质心：总重力的力矩等于各部分重力力矩之和。
For a composite body, let total weight W = W₁ + W₂ + …; then the position x̄ of the centre of mass satisfies W·x̄ = W₁x₁ + W₂x₂ + …
对于组合体，设总重量W = W₁ + W₂ + …；则质心位置x̄满足W·x̄ = W₁x₁ + W₂x₂ + …
8. Solved Example: Uniform Rod on Two Supports | 例题：均匀杆置于两支撑上
Consider a uniform rod AB of length 6 m and weight 120 N, resting horizontally on two supports at points C and D, where AC = 1 m and AD = 5 m. Find the reactions at C and D.
考虑一根均匀杆AB，长度为6 m，重量为120 N，水平放置在C和D两个支撑点上，其中AC = 1 m，AD = 5 m。求C和D处的反力。
Let R_C and R_D be the upward reactions at C and D. The weight of the rod acts at its midpoint, which is 3 m from A.
设R_C和R_D分别为C和D处的向上反力。杆的重力作用于其中点，距A为3 m。
Vertical equilibrium: R_C + R_D = 120 N.
竖直方向平衡：R_C + R_D = 120 N。
Taking moments about A: R_C × 1 + R_D × 5 = 120 × 3 = 360 N·m.
对A点取矩：R_C × 1 + R_D × 5 = 120 × 3 = 360 N·m。
Solving the two equations: from R_C = 120 − R_D, substitute: (120 − R_D) + 5R_D = 360 → 4R_D = 240 → R_D = 60 N, and R_C = 60 N.
联立两方程：由R_C = 120 − R_D，代入得：(120 − R_D) + 5R_D = 360 → 4R_D = 240 → R_D = 60 N，因此R_C = 60 N。
Thus both supports share the load equally because the rod is uniform and the supports are symmetric about the centre.
因此两支撑均匀分担载荷，因为杆是均匀的且支撑关于中心对称。
9. Practical Exam Strategy | 考试解题策略
To solve rigid body equilibrium problems efficiently, follow these steps:
为了高效解决刚体平衡问题，请遵循以下步骤：
- Draw a clear free-body diagram showing all forces with correct directions and points of application.
- 绘制清晰的受力分析图，标出所有力的正确方向和作用点。
- Choose a convenient pivot point — preferably where an unknown force acts.
- 选择方便的转轴——最好是某个未知力的作用点。
- Resolve forces into perpendicular components and write ΣFₓ = 0, ΣFᵧ = 0.
- 将力分解为互相垂直的分量，并列出ΣFₓ = 0，ΣFᵧ = 0。
- Take moments about the pivot and write ΣM = 0, being careful with clockwise/anticlockwise signs.
- 对转轴取矩并列出ΣM = 0，注意顺时针/逆时针的符号。
- If friction is involved, determine whether the body is in limiting equilibrium and use F = μR if needed.
- 若涉及摩擦力，判断物体是否处于极限平衡，必要时使用F = μR。
- Solve the resulting system of equations algebraically. Check that your answers are physically reasonable.
- 用代数方法求解所得方程组，并检查答案是否在物理上合理。
10. Common Mistakes and Tips | 常见错误与提示
One common mistake is forgetting to include the moment of a force whose line of action passes through the pivot; such a moment is zero, so it does not appear in the equation.
一个常见错误是忘记力的作用线通过转轴时的力矩为零，因此该力矩不出现在方程中。
Another mistake is using the distance along a body instead of the perpendicular distance when a force is at an angle. Always measure the perpendicular distance from the line of action to the pivot.
另一个错误是当力成角度时，使用了沿物体方向的距离而非垂直距离。务必测量力的作用线到转轴的垂直距离。
Remember that the normal reaction is not always equal to the weight. In problems with inclined planes or additional vertical forces, resolve carefully and write separate equations for horizontal and vertical components.
记住法向反力不一定等于重力。在斜面问题或有额外竖直力的问题中，需仔细分解并分别列出水平和竖直分量的方程。
When using friction in limiting equilibrium, always verify the direction of the friction force: it must oppose the direction in which the body would otherwise move.
在极限平衡中使用摩擦力时，务必验证摩擦力的方向：它必须与物体本应运动的方向相反。
11. Summary | 总结
Rigid body equilibrium requires both zero resultant force and zero resultant moment. By drawing accurate free-body diagrams, choosing a strategic pivot, and systematically applying ΣFₓ = 0, ΣFᵧ = 0, and ΣM = 0, any equilibrium problem can be solved.
刚体平衡要求合力为零且合力矩为零。通过绘制准确的受力分析图、选择策略性的转轴，并系统地应用ΣFₓ = 0、ΣFᵧ = 0和ΣM = 0，任何平衡问题都可以解决。
Mastering the concept of moments, the correct choice of pivot, and the handling of friction and normal reactions are the core skills tested in A-Level mechanics. Practice with a variety of problems, including ladders, beams, and composite bodies, to build confidence.
掌握力矩的概念、正确选择转轴以及处理摩擦力和法向反力，是A-Level力学考查的核心技能。通过练习各种题型（包括梯子、横梁和组合体）来建立信心。
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