Moments and Equilibrium | 力矩与平衡考点精讲

📚 Moments and Equilibrium | 力矩与平衡考点精讲

In the IGCSE WJEC Mathematics specification, moments and equilibrium form a cornerstone of the mechanics topics. Understanding how forces cause rotation and how objects remain balanced is essential for solving real-world problems involving levers, beams, and supports. This article breaks down every key concept, calculation method, and exam technique you will need to master moments and equilibrium for the WJEC exam.

在IGCSE WJEC数学大纲中，力矩与平衡是力学板块的核心内容。理解力如何引起转动、物体如何保持平衡，对于解决涉及杠杆、横梁和支座的现实问题至关重要。本文拆解了每一个关键概念、计算方法和考试技巧，帮助你彻底掌握WJEC考试中的力矩与平衡考点。

1. Introduction to Moments | 力矩简介

A moment is the turning effect of a force about a fixed point called the pivot or fulcrum. It causes an object to rotate clockwise or anticlockwise. The size of a moment depends on two factors: the magnitude of the applied force and the perpendicular distance from the pivot to the line of action of the force.

力矩是力绕一个固定点（称为支点或转轴）产生的转动效应，它会使物体顺时针或逆时针旋转。力矩的大小取决于两个因素：所施加力的大小，以及从支点到力的作用线的垂直距离。

If you push a door near its hinges, a large force is needed; if you push far from the hinges, a small force produces the same turning effect. This illustrates how moment = force × perpendicular distance.

如果你在靠近门铰链处推门，需要很大的力；如果在远离铰链处推门，很小的力就能产生相同的转动效果。这说明力矩 = 力 × 垂直距离。

2. Moment Calculation: Force × Perpendicular Distance | 力矩计算：力 × 垂直距离

The fundamental formula for a moment is:

力矩的基本公式为：

Moment = Force × Perpendicular Distance

力矩 = 力 × 垂直距离

The force is measured in newtons (N) and the perpendicular distance in metres (m), giving the moment in newton-metres (N m). The distance must be measured at right angles to the direction of the force. If the force is not perpendicular to the lever, use the component of the distance perpendicular to the force, or resolve the force into components.

力以牛顿 (N) 为单位，垂直距离以米 (m) 为单位，力矩的单位是牛顿·米 (N m)。距离必须沿与力方向垂直的方向测量。如果力不与杠杆垂直，需使用力臂的垂直分量，或将力分解。

For a force F applied at an angle θ to a lever at a distance d from the pivot, the moment is given by:

若力 F 与杠杆成 θ 角，作用点距支点 d，则力矩为：

Moment = F × d × sinθ

力矩 = F × d × sinθ

Example: A force of 40 N pulls a spanner at 30° to the handle, 0.25 m from the nut. The moment is 40 × 0.25 × sin30° = 40 × 0.25 × 0.5 = 5 N m.

示例：一个 40 N 的力以与扳手把手成 30° 的方向拉动，距螺母 0.25 m。力矩 = 40 × 0.25 × sin30° = 40 × 0.25 × 0.5 = 5 N m。

3. Principle of Moments | 力矩原理

The principle of moments states that for a system to be in rotational equilibrium, the sum of the clockwise moments about any pivot equals the sum of the anticlockwise moments about that pivot.

力矩原理指出，系统要处于转动平衡，绕任意支点的顺时针力矩之和必须等于绕该支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

When multiple forces act, assign a direction (e.g., clockwise positive) and sum moments accordingly. This principle allows you to find unknown forces or distances in balanced systems such as see-saws, beams, and loaded bars.

有多个力作用时，规定一个正方向（如顺时针为正），并对力矩求和。此原理可用于求解天平、跷跷板、横梁和受载杆等平衡系统中的未知力或距离。

Example: A uniform beam of length 4 m and weight 100 N is supported at its midpoint. A 200 N weight is placed 1 m to the right of the centre. What distance to the left must a 300 N weight be placed to restore balance? Taking moments about the support: clockwise = 200 × 1 = 200 N m; anticlockwise = 300 × x. Setting 300x = 200 gives x = 0.667 m.

示例：一根长 4 m、重 100 N 的均匀梁在中点被支撑。一个 200 N 的重物放在中心右侧 1 m 处。为了恢复平衡，300 N 的重物应放在中心左侧多远？绕支点取矩：顺时针力矩 = 200 × 1 = 200 N m；逆时针力矩 = 300 × x。令 300x = 200，得 x = 0.667 m。

4. Conditions for Equilibrium | 平衡条件

A rigid body is in complete equilibrium when two conditions are satisfied: the resultant force in any direction is zero, and the resultant moment about any point is zero. In WJEC problems, you usually apply the moment condition by choosing a pivot that eliminates an unknown reaction force.

刚体实现完全平衡需满足两个条件：任意方向的合力为零，且对任意点的合力矩为零。在 WJEC 考题中，通常通过选择支点消去未知反力来应用力矩条件。

If a beam is supported at two points, you can take moments about one support to find the reaction at the other. Then use vertical force equilibrium to find the remaining reaction.

若一根梁在两点被支撑，可以绕其中一个支点取矩求出另一个支点的反力，再通过竖直方向力平衡求出剩余反力。

Remember: forces acting through the pivot produce zero moment, so placing the pivot at the line of an unknown force simplifies calculations greatly.

记住：作用线通过支点的力不产生力矩，因此将支点选在未知力的作用线上可极大简化计算。

5. Pivot, Supports and Reaction Forces | 支点、支撑与反作用力

A pivot is a point about which an object can rotate. In many WJEC problems, the pivot is a support, a hinge, or the edge of a surface. The reaction force at a support acts perpendicular to the surface and prevents translation.

支点是物体可以绕其转动的点。在许多 WJEC 题目中，支点是一个支撑、铰链或表面的边缘。支撑处的反力垂直于接触面，并阻止平动。

When drawing free-body diagrams for moment problems, always include: the weight of the object acting at its centre of mass, externally applied loads, and normal reaction forces at supports. Clearly label distances from the chosen pivot.

在画力矩问题的受力图时，务必包含：作用在质心上的物体重力、外部施加的荷载，以及支撑处的法向反力。清晰标出各力到所选支点的距离。

If a beam rests on two supports, the reactions together balance the total downward force. Take moments about one support to find the other, then subtract from total load to get the first support reaction.

若梁放在两个支座上，两个支反力共同平衡总向下力。绕一个支座取矩求出另一个支反力，再从总荷载中减去该值即可得第一个支反力。

6. Uniform Rods: Centre at Midpoint | 均匀杆：重心在中点

A uniform rod or beam has its mass distributed evenly along its length. Its weight acts at the geometric centre, exactly halfway along the rod. This simplifies moment calculations because you can treat the entire weight as a single force acting at the midpoint.

均匀杆或梁的质量沿长度均匀分布。其重力作用在几何中心，即杆的正中点。这简化了力矩计算，因为可以将全部重量视为作用在中点的一个集中力。

Example: A uniform rod of length 5 m and weight 80 N is pivoted at one end. A vertical rope at the other end holds it horizontal. Find the tension in the rope. Taking moments about the pivot: weight moment = 80 × 2.5 = 200 N m clockwise. Tension T provides anticlockwise moment = T × 5. For equilibrium, 5T = 200, so T = 40 N.

示例：一根长 5 m、重 80 N 的均匀杆一端铰接，另一端用竖直绳子拉住使其保持水平。求绳中拉力。绕铰接点取矩：重力力矩 = 80 × 2.5 = 200 N m 顺时针。拉力 T 产生逆时针力矩 = T × 5。平衡时 5T = 200，因此 T = 40 N。

Always remember: for a uniform beam, the perpendicular distance from a pivot at one end to the weight is exactly half the length of the beam.

务必牢记：对于均匀梁，从一端支点到重力的垂直距离正好为梁长的一半。

7. Non-uniform Rods: Finding the Centre of Mass | 非均匀杆：求质心

If a rod is non-uniform, its centre of mass is not at the geometric centre. WJEC questions often ask you to find the centre of mass using the principle of moments. You will be given enough information to set up an equation involving the unknown distance.

若杆不均匀，其质心不在几何中心。WJEC 考题常要求运用力矩原理求解质心位置。题目会提供足够信息以建立包含未知距离的方程。

A typical problem: a non-uniform rod of length 3 m and weight 50 N balances horizontally on a pivot placed 1.2 m from one end. Find the distance of the centre of mass from that end. Let x be the distance. Taking moments about the pivot: 50 × (x – 1.2) = 0 if no other forces, but balance means the weight’s line passes through pivot, so x = 1.2 m. However, more often a supporting force or extra load is added.

典型问题：一根长 3 m 重 50 N 的非均匀杆在距一端 1.2 m 处加支点后水平平衡。求质心到该端的距离。设距离为 x。绕支点取矩：若仅重力作用，则力矩为零意味着重力线通过支点，故 x = 1.2 m。但更多情况下，会加入支承力或额外荷载。

For a rod suspended by one end, and a known force applied at the other to balance, moments give: weight × distance from pivot = applied force × total length. Then calculate the centre of mass.

对于一端悬挂、另一端加已知力使之平衡的杆，力矩方程为：重力 × 质心到支点距离 = 外力 × 总长。然后即可算出质心位置。

8. Tilting and Toppling | 倾斜与倾倒

An object placed on a surface will topple if the vertical line through its centre of mass falls outside the base of support. The moment due to weight then causes rotation rather than restoring equilibrium.

当物体重心所在的竖直线落在支撑底面之外时，物体会发生倾倒。此时重力产生的力矩将导致转动，而非维持平衡。

For a uniform cuboid of width w and height h, the critical angle of tilt θ at the point of toppling satisfies:

对于一个宽为 w、高为 h 的均匀长方体，发生倾倒时的临界倾斜角 θ 满足：

tanθ = w / h

This arises because the centre of mass is at height h/2 and half-width w/2 from the pivot edge. When the centre of mass is directly above the edge, tanθ = (w/2) / (h/2) = w/h.

这一关系来源于质心高度为 h/2，距倾倒边缘的水平距离为 w/2。当质心刚好位于边缘正上方时，tanθ = (w/2) / (h/2) = w/h。

WJEC questions may ask: ‘How far can a person lean over an edge?’ or ‘What minimum force is needed to just tilt a block?’ Always identify the pivotal edge and apply the principle of moments to the point where the normal reaction acts at the edge.

WJEC 题目可能问：“一个人向外倾斜多远而不会跌落？”或“使一个物块刚好倾斜的最小力是多少？”始终要找准转动边缘，并应用力矩原理，此时法向反力恰好作用在边缘上。

9. Multi-support Beams and Load Distribution | 多支撑梁与荷载分布

When a beam rests on more than two supports, or carries multiple loads between two supports, you will need to use both force equilibrium and moment equilibrium. Usually, the problem provides dimensions and loads, and asks for the reactions at the supports.

当一根梁由多于两个支座支撑，或在两个支座间承载多个荷载时，需要同时使用力平衡和力矩平衡。通常题目会给出尺寸和荷载，要求计算支座反力。

Consider a uniform beam of length 6 m, weight 200 N, supported at points 1 m from the left end and 1 m from the right end. A 300 N load sits at the centre. Find the two reactions. Take moments about left support: weight acts at 3 m from left support (midpoint is at 3 m, left support at 1 m, so distance = 2 m), giving 200 × 2 = 400 N m clockwise. Load 300 N at centre (3 m from left end, so 2 m from left support) gives 300 × 2 = 600 N m clockwise. Total clockwise = 1000 N m. Right support is 4 m from left pivot; its anticlockwise moment = R2 × 4. So 4R2 = 1000, R2 = 250 N. Then vertical equilibrium: R1 + R2 = 200 + 300 = 500, thus R1 = 250 N.

设想一根均匀梁长 6 m、重 200 N，左端 1 m 处和右端 1 m 处被支撑。中心处有一 300 N 的荷载。求两个支反力。绕左支座取矩：重力作用在梁中点（3 m 处），左支座在 1 m 处，力臂 = 2 m，产生顺时针力矩 200 × 2 = 400 N m。300 N 荷载在中心（3 m 处，距离左支座也是 2 m）产生 300 × 2 = 600 N m 顺时针。顺时针总和 = 1000 N m。右支座距左支点 4 m，其逆时针力矩 = R2 × 4。因此 4R2 = 1000，R2 = 250 N。竖直方向力平衡：R1 + R2 = 200 + 300 = 500，所以 R1 = 250 N。

Always check that the sum of reactions equals the total downward force. Using symmetry can sometimes save time.

务必检验支座反力之和等于向下总力。有时对称性可以简化计算，节省时间。

10. Problem-solving Strategies and Exam Tips | 解题策略与应试技巧

To confidently tackle WJEC moments and equilibrium questions, follow a structured approach:

要自信地应对WJEC力矩与平衡试题，请遵循系统化的步骤：

1. Draw a clear diagram showing all forces, distances, and the pivot. Indicate directions of rotation (clockwise/anticlockwise).

1. 画出清晰的受力图，标出所有力、距离和支点，并注明转动方向（顺时针/逆时针）。

2. Choose a pivot that eliminates an unknown force (often a support or reaction).

2. 选择能消去未知力（通常是支反力）的支点。

3. Write the principle of moments equation, carefully measuring perpendicular distances. Use consistent units.

3. 写出力矩原理方程，仔细测量垂直距离。单位统一使用 N 和 m。

4. Solve for the unknown, then use force equilibrium if necessary for other reactions.

4. 解出未知量，然后必要时利用力平衡求其他反力。

5. Check your answer: sum of clockwise moments should equal sum of anticlockwise moments about any other point for verification.

5. 检验答案：绕任意其他点，顺时针力矩总和应等于逆时针力矩总和。

Common pitfalls include using the sloping distance instead of the perpendicular distance, forgetting to include the beam’s own weight, and mixing units. Practise with past WJEC papers to familiarise yourself with the wording and typical setups.

常见错误包括误用斜距而非垂直距离，忘记考虑梁的自重，以及单位混用。多练WJEC历年真题，熟悉题型表述和典型装置。

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Moments and Equilibrium | 力矩与平衡 考点精讲