IGCSE Maths: Moments and Equilibrium Exam Focus | IGCSE 数学：力矩与平衡 考点精讲

📚 IGCSE Maths: Moments and Equilibrium Exam Focus | IGCSE 数学：力矩与平衡 考点精讲

In IGCSE Mathematics, the topic of moments and equilibrium bridges algebra and mechanics, requiring you to model real-world balancing scenarios with precise equations. This article unpacks every essential concept — from the definition of a moment to solving multi‑force equilibrium problems — giving you the confidence to tackle any exam question.

在 IGCSE 数学中，力矩与平衡是连接代数与力学的桥梁，要求你用精确的方程来模拟现实中的平衡场景。本文逐一讲解力矩的定义、多力平衡问题的求解等每个核心概念，帮助你自信应对任何考题。

1. What is a Moment? | 什么是力矩？

A moment is the turning effect of a force about a 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. Imagine opening a door — pushing near the hinges requires a much larger effort than pushing at the handle, because the distance from the pivot is smaller.

力矩是力对支点产生的转动效应。它取决于两个因素：力的大小以及支点到力作用线的垂直距离。想象一下开门——在靠近铰链处推门比在门把手处推需要更大的力，因为此时力到支点的距离更小。

2. Calculating Moments: Formula and Units | 计算力矩：公式与单位

The moment M of a force F about a pivot is given by M = F × d, where d is the perpendicular distance from the pivot to the line of action of the force. In IGCSE questions, d is always the shortest distance, so you can often read it directly from the diagram. The SI unit of a moment is the newton metre (N m).

力 F 关于支点的力矩 M 由公式 M = F × d 给出，其中 d 是支点到力作用线的垂直距离。在 IGCSE 考题中，d 总是最短距离，因此你通常可以直接从图中读取。力矩的国际单位是牛顿米（N m）。

M = F × d    (d ⟂ F)

3. Clockwise and Anticlockwise Moments | 顺时针力矩与逆时针力矩

A force can cause a clockwise or an anticlockwise rotation. By convention, we treat one direction as positive and the other as negative when writing equilibrium equations. You must always state clearly which direction you take as positive. In the exams, ‘moments about a point’ are often labelled clockwise or anticlockwise to help you set up the equation.

力可以引起顺时针或逆时针转动。习惯上，我们在列平衡方程时将其中一个方向视为正，另一个方向视为负。你必须清楚地说明选定哪一个方向为正。考试中，“关于某点的力矩”常被标注为顺时针或逆时针，帮助你建立方程。

4. The Principle of Moments | 力矩原理

For a body in rotational equilibrium, the sum of the clockwise moments about any pivot equals the sum of the anticlockwise moments about that same pivot. This is the principle of moments. It is the foundation for solving all static equilibrium problems where an object is not turning.

对于处于转动平衡的物体，关于任意支点的顺时针力矩之和等于关于同一点的逆时针力矩之和。这就是力矩原理。它是求解所有不转动的静力平衡问题的基础。

5. Equilibrium Conditions – Translational and Rotational | 平衡条件——平动平衡与转动平衡

A rigid body in complete equilibrium must satisfy two separate conditions: (1) the resultant force in any direction is zero, and (2) the resultant moment about any point is zero. In IGCSE Maths, you will primarily use the moment condition to find an unknown force or distance, and then sometimes check vertical force equilibrium to find a reaction force at the pivot.

一个处于完全平衡的刚体必须满足两个独立条件：（1）任意方向的合力为零；（2）关于任意点的合力矩为零。在 IGCSE 数学中，你主要利用力矩条件求出未知力或距离，然后有时会用竖直方向力的平衡来求支点处的反作用力。

6. Choosing the Right Pivot | 选择合适的支点

A powerful strategy is to take moments about a point where an unknown force acts. Since the moment of that unknown force about that point is zero (because d = 0), it will not appear in your equation. This simplifies the algebra considerably and is especially useful when a question involves a pivot reaction that you are not asked to find.

一个有效的策略是：对某个未知力作用点取力矩。由于该未知力关于此点的力矩为零（因为 d = 0），它就不会出现在你的方程中。这极大地简化了代数运算，当题目中的支点反力不是所求量时尤其管用。

7. Uniform and Non‑Uniform Rods | 均匀杆与非均匀杆

For a uniform rod, the weight acts exactly at its centre. For a non‑uniform rod, the centre of mass may be at a different position, and a typical exam question will ask you to find this position using moments. Always draw the weight arrow from the centre of mass, and label its distance from the pivot carefully.

对于均匀杆，重力恰好作用在杆的中心。对于非均匀杆，质心可能在不同位置，典型的考题会要求你用力矩求出该位置。始终从质心画出重力的箭头，并仔细标出它到支点的距离。

8. Worked Example: Uniform Beam with a Load | 实例：均匀横梁加载重物

A uniform beam AB of length 4 m and weight 100 N rests on a pivot at C, where AC = 1 m. A weight of 150 N is hung from end A. Find the distance from B where an upward force of 200 N must be applied to keep the beam horizontal. Take moments about C. Clockwise: moment of 150 N about C = 150 × 1. Anticlockwise: moment of beam’s weight about C = 100 × 1 (since centre is 2 m from A, so 1 m from C on the other side), and the upward 200 N at distance x from B, so its distance from C is (3 − x). Set sum clockwise = sum anticlockwise and solve: 150 = 100 + 200(3 − x) gives x = 2.75 m from B.

一根均匀横梁 AB 长 4 m，重 100 N，支在 C 点，其中 AC = 1 m。在 A 端悬挂 150 N 重物。欲使横梁保持水平，需在距 B 端多远的位置施加一个 200 N 的向上力？对 C 点取力矩。顺时针：150 N 的力矩 = 150 × 1。逆时针：横梁自身重力的力矩 = 100 × 1（因为重心距 A 端 2 m，故在 C 点另一侧 1 m 处），以及向上 200 N 的力，设它距 B 端为 x，则距 C 点为 (3 − x)。令顺时针力矩和等于逆时针力矩和：150 = 100 + 200(3 − x)，解得 x = 2.75 m（距 B 端）。

9. Non‑Uniform Rod – Finding Centre of Mass | 非均匀杆——求质心位置

A non‑uniform rod PQ of length 5 m is balanced horizontally on a pivot 2 m from P. When a mass of 8 kg is hung from P, the rod balances when the pivot is 1.8 m from P. Find the distance of the centre of mass from P. Let the weight of the rod be W and its centre of mass at distance d from P. First case: taking moments about pivot (2 m from P), W × (d − 2) = 0? No, rod alone balances at 2 m from P, so the pivot is directly under the centre of mass: d = 2 m. Wait, careful: ‘balanced horizontally on a pivot 2 m from P’ means the rod alone balances there, so d = 2 m. Second case with mass: total clockwise moments = 8g × 1.8, anticlockwise = W × (1.8 − d) but d = 2 m, so W × (−0.2) — this suggests a sign issue. The better interpretation: the rod alone has d = 2 m. With the 8 kg mass at P, the new balance pivot is 1.8 m from P. Take moments about this new pivot: clockwise: moment of 8g = 8g × 1.8. Anticlockwise: moment of weight W = W × (d − 1.8) = W × (2 − 1.8) = 0.2W. Set equal: 8g × 1.8 = 0.2W ⇒ W = 72g. The question might ask for d if unknown: suppose first balance alone gives pivot at x from P unknown, but the wording suggests first balance reveals d = 2 m. Let’s restructure: If the rod balances at 2 m from P, then its centre of mass is at that point, d = 2 m. So the example already gave d. We’ll present a simpler scenario: A non‑uniform rod of weight 50 N balances at a point 1.2 m from end A. When a 20 N weight is hung from A, the rod balances at a point 0.8 m from A. Find distance of centre of mass from A. Let d be distance from A. First balance: d = 1.2 m (since pivot lies under centre of mass when no other forces). Second: take moments about new pivot 0.8 m from A. Clockwise: moment of 20 N = 20 × 0.8 = 16 Nm. Anticlockwise: moment of weight 50 N = 50 × (1.2 − 0.8) = 50 × 0.4 = 20 Nm. They are not equal, so d is not 1.2 m. Actually, the ‘balanced horizontally on a pivot 2 m from P’ means that pivot is the balance point, so the centre of mass must be directly above that pivot because the weight acts vertically down and the pivot reaction is upward. The rod alone balanced means the moment of weight about that pivot is zero, so the weight line passes through the pivot, hence centre of mass is 2 m from P. So d = 2 m. The second case confirms the weight. That’s a common exam type: use the two situations to find the weight of the rod. I’ll rewrite the example accordingly.

一根非均匀杆 PQ 长 5 m，在距 P 端 2 m 处的支点上能单独水平平衡。当在 P 端悬挂 8 kg 的物体时，杆在支点距 P 端 1.8 m 处平衡。求杆的重心到 P 端的距离。由单独平衡可知，重心就在支点正上方，因此 d = 2 m。再对第二情形的新支点（距 P 端 1.8 m）取力矩：顺时针——8g × 1.8；逆时针——杆的重力 W × (2 − 1.8) = 0.2W。令两者相等得 W = 72g。此题巧妙地利用力矩原理求出了杆的重量，而重心位置则直接由第一次平衡条件得出。

10. Multiple Forces and the Moment Equation | 多个力与力矩方程

When a beam is under several parallel forces, write one moment equation about a chosen pivot. Sum all clockwise moments and equate to the sum of all anticlockwise moments. Remember to include the weight of the beam itself acting at its centre. If there are two unknown forces, you will also need to use the vertical force equilibrium: total upward forces = total downward forces.

当横梁受到多个平行力作用时，针对选定的支点写出一个力矩方程。将所有顺时针力矩相加，并令其等于所有逆时针力矩之和。务必把横梁自身的重量考虑在内，作用在它的中心。若出现两个未知力，你还需要利用竖直方向的力平衡：向下的总力等于向上的总力。

11. Common Pitfalls and How to Avoid Them | 常见错误与避错技巧

Pitfall 1: using a distance that is not perpendicular to the force. Always draw a dashed perpendicular line from the pivot to the force’s line of action. Pitfall 2: forgetting the weight of the beam. Uniform beam – weight acts at the centre. Pitfall 3: sign errors when taking moments on both sides of the pivot. Adopt a consistent sign convention and label clockwise/anticlockwise on your diagram.

错误 1：使用的距离不与力垂直。务必从支点向力的作用线画出垂直虚线。错误 2：忘记横梁自身的重量。均匀横梁——重力作用在中心。错误 3：在支点两侧取力矩时出现符号错误。采用一致的符号约定，并在图上标出顺时针/逆时针。

12. Exam‑style Question Walkthrough | 考题实战演练

Question: A uniform plank of length 6 m and mass 20 kg rests on two supports at its ends. A painter of mass 80 kg stands 2 m from the left end. Find the reactions at the supports. Solution: Let reactions be R_L and R_R. Weight of plank = 20g at 3 m from left. Painter’s weight = 80g at 2 m from left. Vertical equilibrium: R_L + R_R = 20g + 80g = 100g. Moments about left end: clockwise moments = (20g × 3) + (80g × 2) = 60g + 160g = 220g. Anticlockwise moment = R_R × 6. So 6 R_R = 220g ⇒ R_R = (220/6)g ≈ 35.9g N ≈ 352 N (taking g = 9.8). Then R_L = 100g − R_R ≈ 64.1g N ≈ 628 N.

题目：一块长 6 m、质量 20 kg 的均匀木板，两端支在两个支座上。一位质量 80 kg 的油漆工站在距左端 2 m 处。求两支座的反作用力。解答：设反力为 R_L 和 R_R。木板重 20g，作用在距左端 3 m 处；油漆工重 80g，作用在距左端 2 m 处。竖直平衡：R_L + R_R = 20g + 80g = 100g。对左端取力矩：顺时针力矩 = (20g × 3) + (80g × 2) = 60g + 160g = 220g；逆时针力矩 = R_R × 6。于是 6 R_R = 220g ⇒ R_R ≈ 35.9g N ≈ 352 N（取 g = 9.8）。则 R_L = 100g − R_R ≈ 628 N。

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