Moments and Equilibrium – AQA IGCSE Mathematics Revision | 力矩与平衡考点精讲

📚 Moments and Equilibrium – AQA IGCSE Mathematics Revision | 力矩与平衡考点精讲

In IGCSE Mathematics, the topic of moments and equilibrium applies proportional reasoning and equation-solving to real-world contexts such as levers, seesaws, and beams. A moment measures the turning effect of a force about a pivot. When an object is balanced, the clockwise and anticlockwise moments are equal. This article covers the essential definitions, the principle of moments, worked examples, and common pitfalls, providing a clear revision guide for AQA candidates.

在 IGCSE 数学中，力矩与平衡这一主题将比例推理和方程求解应用到杠杆、跷跷板和横梁等真实情境中。力矩衡量力对支点的转动效应。当物体平衡时，顺时针力矩与逆时针力矩相等。本文涵盖了基本定义、力矩原理、典型例题以及常见错误，为 AQA 考生提供一份清晰的复习指南。

1. What is a Moment? | 什么是力矩？

A moment is the turning effect produced when a force is applied to an object at a distance from a pivot. In everyday life, using a spanner to loosen a nut or sitting on a seesaw both involve moments. The turning effect depends on two factors: the size of the force and the perpendicular distance from the pivot to the line of action of the force.

力矩是力作用在物体上、距支点一定距离时产生的转动效应。在日常生活中，用扳手拧松螺母或坐在跷跷板上都涉及力矩。转动效应取决于两个因素：力的大小以及支点到力的作用线的垂直距离。

In IGCSE Mathematics problems, the pivot is often a fixed point, such as a fulcrum under a beam or the centre of a see-saw. You must identify the pivot and then consider all forces and their distances, ensuring these distances are measured at right angles to the force direction.

在 IGCSE 数学问题中，支点通常是一个固定点，比如横梁下的支点或跷跷板中心。你必须确定支点，然后考虑所有力及其距离，确保这些距离是与力的方向成直角的距离。

2. The Moment Formula | 力矩公式

The moment of a force is calculated using the formula:

力的力矩用以下公式计算：

M = F × d

where M is the moment (measured in newton-metres, N m), F is the force in newtons (N), and d is the perpendicular distance from the pivot to the line of action of the force in metres (m). It is crucial to use perpendicular distance, not the direct distance along the object if the force is not applied at a right angle.

其中 M 为力矩（单位牛顿米，N m），F 为力（牛顿，N），d 为支点到力的作用线的垂直距离（米，m）。务必使用垂直距离，如果力不是以直角施加，不能直接使用沿物体的距离。

If a force is applied at an angle, you may need to resolve it into components. However, most IGCSE questions involve forces acting perpendicular to the beam or lever, so the distance along the object is the perpendicular distance.

如果力以某个角度施加，你可能需要将其分解为分量。然而，大多数 IGCSE 问题涉及垂直于横梁或杠杆的力，因此沿物体的距离就是垂直距离。

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

Every moment has a direction: it can turn an object clockwise or anticlockwise about the pivot. To apply the principle of moments, we must label each moment as clockwise or anticlockwise. A convenient way is to imagine whether the force would rotate the object in the same direction as the hands of a clock or opposite to them.

每个力矩都有一个方向：它可以使物体绕支点顺时针或逆时针转动。为了应用力矩原理，我们必须将每个力矩标记为顺时针或逆时针。一种简便的方法是想象该力会使物体按钟表指针方向还是相反方向转动。

For example, if a downward force acts to the right of a pivot on a beam, it tends to turn the beam clockwise. The same downward force to the left of the pivot gives an anticlockwise moment. Always check from the pivot’s perspective.

例如，如果一个向下的力作用在横梁支点右侧，它往往会使横梁顺时针转动。同一个向下的力在支点左侧则产生逆时针力矩。务必从支点的视角来判断。

4. Principle of Moments (Equilibrium) | 力矩原理（平衡条件）

When a system is in equilibrium – that is, it is balanced and not rotating – the sum of the clockwise moments equals the sum of the anticlockwise moments. This is the principle of moments:

当系统处于平衡状态（即平衡且不转动）时，顺时针力矩的总和等于逆时针力矩的总和。这就是力矩原理：

Sum of clockwise moments = Sum of anticlockwise moments

In equation form, for multiple forces this can be written as:

用方程形式表示，对于多个力可写为：

Σ Fᵢ dᵢ (clockwise) = Σ Fⱼ dⱼ (anticlockwise)

This equation is the foundation for solving nearly all moment problems in IGCSE Mathematics. You simply substitute known values and solve for the unknown, often a distance or a force.

这个方程是解决 IGCSE 数学中几乎所有力矩问题的基础。你只需代入已知值，求解未知量，通常是距离或力。

5. Calculating an Unknown Force | 计算未知力

Many questions involve a beam pivoted at one end or at its centre, with known distances and forces on one side, and an unknown force on the other side that keeps the beam horizontal. Start by taking moments about the pivot. This eliminates any reaction force at the pivot from the equation, as its distance is zero.

许多问题涉及一根梁在其一端或中心被支起，其中一侧已知距离和力，另一侧有一个未知力保持梁水平。首先对支点取矩。这样由于支点处的反作用力距离为零，它就不会出现在方程中。

For example, a uniform beam of length 4 m is pivoted at its centre. A 30 N weight is placed 0.8 m to the left of the pivot. What upward force must be applied 1.2 m to the right to balance the beam? The weight produces an anticlockwise moment (if we define left as anticlockwise) of 30 × 0.8 = 24 N m. The unknown force F must produce a clockwise moment of F × 1.2. Setting them equal: F × 1.2 = 24, hence F = 20 N.

例如，一根长 4 m 的均匀梁在其中心被支起。一个 30 N 的重物放在支点左侧 0.8 m 处。那么必须在右侧 1.2 m 处施加多大的向上力才能使梁平衡？重物产生一个逆时针力矩（假设左侧为逆时针）：30 × 0.8 = 24 N m。未知力 F 必须产生顺时针力矩 F × 1.2。令两者相等：F × 1.2 = 24，因此 F = 20 N。

6. Finding an Unknown Distance | 求未知距离

Similarly, you might be asked to determine at what distance from the pivot a known force must act to achieve equilibrium. Use the same principle: equate clockwise and anticlockwise moments and solve for the distance.

类似地，你可能会被问到，已知的力必须作用在离支点多远的地方才能达到平衡。运用相同的原理：令顺时针与逆时针力矩相等，然后求解距离。

Suppose a 50 N child sits 2.0 m from the centre of a seesaw. A 40 N child sits on the other side. How far from the centre must the 40 N child sit? Taking moments about the centre: anticlockwise moment = 50 × 2.0 = 100 N m. Clockwise moment = 40 × d. Equating: 40d = 100 ⇒ d = 2.5 m. The answer is 2.5 m.

假设一个 50 N 的儿童坐在跷跷板中心 2.0 m 处。一个 40 N 的儿童坐在另一侧。那么 40 N 的儿童必须坐在离中心多远的地方？对中心取矩：逆时针力矩 = 50 × 2.0 = 100 N m。顺时针力矩 = 40 × d。令两者相等：40d = 100 ⇒ d = 2.5 m。答案是 2.5 m。

7. Uniform Rods and Centre of Mass | 均匀杆与重心

When a rod or beam is described as ‘uniform’, its mass is evenly distributed. For moment calculations, the entire weight of a uniform rod acts through its centre of mass, which is at the geometric centre of the rod. So if you have a uniform plank of length L and weight W, you can treat its weight as a single force W acting downwards at a distance L/2 from either end, regardless of where the pivot is.

当一根杆或梁被描述为“均匀”时，其质量均匀分布。在进行力矩计算时，均匀杆的全部重力作用在其重心上，即杆的几何中心。因此，如果你有一块长度为 L、重量为 W 的均匀木板，不论支点在哪里，你都可以将其重量视为一个单独向下的力 W，作用在距任意一端 L/2 处。

In problems where a uniform beam is supported at two points or has additional loads, you must include the beam’s own weight as a concentrated force at its centre when taking moments about any point. This is a common requirement in IGCSE questions.

在均匀梁有两处支撑或有额外负载的问题中，对任意点取矩时，你必须将梁自身的重量作为一个集中力包含在内，作用在其中心。这是 IGCSE 问题中的常见要求。

8. Multiple Forces and Support Reactions | 多力与支持力

More advanced equilibrium problems involve several forces, including unknown reaction forces at supports. To find an unknown reaction, you can take moments about a point where another unknown acts, so that unknown is eliminated. Alternatively, you can use the condition that the total upward forces equal the total downward forces (vertical equilibrium) together with the moment equation.

更复杂的平衡问题涉及多个力，包括支撑点处未知的反作用力。为求出一个未知反作用力，你可以对另一个未知力作用的点取矩，这样那个未知力就会被消去。或者，你也可以利用向上的总力等于向下的总力（竖直平衡）这一条件，并结合力矩方程来求解。

For instance, a uniform beam weighing 100 N rests on two supports at its ends. A 200 N weight is placed 1 m from the left end. If the beam is 4 m long, find the reaction forces at the supports. By taking moments about the left support, you eliminate the left reaction and solve for the right reaction. Then use vertical equilibrium to find the left reaction.

例如，一根重 100 N 的均匀梁两端搁在两个支座上。一个 200 N 的重物放在距左端 1 m 处。如果梁长 4 m，求支座的反作用力。通过对左支座取矩，可消去左支座反力，求出右支座反力。然后利用竖直方向平衡求出左支座反力。

9. Worked Example | 典型例题

Let us solve a typical IGCSE AQA-style problem in full:

让我们完整地求解一道典型的 IGCSE AQA 风格的题目：

Problem: A uniform plank of length 5 m and weight 80 N is placed on a pivot 2 m from one end. A 120 N load is hung from the shorter end. What force must be applied at the opposite end to balance the plank horizontally?

题目：一块长 5 m、重 80 N 的均匀木板，支点设在距一端 2 m 处。一个 120 N 的重物挂在较短的一端。需要在另一端施加多大的力才能使木板水平平衡？

Solution: First, identify distances from the pivot. Shorter end is 2 m from pivot, longer end is 3 m from pivot. The weight of the plank (80 N) acts at its centre, which is 2.5 m from either end. The centre is 0.5 m from the pivot on the longer side (since pivot is 2 m from one end, centre is 2.5 m from that end, so centre is 0.5 m beyond the pivot towards the longer end).

解答：首先，确定各力到支点的距离。较短一端距支点 2 m，较长一端距支点 3 m。木板的重力（80 N）作用在其中心，中心距任意一端 2.5 m。中心位于支点较长一侧 0.5 m 处（因为支点距一端 2 m，中心距该端 2.5 m，所以中心在越过支点向较长端 0.5 m 处）。

Take moments about the pivot. Clockwise moments are produced by the plank’s weight (tending to rotate clockwise from this pivot’s perspective) and possibly by the applied force. Let the required force at the longer end be F, acting downwards or upwards? To balance, the force must act upwards to create an anticlockwise moment, opposing the clockwise moments. We define anticlockwise as positive.

对支点取矩。顺时针力矩由木板重力（从该支点看，趋向于顺时针转动）产生，还可能有施加的力。设较长端所需的力为 F，方向向上还是向下？为达到平衡，该力必须向上以产生逆时针力矩，与顺时针力矩对抗。我们定义逆时针为正。

Moments about pivot:

对支点的力矩：

120 N load at 2 m from pivot (on the left) produces an anticlockwise moment = 120 × 2 = 240 N m.

距支点 2 m 处的 120 N 负载（左侧）产生逆时针力矩 = 120 × 2 = 240 N m。
80 N weight of plank acts 0.5 m to the right of pivot, producing a clockwise moment = 80 × 0.5 = 40 N m.

80 N 的木板重力作用在支点右侧 0.5 m 处，产生顺时针力矩 = 80 × 0.5 = 40 N m。
Upward force F at the longer end (3 m from pivot, on the right) produces an anticlockwise moment? Wait, upward force on the right would tend to rotate anticlockwise from pivot’s view, so it is anticlockwise moment = F × 3.

较长端（距支点 3 m，右侧）向上的力 F 产生逆时针力矩？等一下，右侧向上的力从支点看会趋向于逆时针转动，因此是逆时针力矩 = F × 3。

For equilibrium: sum of anticlockwise moments = sum of clockwise moments.

对于平衡：逆时针力矩总和 = 顺时针力矩总和。

240 + 3F = 40

This equation gives 3F = 40 – 240 = -200, so F = -66.7 N. The negative sign indicates our assumed direction (upwards) was wrong; the force must actually be 66.7 N downwards to provide a clockwise moment. Checking: if F is 66.7 N downwards on the right, that produces a clockwise moment of 66.7 × 3 = 200 N m. Total clockwise = 40 + 200 = 240 N m, which balances the anticlockwise 240 N m. So the answer is a downward force of 66.7 N (or 200/3 N).

这个方程得出 3F = 40 – 240 = -200，因此 F = -66.7 N。负号表明我们假设的方向（向上）错了；该力实际上必须是 66.7 N 向下，以提供顺时针力矩。检验：如果 F 为右侧向下的 66.7 N，则产生顺时针力矩 66.7 × 3 = 200 N m。总顺时针力矩 = 40 + 200 = 240 N m，与逆时针的 240 N m 平衡。所以答案是向下的力 66.7 N（或 200/3 N）。

10. Common Mistakes and Tips | 常见错误与提示

Many students lose marks by using the wrong distance. Always ensure you use the perpendicular distance from the pivot to the force. If a force acts at an angle, you must either resolve the force into components perpendicular and parallel to the beam, or multiply the force by the perpendicular distance from the pivot to the line of action (the lever arm). In IGCSE, most forces are perpendicular, so the distance along the beam is correct.

许多学生因使用错误的距离而失分。务必使用支点到力的垂直距离。如果力以某个角度作用，你必须要么将力分解为垂直于梁和平行于梁的分量，要么将力乘以支点到作用线的垂直距离（力臂）。在 IGCSE 中，大多数力是垂直的，因此沿梁的距离是正确的。

Another common error is forgetting to include the weight of a uniform object acting at its centre. Whenever a question states ‘uniform rod’ or ‘uniform plank’, you must include its weight at the midpoint unless the pivot is exactly at that midpoint (then its moment is zero). Also, be careful with units: distances in metres, forces in newtons, moments in newton-metres.

另一个常见错误是忘记计入均匀物体作用在中心的重量。每当题目说明“均匀杆”或“均匀木板”时，除非支点恰好在中心（此时其力矩为零），否则必须将重量包含在支点处。此外，注意单位：距离用米，力用牛顿，力矩用牛顿米。

Finally, always define your positive direction (clockwise or anticlockwise) and stick to it consistently. Writing a clear equation avoids sign errors. Practice drawing a clear diagram and labelling all forces and distances. This will help you visualise the moments and check your equation before solving.

最后，务必定义你的正方向（顺时针或逆时针），并始终如一地遵守。写出清晰的方程可以避免符号错误。多练习绘制清晰的示意图，标注所有力和距离。这将帮助你直观想象力矩，并在求解前检查你的方程。

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Moments and Equilibrium – AQA IGCSE Mathematics Revision | 力矩与平衡考点精讲