📚 Moments and Equilibrium for GCSE WJEC Mathematics | WJEC GCSE 数学：力矩与平衡考点精讲
In the WJEC GCSE Mathematics specification, the topic of moments and equilibrium bridges practical physics with algebraic problem-solving. You are expected to calculate turning effects, apply the principle of moments, and solve for unknown forces or distances when an object is balanced. Mastering these skills will not only boost your exam performance but also give you a solid foundation for mechanics in A-level studies.
在 WJEC GCSE 数学大纲中，力矩与平衡这一主题将实际物理问题与代数解题技巧结合在一起。你需要计算转动效应，运用力矩原理，并在物体平衡时求出未知力或距离。掌握这些技能不仅会提升你的考试成绩，还能为 A-level 力学学习打下扎实的基础。
1. What is a Moment? | 什么是力矩？
A moment is the turning effect of a force about a specific point or pivot. It measures how strongly a force tries to rotate an object.
力矩是力绕某个特定点或支点产生的转动效应，它度量了一个力试图使物体转动的强弱程度。
The 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.
该效应取决于两个因素：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, it is hard to open. Push far from the hinge, and the same force produces a much greater turning effect — that is a larger moment.
如果你在门轴附近推门，很难打开。远离铰链用同样大小的力去推，就会产生更大的转动效应——也就是更大的力矩。
2. The Moment Formula | 力矩公式
The moment M is calculated using the equation:
力矩 M 用以下公式计算：
M = F × d
M = F × d
where 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).
其中 F 是力（单位为牛顿，N），d 是从支点到力的作用线的垂直距离（单位为米，m）。
The standard unit for a moment is the newton-metre (N m). It is essential that the distance used is always perpendicular; if the force is not already perpendicular, you must find the perpendicular component of the distance or the force.
力矩的标准单位是牛·米（N m）。关键在于所使用的距离必须始终是垂直距离；如果力本身并非垂直，你必须求出距离的垂直分量或力的垂直分力。
3. The Principle of Moments | 力矩原理
When an object is in equilibrium (no rotation), the sum of clockwise moments about any pivot equals the sum of anticlockwise moments about the same pivot.
当物体处于平衡状态（不转动）时，对任意支点的顺时针力矩之和等于对该支点的逆时针力矩之和。
∑ Mclockwise = ∑ Manticlockwise
∑ M顺时针 = ∑ M逆时针
This principle is the cornerstone of solving most exam questions. You will often set up one equation by taking moments about a chosen pivot, then solve for the unknown force or distance.
这一原理是解决大多数考题的基石。你通常会选择一个合适的支点，用它建立一个力矩方程，然后解出未知力或距离。
4. Conditions for Equilibrium | 平衡条件
For a rigid body to be in complete static equilibrium, two conditions must be satisfied:
一个刚体要处于完全静力平衡，必须满足两个条件：
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The resultant force in any direction is zero (∑ F = 0).
任何方向上的合力为零（∑ F = 0）。
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The resultant moment about any point is zero (∑ M = 0).
对任何点的合力矩为零（∑ M = 0）。
In WJEC GCSE questions, vertical forces are the most common. You will often use ∑ upward forces = ∑ downward forces, and then use the principle of moments to find an unknown reaction force at a support.
在 WJEC GCSE 题目中，最常见的是竖直方向上的力。你通常会使用向上的合力 = 向下的合力，然后再用力矩原理求出支座处未知的反作用力。
5. Uniform Rods and the Centre of Mass | 均匀杆与质心
A uniform rod has its weight distributed evenly along its length. As a result, its centre of mass is exactly at the geometrical centre of the rod.
均匀杆的重量沿杆长均匀分布。因此，它的质心恰好位于杆的几何中心。
When you take moments for a uniform rod, you can treat its entire weight W as acting downwards from the centre. For a rod of length L, the weight acts at a distance L/2 from either end.
当你对一个均匀杆取矩时，可以将其全部重量 W 视为作用在杆的中心点处竖直向下。对于长度为 L 的杆，重量的作用点距离两端均为 L/2。
This simplification is a favourite in exam questions, so always identify if the rod or beam is described as ‘uniform’.
这种简化是考试中的经典考法，因此一定要留意题目是否将杆或梁描述为“均匀的”。
6. Choosing a Pivot to Take Moments | 选择支点取矩
A powerful strategy is to take moments about a point where an unknown force acts. Because the moment of that unknown force about that point is zero (its distance is zero), the force is eliminated from the equation.
一个非常有效的策略是，选择一个有未知力作用的点作为支点来取矩。因为该未知力对该点的力矩为零（距离为零），该力就从方程中消掉了。
This reduces the number of unknowns and allows you to find another force directly. After finding one support force, you can then use the vertical force balance to find the remaining force.
这减少了未知数的个数，让你能够直接求出另一个力。求出一个支座力后，你就可以利用竖直方向力的平衡求出剩下的那个力。
7. Problems with Two Supports | 双支撑问题
A common WJEC exam style presents a uniform beam supported by two vertical supports at its ends, with additional loads placed along the beam.
WJEC 考试中一种常见题型是：一根均匀梁由两端的竖直支座支撑，梁上不同位置还放置有其他荷载。
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Example approach: 例题思路： |
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1. Draw the beam with all forces: the weight of the beam acting at its centre, additional loads, and the two upward reaction forces R1 and R2. 1. 画出梁并标出所有力：作用在中心的梁自重、额外荷载以及两个向上的支反力 R1 和 R2。 2. Take moments about one support (say R1) to find the other support force R2. 2. 对其中一个支座（如 R1）取矩，求出另一个支座力 R2。 3. Use ∑ upward = ∑ downward to find R1. 3. 利用向上的合力 = 向下的合力求出 R1。 |
And always check that the two reaction forces add up to the total downward load. This provides a quick sense-check.
而且一定要检查两个支反力之和是否等于全部向下的荷载之和，这是一个快速的验算方法。
8. Inclined Forces and Perpendicular Distance | 倾斜力与垂直距离
If a force does not act perpendicular to the lever arm, you cannot simply use the given distance. You must use either the perpendicular component of the force or the perpendicular distance from the pivot to the line of action.
如果一个力并非垂直于力臂作用，你就不能直接使用给出的距离。你必须使用力的垂直分力，或者从支点到力作用线的垂直距离。
For a force F applied at an angle θ to the beam, the perpendicular distance is d sin θ, giving:
对于与杆成角度 θ 的力 F，垂直距离为 d sin θ，因此：
M = F × d × sin θ
M = F × d × sin θ
Alternatively, you can resolve the force into components and use the component perpendicular to the beam. Both methods produce the same result, as long as you are consistent.
另一种方法是把力进行分解，然后使用垂直于杆的分力。两种方法结果一致，只要保持前后一致即可。
9. Beams with Multiple Loads | 多荷载的梁
When several weights act at different points on a beam, list all clockwise and anticlockwise moments separately about your chosen pivot.
当一根梁上有多个重物作用在不同位置时，要把所选支点的所有顺时针力矩和逆时针力矩分别列出。
A clear diagram with all distances marked is essential. Label distances from the pivot to each force, and carefully check the direction of rotation each force would cause if it acted alone.
画一张清晰的示意图并标出所有距离是至关重要的。标出从支点到每个力的距离，并仔细检查每个力单独作用时会造成什么方向的转动。
Write the principle of moments equation, substitute all values, and solve step by step. Avoid skipping steps—examiners award method marks for correct equations even if the final answer contains a small slip.
写出力矩原理方程，代入所有数值，然后一步步解答。不要跳步——即使最终答案有小小的笔误，只要方程正确，考官也会给步骤分。
10. Common Mistakes and Exam Tips | 常见错误与考试技巧
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Forgetting to use perpendicular distance is the most frequent error. Always double-check whether the force is perpendicular.
最常见的错误是忘记使用垂直距离。一定要反复检查该力是否垂直于距离方向。
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Using wrong signs for clockwise/anticlockwise. Choose a consistent convention, e.g. clockwise positive, and stick to it.
顺时针/逆时针方向符号用错。选择一致的符号规则（例如设顺时针为正）并坚持到底。
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Omitting the weight of the beam when it is uniform. Always include the weight at the centre unless the question says ‘light’ or ‘weightless’.
对于均匀梁，忘记计入梁的自重。除非题目明确说明是“轻质”或“不计重”，否则始终要在中心点加上梁的重量。
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Not converting all distance units to metres. When the moment unit is N m, ensure distances are in m.
没有把所有距离单位转换为米。由于力矩单位为 N m，务必确保距离以米为单位。
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Relying on intuition rather than writing the equation. Always set up ∑ Mcw = ∑ Macw explicitly.
凭直觉做题而不列方程。一定要明确写出 ∑ Mcw = ∑ Macw。
11. Worked Example | 例题解析
Question: A uniform plank of length 4 m and weight 80 N rests on two supports at its ends. A 120 N weight is placed 1 m from the left end. Find the upward reaction forces at the left and right supports.
题目：一根均匀木板长 4 m，重 80 N，两端由支座支撑。一个 120 N 的重物放在距左端 1 m 处。求左右两个支座的向上反力。
Solution: Draw the plank. Weight of plank 80 N acts at 2 m from either end (the centre). Let left support reaction be RL, right support be RR. Take moments about the left support to eliminate RL:
解答：画图。木板自重 80 N 作用在距离两端 2 m 处（中心）。设左支座反力为 RL，右支座为 RR。对左支座取矩以消去 RL：
Clockwise moments = (80 N × 2 m) + (120 N × 1 m)
顺时针力矩 = (80 N × 2 m) + (120 N × 1 m)
Anticlockwise moment = RR × 4 m
逆时针力矩 = RR × 4 m
Set them equal: 160 + 120 = 4RR → 280 = 4RR → RR = 70 N.
令其相等：160 + 120 = 4RR → 280 = 4RR → RR = 70 N。
Then ∑ upward = ∑ downward gives RL + 70 = 80 + 120 → RL = 130 N. Check: total upward = 200 N, total downward = 200 N. Correct.
再利用向上的合力 = 向下的合力，得 RL + 70 = 80 + 120 → RL = 130 N。校验：向上总力 200 N，向下总力 200 N，正确。
12. Summary and Key Formulae | 总结与核心公式
| Key idea / 核心要点 | Formula or method / 公式或方法 |
| Moment / 力矩 | M = F × d (perpendicular / 垂直) |
| Principle of moments / 力矩原理 | ∑ Mcw = ∑ Macw |
| Equilibrium / 平衡 | ∑ F = 0 and ∑ M = 0 |
| Uniform rod weight / 均匀杆自重 | Acts at centre / 作用于中心 |
| Inclined force / 倾斜力 | Use M = F × d × sin θ |
Keep these principles organised, practise with past WJEC papers, and always draw clear, labelled diagrams. With careful method, moments questions become very manageable and are excellent marks to secure.
将这些原理梳理清楚，利用 WJEC 历年真题加以练习，并始终画出清晰、标有注释的示意图。只要方法得当，力矩题目会变得非常容易把握，是拿分的好机会。
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