Moments and Equilibrium | 力矩与平衡

📚 Moments and Equilibrium | 力矩与平衡

In GCSE CCEA Mathematics, the topic of moments and equilibrium brings together practical problem‑solving with the principles of turning forces. You will learn how to calculate a moment, how to decide whether an object will rotate, and how to use the condition for equilibrium to find unknown forces or distances. This revision guide walks you through the key ideas step by step, with clear examples and tips for the exam.

在GCSE CCEA数学中，力矩与平衡这一主题将实际问题求解与转动力的原理结合在一起。你将学会如何计算力矩、判断物体会不会转动，以及如何利用平衡条件来求未知力或距离。这份复习指南带你一步步理清核心概念，配有清晰的例子和考试技巧。

1. What is a Moment? | 什么是力矩？

A moment is the turning effect of a force about a pivot (or fulcrum). 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. Mathematically, the moment M is given by M = F × d, where F is the force in newtons (N) and d is the perpendicular distance in metres (m). The SI unit for a moment is the newton‑metre (Nm).

力矩是力绕支点（或转动中心）产生的转动效果。它取决于两个因素：力的大小，以及从支点到力的作用线的垂直距离。力矩 M 的公式为 M = F × d，其中 F 是力（牛顿，N），d 是垂直距离（米，m）。力矩的国际单位是牛顿·米（Nm）。

2. Perpendicular Distance – The Key | 垂直距离——关键所在

Only the perpendicular distance counts. If a force is applied at an angle, you must either resolve the force to find the component perpendicular to the lever, or calculate the perpendicular distance from the pivot to the line of the force. In most GCSE problems, forces act vertically and levers are horizontal, so the distance is simply the horizontal length along the lever.

只有垂直距离才算数。如果力的方向不是垂直于杆，你就必须分解力，找出垂直于杆的分力，或计算由支点到力作用线的垂直距离。在大多数GCSE题目中，力竖直作用，杆水平，因此垂直距离就是沿着杆的水平长度。

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

Moments can cause rotation in two directions. By convention, a moment that would cause a clockwise turn is called a clockwise moment, and one causing an anticlockwise turn is an anticlockwise moment. In equilibrium problems, we compare the total clockwise moment with the total anticlockwise moment about the same pivot.

力矩可以引起两个方向的转动。习惯上，能使物体顺时针转动的力矩称为顺时针力矩，能使物体逆时针转动的称为逆时针力矩。在平衡问题中，我们把对同一个支点的总顺时针力矩与总逆时针力矩进行比较。

4. The Principle of Moments | 力矩原理

For an object to be in rotational equilibrium (balanced, not turning), the sum of the clockwise moments about any point must equal the sum of the anticlockwise moments about that point. This is the principle of moments. It can be written as: ∑ clockwise moments = ∑ anticlockwise moments. This condition allows you to find one unknown quantity, whether it is a force or a distance.

物体要处于转动平衡（平衡、不转动），绕任何一点的总顺时针力矩必须等于绕该点的总逆时针力矩。这就是力矩原理，可写作：顺时针力矩之和 = 逆时针力矩之和。利用这个条件，你可以求出一个未知量，无论是力还是距离。

5. Worked Example – Seesaw | 例题——跷跷板

A uniform seesaw of length 4 m pivots at its centre. A child of weight 300 N sits 1.2 m to the right of the pivot. Where must a second child of weight 400 N sit on the left to balance the seesaw? Let the required distance be x metres. Anticlockwise moment = 400 × x, clockwise moment = 300 × 1.2. Set them equal: 400x = 360, so x = 0.9 m. The second child sits 0.9 m to the left of the pivot.

一根长4 m的均匀跷跷板在其中点支起。一个重300 N的孩子坐在支点右侧1.2 m处。另一个重400 N的孩子应坐在左侧何处才能使跷跷板平衡？设所求距离为 x 米。逆时针力矩 = 400 × x，顺时针力矩 = 300 × 1.2。令它们相等：400x = 360，得 x = 0.9 m。第二个孩子坐在支点左侧0.9 m处。

6. Dealing with Multiple Forces | 处理多个力

When more than two forces act on a lever, simply calculate each moment and assign a direction (clockwise or anticlockwise). Sum the clockwise moments, sum the anticlockwise moments, and apply the principle of moments. Sometimes you may also need to consider the weight of the lever itself if it is non‑uniform or the pivot is not at the centre of mass.

当杠杆上作用有两个以上的力时，只需分别计算每个力矩并指定方向（顺时针或逆时针）。求出顺时针力矩之和、逆时针力矩之和，然后应用力矩原理。有时如果杠杆不均匀或支点不在质心，还需要考虑杠杆自身的重量。

7. Uniform Beams and the Weight Acting at the Centre | 均匀梁及其重量作用在中心

For a uniform beam, the entire weight can be considered to act at its centre. This is a crucial simplification. For example, a uniform metre rule weighing 1 N pivoted at its centre has its weight acting at the pivot, producing zero moment about the pivot. If the pivot is moved to 30 cm from one end, the weight still acts at the 50 cm mark, so the moment about the pivot can be calculated.

对于均匀的梁，其全部重量可以认为作用在它的中心。这是一个关键简化。例如，一根重1 N的均匀米尺在其中点支起，其重量作用在支点处，对支点的力矩为零。如果支点移到距一端30 cm处，重量仍作用在50 cm刻度处，则可以对支点计算该重量的力矩。

8. Reaction Forces and Equilibrium | 支反力与平衡

In many questions, a beam is supported at two points. As well as rotational equilibrium, you must consider vertical equilibrium: the sum of upward forces equals the sum of downward forces. You can take moments about one support to find the reaction at the other support, then use vertical equilibrium to find the remaining unknown. Choose the pivot point that eliminates one unknown to make the algebra simpler.

在许多题目中，梁由两点支撑。除了转动平衡外，还必须考虑竖直方向力的平衡：向上的力之和等于向下的力之和。你可以对其中一个支点取矩，求出另一个支点的支反力，然后利用竖直方向力平衡求出剩下的未知量。选择能消去一个未知力的点作为取矩中心，能够简化计算。

9. Moments When Forces Are Not Perpendicular | 力不垂直时的力矩

If a force acts at an angle, you must use the perpendicular distance from the pivot to the line of action. This distance is d sin θ, where θ is the angle between the lever and the line of action, or you can resolve the force into components. In GCSE, such situations often involve a string or cable pulling a beam at an angle. Always draw a clear diagram and label the right‑angled triangle to find the perpendicular distance accurately.

如果力以某一角度作用，你必须使用支点到力作用线的垂直距离。这个距离等于 d sin θ，其中 θ 是杆与力作用线之间的夹角；你也可以将力分解为分力。在GCSE中，这种情况常涉及绳子或缆绳以一定角度拉着一根梁。务必画出清晰的示意图，并标出直角三角形，以便准确求出垂直距离。

10. Exam Technique and Common Pitfalls | 考试技巧与常见错误

Always state the principle of moments clearly in your working. Write the equation: sum of clockwise moments = sum of anticlockwise moments. Show substitution of numbers, and include units throughout. Common mistakes include forgetting to convert cm to m, mixing up clockwise and anticlockwise directions, and using the wrong distance when forces are not perpendicular. Double‑check your diagram before calculating.

在解题过程中，始终要清楚地写出力矩原理：顺时针力矩之和 = 逆时针力矩之和。展示代入数字的过程，并自始至终带上单位。常见错误包括忘记把厘米转换为米，混淆顺时针与逆时针方向，以及当力不垂直时使用了错误的距离。在计算前，先仔细检查你的示意图。

11. Practice Problem with Scaffolding | 框架性练习题

A uniform plank of length 5 m and weight 80 N rests on two trestles, each 1 m from an end. A load of 120 N is placed 1.5 m from the left end. By taking moments about the left trestle, calculate the upward reaction force at the right trestle. (Hint: identify all forces, their distances from the left trestle, and apply the principle of moments. Then use vertical equilibrium to find the other reaction.)

一根长5 m、重80 N的均匀木板搁在两个支架上，每个支架距一端1 m。一个120 N的重物放在距左端1.5 m处。通过绕左支架取矩，计算右支架向上的支反力。（提示：找出所有力及其到左支架的距离，应用力矩原理。然后利用竖直方向力平衡求出另一个支反力。）

12. Linking Moments to Real‑Life Contexts | 将力矩与实际生活联系起来

The principle of moments explains why a longer spanner makes it easier to undo a tight nut, why a crane’s counterweight is placed far from the pivot, and why doors have handles far from the hinges. Recognising these applications helps you understand why moments matter and can help you set up problems correctly in exams.

力矩原理能解释为什么长扳手更容易拧松紧螺母，为什么起重机的配重放在离转动中心很远的地方，以及为什么门把手总是装在远离铰链的位置。认识这些应用有助于理解力矩的重要性，也能帮助你在考试中正确地建立问题模型。

Published by TutorHao | GCSE CCEA Mathematics Revision Series | aleveler.com

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