Moments and Equilibrium in GCSE CIE Mathematics | GCSE CIE 数学：力矩与平衡考点精讲

📚 Moments and Equilibrium in GCSE CIE Mathematics | GCSE CIE 数学：力矩与平衡考点精讲

Moments describe the turning effect of a force about a pivot. In GCSE CIE Mathematics, you must be able to calculate moments, apply the principle of moments, and solve problems involving beams, rods, and equilibrium. Understanding these concepts will also support your physics studies. This revision guide covers key definitions, formulas, and exam-style worked examples to help you master moments and equilibrium.

力矩描述力对支点的转动效应。在GCSE CIE数学中，你需要会计算力矩、应用力矩原理，并解决涉及梁、杆和平衡的问题。掌握这些概念也将为物理学习提供支持。本复习指南覆盖关键定义、公式和考试风格的例题，助你掌握力矩与平衡。

1. Definition of a Moment | 力矩的定义

A moment is the turning effect of a force about a point, called the pivot or fulcrum. The size of a moment depends on two factors: the magnitude of the force and the perpendicular distance from the pivot to the line of action of the force. Moment (M) = Force (F) × Perpendicular distance (d). The SI unit is newton-metre (N m). Moments can be clockwise or anticlockwise.

力矩是力绕某一点（称为支点或转轴）产生的转动效应。力矩的大小取决于两个因素：力的大小，以及支点到力的作用线的垂直距离。力矩 (M) = 力 (F) × 垂直距离 (d)。国际单位是牛顿·米 (N·m)。力矩可以是顺时针或逆时针方向。

2. Calculating Moments | 力矩的计算

To calculate the moment, you must use the perpendicular distance. If the force is not perpendicular to the lever, you need to resolve it or use the perpendicular component. For a force F acting at an angle θ to the lever, moment = F × d × sin θ, where d is the distance from pivot to point of application. However, at GCSE level, most forces are perpendicular, so moment = F × d. Always state the direction (clockwise or anticlockwise).

计算力矩必须使用垂直距离。如果力不与杆垂直，你需要分解力或使用垂直分量。对于与杆成θ角的力F，力矩 = F × d × sin θ，其中d是支点到作用点的距离。但在GCSE层面，大部分力是垂直的，因此力矩 = F × d。始终标明方向（顺时针或逆时针）。

Moment = Force × Perpendicular distance from pivot

力矩 = 力 × 支点到力作用线的垂直距离

3. The Principle of Moments | 力矩原理

For an object in equilibrium (not turning), the sum of clockwise moments about any pivot equals the sum of anticlockwise moments about that same pivot. This is known as the principle of moments. Mathematically: Σ clockwise moments = Σ anticlockwise moments. You can choose the pivot point arbitrarily to simplify calculations.

对于处于平衡状态（不发生转动）的物体，绕任意支点的顺时针力矩之和等于逆时针力矩之和。这就是力矩原理。数学表达为：总顺时针力矩 = 总逆时针力矩。你可以任意选择支点位置以简化计算。

4. Conditions for Equilibrium | 平衡条件

For a rigid body to be in static equilibrium, two conditions must be met: 1) The resultant force in any direction is zero (translational equilibrium). 2) The resultant moment about any point is zero (rotational equilibrium). In many GCSE problems, forces act vertically and the only turning effects are from moments; you mainly apply Σ clockwise moments = Σ anticlockwise moments. Also consider upward forces = downward forces to solve for unknown reactions.

刚体要保持静力平衡，必须满足两个条件：1) 任意方向的合力为零（平动平衡）。2) 对任意点的合力矩为零（转动平衡）。在许多GCSE问题中，力是垂直作用的，唯一的转动效应来自力矩；主要应用总顺时针力矩 = 总逆时针力矩。同时还要考虑向上力 = 向下力来求解未知反力。

5. Uniform Beams and Rods | 均匀梁与均匀杆

A uniform beam has its weight acting at its centre. When solving problems, model the weight as a single force acting at the midpoint. For a uniform rod of length L, the weight acts at distance L/2 from either end. This simplifies moment calculations. For example, a uniform beam of weight W resting on a pivot at its centre is already balanced; with additional loads, use the principle of moments.

均匀梁的重力作用于其中心。解题时，将重量视为作用于中点的一个集中力。长度为L的均匀杆，重量作用在距两端L/2处。这简化了力矩计算。例如，一根重量为W的均匀梁支在中心，已经平衡；若有额外负载，则用力矩原理。

6. Non-Uniform Rods and Centre of Mass | 非均匀杆与质心

A non-uniform rod’s weight does not act at the geometrical centre. You may need to find the centre of mass using the principle of moments. For instance, balance the rod on a pivot, or suspend it to locate the centre of mass. In exam questions, sometimes the distance of the centre of mass from one end is given, and you calculate unknown forces. Remember: weight always acts vertically downwards through the centre of mass.

非均匀杆的重量并不作用在几何中心。可能需要用力矩原理找到质心位置。例如，将杆支起平衡或悬挂以确定质心。在考题中，有时会给出质心距一端的距离，要求计算未知力。记住：重力总是竖直向下通过质心作用。

7. Resolving Forces Not Perpendicular to the Beam | 分解不垂直于梁的力

If a force acts at an angle, only its perpendicular component produces a moment about the pivot. To find the moment, you can either find the perpendicular distance from the pivot to the line of action, or resolve the force into perpendicular and parallel components. Use trigonometry. For a force F at angle θ to the beam, the perpendicular component is F sin θ (if θ is between force and beam). Moment = F sin θ × distance along the beam from pivot.

如果力以一定角度作用，只有其垂直分量才会对支点产生力矩。可以求出支点到力作用线的垂直距离，或将力分解为垂直和平行分量。使用三角学。若力F与梁成θ角，垂直分量为F sin θ（θ为力与梁的夹角）。力矩 = F sin θ × 沿梁从支点起的距离。

8. Step-by-Step Approach to Equilibrium Problems | 平衡问题的分步解法

1. Draw a clear diagram showing all forces and their directions. 2. Choose a pivot. Often it’s convenient to choose the point where an unknown force acts, so that force has zero moment. 3. Mark perpendicular distances from pivot to each force. 4. Identify clockwise and anticlockwise moments. 5. Write the equation: Σ clockwise moments = Σ anticlockwise moments. 6. Solve for the unknown. Also apply Σ upward forces = Σ downward forces if needed. 7. Check units and direction.

1. 画清楚图示，标出所有力及其方向。2. 选择一个支点。通常选择未知力作用点作为支点，这样该力的力矩为零，简化计算。3. 标出支点到每个力的垂直距离。4. 区分顺时针和逆时针力矩。5. 列方程：总顺时针力矩 = 总逆时针力矩。6. 求解未知量。必要时再应用向上合力 = 向下合力。7. 检查单位和方向。

9. Common Types of Exam Questions | 常见考试题型

Typical GCSE CIE questions include: a) A uniform beam supported at its centre with loads placed on either side. b) A beam pivoted at one end with a load and a support. c) A non-uniform rod suspended by strings; find tension or centre of mass. d) A person standing on a plank supported by two trestles; find reaction forces. e) A seesaw problem with children of different weights. Always apply the principle of moments and force balance.

典型的GCSE CIE考题包括：a) 均匀梁支在中心，两侧加负载。b) 梁一端支起，另端有负载和支撑。c) 非均匀杆用细绳悬挂；求张力或质心。d) 人站在由两个支架支撑的木板上；求支撑反力。e) 不同体重儿童玩跷跷板问题。始终应用力矩原理和力的平衡。

10. Worked Example: Balanced See-saw | 例题：平衡的跷跷板

A uniform see-saw of weight 200 N is 4 m long, pivoted at its centre. A child of weight 300 N sits 1.5 m to the left of the pivot. Where must a child of weight 400 N sit on the right to balance it? Solution: The see-saw’s weight acts at centre, no moment about pivot. Take clockwise as the moment of right-side forces. Left child: 300 N × 1.5 m = 450 Nm anticlockwise. For equilibrium, clockwise moment must equal 450 Nm. So 400 N × d = 450, thus d = 450/400 = 1.125 m. The second child must sit 1.125 m to the right of pivot.

一个均匀跷跷板重200 N，长4 m，支在中心。一个重300 N的小孩坐在支点左侧1.5 m处。一个重400 N的小孩需坐在右侧何处才能平衡？解：跷跷板自重作用于中心，对支点不产生力矩。取右侧力的力矩为顺时针。左侧小孩：300 N × 1.5 m = 450 Nm 逆时针。平衡时顺时针力矩须等于450 Nm。故400 N × d = 450，d = 450÷400 = 1.125 m。第二个小孩须坐在支点右侧1.125 m处。

11. Worked Example: Non-uniform Rod Suspended | 例题：悬挂的非均匀杆

A non-uniform rod AB of length 2 m and weight 50 N hangs horizontally by two vertical strings attached at A and B. The string at A has tension 30 N. Find the tension in the string at B and the distance of the centre of mass from A. Solution: Vertical equilibrium: T_A + T_B = 50, so 30 + T_B = 50 ⇒ T_B = 20 N. Take moments about A: Weight (50 N) acts at unknown distance x from A, creating clockwise moment (say). T_B creates anticlockwise moment about A: 20 N × 2 m = 40 Nm anticlockwise. 50 N × x clockwise. 50x = 40, x = 0.8 m. Thus centre of mass is 0.8 m from A.

一根非均匀杆AB长2 m，重50 N，通过系于A和B的两根竖直细绳水平悬挂。A处绳的张力为30 N。求B处绳的张力和质心距A的距离。解：竖直方向平衡：T_A + T_B = 50，30 + T_B = 50 ⇒ T_B = 20 N。对A取矩：重力(50 N)作用于距A为x处，产生顺时针力矩。T_B对A产生逆时针力矩：20 N × 2 m = 40 Nm 逆时针。50 N × x 顺时针。50x = 40，x = 0.8 m。质心距A 0.8 m。

12. Tips and Common Mistakes | 技巧与常见错误

Always use perpendicular distance; do not use the slanted distance unless the force is perpendicular to that distance.
Choose the pivot wisely to eliminate unknown forces from the moment equation.
Remember that the weight of a uniform beam acts at its centre.
Check the direction of moments consistently (cw and acw).
For non-uniform objects, the centre of mass is not at the midpoint.
If a force passes through the pivot, its moment is zero.
Convert all units to metres and newtons before calculating.

始终使用垂直距离；不要使用斜向距离，除非力与该距离垂直。
巧妙选择支点，以消除力矩方程中的未知力。
记住均匀梁的重量作用于中心。
前后一致地检查力矩方向（顺时针和逆时针）。
对于非均匀物体，质心不在中点。
如果力的作用线通过支点，其力矩为零。
计算前将所有单位转换为米和牛顿。

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Moments and Equilibrium in GCSE CIE Mathematics | GCSE CIE 数学：力矩与平衡 考点精讲