📚 The Turning Effect of a Force | 力的转动效应
In physics, forces do not only push or pull objects along a straight line; they can also cause objects to rotate. The turning effect produced by a force about a pivot is called the moment of a force, or simply torque. Mastering the concept of moments is essential for understanding the equilibrium of extended bodies, analysing structures such as bridges and cranes, and explaining how simple machines like levers and spanners multiply our effort. This article covers the key ideas required for CIE A-Level Physics, from the definition and calculation of moments to the principle of moments, couples, centres of gravity, and real-world applications.
在物理学中，力不仅能沿直线推拉物体，还能使物体转动。力绕某一支点产生的转动效应称为力矩（moment of a force），有时也叫转矩。掌握力矩的概念对于理解延伸物体的平衡、分析桥梁与起重机等结构、以及解释杠杆和扳手等简单机械如何放大作用力至关重要。本文涵盖 CIE A-Level 物理需要掌握的核心内容，从力矩的定义与计算，到力矩原理、力偶、重心以及实际应用。
1. Defining the Moment of a Force | 力矩的定义
The moment of a force about a point (often called a pivot or fulcrum) is a measure of its ability to cause rotation about that point. It is defined as the product of the force and the perpendicular distance from the pivot to the line of action of the force.
力对某一点（常称为支点或 pivot）的力矩，是衡量该力使物体绕该点转动能力的物理量。它定义为力的大小与从支点到力作用线的垂直距离的乘积。
In equation form, if a force F acts at a distance d from the pivot, where d is measured perpendicular to the force's line of action, the magnitude of the moment M is given by:
用公式表示，若力 F 作用于距支点垂直距离为 d 的位置，力矩大小 M 由下式给出：
M = F × d
The perpendicular distance is the shortest distance from the pivot to the line along which the force acts. If the force is not applied at right angles to the lever arm, only the perpendicular component of the force contributes to the moment.
垂直距离是指从支点到力作用线的最短距离。如果力不是垂直于力臂施加，则只有力的垂直分量才会产生力矩。
2. Calculating Moments in Different Situations | 不同情况下力矩的计算
When a force is applied at an angle, the moment can be expressed using trigonometry. For a force F acting at a distance r from the pivot, where the angle between the force and the lever arm is θ, the perpendicular distance is d = r sin θ. The moment then becomes:
当力以一定角度施加时，可用三角函数表示力矩。对于距支点为 r 处、与力臂夹角为 θ 的力 F，垂直距离为 d = r sin θ。此时力矩为：
M = F × r × sin θ
In many practical problems, it is easier to resolve the force into components parallel and perpendicular to the lever arm and then use only the perpendicular component for the moment calculation.
在许多实际问题中，将力分解为平行于力臂和垂直于力臂的两个分量，并只用垂直分量来计算力矩，往往更为简便。
The direction of the moment (clockwise or anticlockwise) must also be considered when applying the principle of moments, but the magnitude is always given by the product above.
在应用力矩原理时，还需要考虑力矩的方向（顺时针或逆时针），但其大小始终由上述乘积给出。
3. Units of Moment | 力矩的单位
The SI unit of moment is the newton metre, written as N m ugh this combination is dimensionally equivalent to the joule (J), moment is not a form of energy and is never expressed in joules. The distinction is important: moment is a vector quantity associated with rotation, whereas energy is a scalar.
力矩的国际单位是牛顿米，写作 N m。由于力矩是力与距离的乘积，其单位是 N × m。尽管这一组合在量纲上与焦耳（J）相同，但力矩并非能量的一种形式，因此绝不能以焦耳表示。这一区别至关重要：力矩是与转动相关的矢量，而能量是标量。
In written work, always use 'N m' (with a space) or 'Nm' according to the convention required by your examination board. CIE typically accepts 'N m' or 'Nm'.
在书写时，根据考试局要求使用 'N m'（中间有空格）或 'Nm'。CIE 通常接受 'N m' 或 'Nm'。
4. Clockwise and Anticlockwise Moments | 顺时针力矩与逆时针力矩
Moments can cause rotation in two directions: clockwise (CW) and anticlockwise (ACW). By convention, we often assign a positive sign to anticlockwise moments and a negative sign to clockwise moments, or vice versa. The key idea is that for an object to be in rotational equilibrium, the total clockwise moment must exactly balance the total anticlockwise moment.
力矩可以引起两个方向的转动：顺时针（CW）和逆时针（ACW）。我们通常人为规定逆时针力矩为正、顺时针力矩为负，或者相反。核心思想是，要使物体处于转动平衡，顺时针总力矩必须恰好与逆时针总力矩平衡。
When solving problems, it is essential to clearly state which direction you take as positive and to consistently apply that sign convention when summing moments.
在解题时，必须明确说明取哪个方向为正，并在求和力矩时始终贯彻这一符号约定。
5. Principle of Moments | 力矩原理
The principle of moments states that for a system to be in rotational equilibrium about a given point, the sum of the clockwise moments about that point must equal the sum of the anticlockwise moments about the same point. This can be written as:
力矩原理指出，要使系统对某一点处于转动平衡，则对该点的顺时针力矩之和必须等于逆时针力矩之和。这可以写作：
∑ Mclockwise = ∑ Manticlockwise
This principle is a direct consequence of Newton' ing on it. It is the rotational analogue of the equilibrium condition for forces (∑F = 0).
该原理是牛顿第一定律应用于转动的直接结果：一个静止或匀速转动的物体，所受的外力矩之和为零。它是力的平衡条件（∑F = 0）在转动中的类比。
The principle is extremely powerful because it allows us to find unknown forces or distances in static systems without needing to consider all the internal forces.
该原理非常强大，因为它可以让我们在静态系统中求解未知的力或距离，而不需要考虑所有的内力。
6. Conditions for Complete Equilibrium | 完整平衡的条件
For an extended rigid body to be in complete static equilibrium, two conditions must be satisfied simultaneously:
要使一个延伸刚体处于完整的静态平衡，必须同时满足两个条件：
The net force in any direction is zero (translational equilibrium).
任意方向上的合力为零（平动平衡）。
The net moment about any point is zero (rotational equilibrium).
对任意点的合力矩为零（转动平衡）。
In most A-Level problems, you are required to apply both conditions. Typically, the force condition is used to find unknown reaction forces after the moment condition has given one unknown.
在大多数 A-Level 问题中，你需要同时应用这两个条件。通常，先用力矩条件求出一个未知量，再通过力的条件求出其他未知反力。
When choosing the point about which to take moments, it is wise to select a point where one or more unknown forces act, so that those forces have zero moment about that point and are eliminated from the moment equation.
在选择取矩点时，明智的做法是选一个或多个未知力通过的点，这样这些力对该点的力矩为零，从而在力矩方程中被消去。
7. Centre of Gravity and Stability | 重心与稳定性
The centre of gravity (C.G.) of a body is the point through which the entire weight of the body appears to act, regardless of its orientation. For a uniform symmetrical object, the centre of gravity lies at its geometrical centre. For irregular shapes, it can be found experimentally.
物体的重心（C.G.）是指无论物体朝向如何，其全部重量看似都作用于此的那一点。对于均匀对称的物体，重心位于其几何中心。对于不规则形状，可通过实验确定。
The position of the centre of gravity is crucial for understanding stability. If the vertical line through the centre of gravity falls within the base of an object, the object is stable; if it falls outside the base, the object will topple.
重心的位置对于理解稳定性至关重要。如果通过重心的铅垂线落在物体支撑面之内，物体是稳定的；若落在支撑面之外，物体将倾倒。
When calculating moments due to weight, we can treat the entire weight as acting at the centre of gravity, which greatly simplifies calculations for extended bodies.
在计算由重力产生的力矩时，我们可以认为整个重量都作用在重心上，这极大地简化了延伸物体的计算。
8. Couples: A Special Pair of Forces | 力偶：一对特殊的力
A couple consists of two equal and opposite parallel forces whose lines of action do not coincide. The turning effect of a couple is called a torque. The magnitude of the torque produced by a couple is given by the product of one of the forces and the perpendicular distance between their lines of action:
力偶由大小相等、方向相反、作用线不重合的两个平行力组成。力偶产生的转动效应称为转矩。力偶产生的转矩大小等于其中一个力与两力作用线之间垂直距离的乘积：
Torque of a couple = F × d
where d is the perpendicular separation between the forces. A couple produces pure rotation without any net translational force, which makes it fundamentally different from a single force.
其中 d 是两力之间的垂直距离。力偶产生纯转动而无任何净平动力，这与单个力有本质区别。
Examples of couples include turning a steering wheel, using a tap, or the forces acting on a magnetic dipole in a uniform magnetic field. In A-Level physics, couples often appear when analysing rotating rigid bodies.
力偶的例子包括转动方向盘、拧水龙头，或均匀磁场中作用在磁偶极子上的力。在 A-Level 物理中，分析转动刚体时常会出现力偶。
9. Levers and Mechanical Advantage | 杠杆与机械利益
A lever is a simple machine that uses the principle of moments to amplify an applied force. It consists of a rigid bar pivoted at a fixed point (the fulcrum). By applying an effort force at one point, a larger load force can be overcome at another point.
杠杆是利用力矩原理来放大作用力的简单机械。它由一根绕固定支点转动的刚性杆构成。通过在一点施加动力（effort），可以在另一点克服更大的负载阻力（load）。
istance from fulcrum to effort = load × distance from fulcrum to load. Hence, by making the effort arm longer than the load arm, a small effort can lift a heavy load.
杠杆的机械利益（MA）是负载与动力的比值。对于无摩擦的理想杠杆，由力矩原理可得：动力 × 动力臂 = 负载 × 负载臂。因此，通过使动力臂长于负载臂，就可以用小动力举起重负载。
Three classes of levers exist, depending on the relative positions of fulcrum, effort, and load. In a Class 1 lever (e.g. crowbar, seesaw), the fulcrum is between effort and load. In Class 2 (e.g. wheelbarrow), the load is between fulcrum and effort. In Class 3 (e.g. fishing rod), the effort is between fulcrum and load.
根据支点、动力与负载的相对位置，杠杆可分为三类。第一类杠杆（如撬棍、跷跷板），支点在动力与负载之间。第二类杠杆（如手推车），负载在支点与动力之间。第三类杠杆（如钓鱼竿），动力在支点与负载之间。
10. Experimental Determination of Centre of Gravity | 实验测定重心
The centre of gravity of an irregularly shaped lamina can be found using a simple suspension method. The lamina is freely suspended from a point near its edge. When it comes to rest, a plumb line is used to mark the vertical line through the point of suspension. The procedure is repeated with a different suspension point. The intersection of the two marked lines gives the centre of gravity.
不规则薄板的重心可通过简单的悬挂法求得。将薄板从靠近边缘的一点自由悬挂，待其静止后，用铅垂线画出通过悬挂点的竖直线。再选择另一个悬挂点重复这一步骤。两条标明线的交点即为重心。
Alternatively, if the object has a symmetrical shape or if it can be balanced on a pivot, the centre of gravity lies directly above the pivot point when balanced. This principle is often used in laboratory experiments to verify the position of the C.G.
另外，如果物体具有对称形状，或者能在一个支点上平衡，那么平衡时重心就位于支点正上方。这一原理常用于实验室中验证重心的位置。
11. Solving Problems with Beams and Supports | 求解梁与支点问题
A typical A-Level problem involves a uniform beam supported at one or two points, with additional loads placed along its length. To find reaction forces at the supports, you apply the conditions for equilibrium: upward forces equal downward forces, and clockwise moments equal anticlockwise moments about a chosen pivot.
典型的 A-Level 问题涉及一根均匀梁，在一处或两处支撑，并沿长度方向加上额外荷载。为求出支点反力，需要应用平衡条件：向上的力等于向下的力，且对所选支点的顺时针力矩等于逆时针力矩。
For example, a uniform beam of length L and weight W is supported at its ends. A load P is placed at a distance x from one end. Taking moments about one support eliminates the reaction at that support from the equation, allowing the other reaction to be found directly. Then vertical equilibrium gives the remaining reaction.
例如，一根长为 L、重为 W 的均匀梁在两端支撑，距一端 x 处放置负载 P。对该端支点取矩，方程中将消去该支点反力，从而直接求出另一端反力。再由竖直方向力的平衡求出剩下的那个支点反力。
Remember to include the weight of the beam itself, acting at its centre, unless the beam is described as 'light' (negligible weight).
务必包含梁自身的重量，作用在其中心，除非题目说明该梁为“轻质”梁（重量可忽略）。
12. Everyday Applications of the Turning Effect | 转动效应在日常生活中的应用
Moments are everywhere in daily life. A spanner provides a larger turning effect by increasing the perpendicular distance from the nut, which is why long-handled spanners are easier to use. Door handles are placed far from the hinge to maximise the moment produced by a given push, making the door easier to open.
力矩在日常生活中随处可见。扳手通过增大与螺母的垂直距离来提供更大的转动效果，这就是长柄扳手更省力的原因。门把手安装在远离铰链的位置，以最大化同一推力所产生的力矩，使开门更加轻松。
Seesaws (teeter-totters) are a classic demonstration of the principle of moments: a heavier child must sit closer to the pivot to balance a lighter child further away. Cranes use counterweights to keep the centre of gravity over the base, preventing toppling. Even the human forearm acts as a Class 3 lever, with the biceps providing the effort close to the elbow joint.
跷跷板是力矩原理的经典演示：较重的孩子必须坐得离支点更近，才能与坐在较远处的较轻孩子平衡。起重机利用配重使重心保持在基座上方，以防倾倒。就连人类的前臂也充当着一个第三类杠杆，肱二头肌在靠近肘关节处提供动力。
Understanding the turning effect of force not only helps us solve textbook problems but also allows engineers to design safer buildings, stable vehicles, and more efficient machines.
理解力的转动效应不仅有助于我们解决课本问题，还能让工程师设计出更安全的建筑、更稳定的车辆以及更高效的机械。
Published by TutorHao | Physics Revision Series | aleveler.com
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