📚 The Turning Effect of a Force | 力的转动效应
In everyday life, forces do not only push or pull objects in straight lines; they can also make objects rotate. The turning effect of a force, called its moment, is a core concept in CIE A Level Physics because it explains how levers, beams, doors and bridges remain balanced or topple.
在日常生活中，力不仅能沿直线推拉物体，还能使物体转动。力的转动效应——即力矩——是 CIE A Level 物理的核心概念，因为它解释了杠杆、横梁、门和桥梁如何保持平衡或发生倾倒。
1. Defining Moment of a Force | 力矩的定义
The moment of a force about a pivot is defined as the product of the force and the perpendicular distance from the pivot to the line of action of the force.
力对某转轴的力矩定义为力与从转轴到力作用线的垂直距离的乘积。
M = F × d
Here, M is the moment, F is the applied force, and d is the perpendicular distance from the pivot to the line of action. The SI unit of moment is the newton metre (N m).
其中 M 是力矩，F 是施加的力，d 是从转轴到力作用线的垂直距离。力矩的国际单位是牛·米（N m）。
It is essential that the distance d is measured at 90° to the force direction, not simply the distance along the object.
关键在于距离 d 必须沿与力方向成 90° 的方向测量，而不是简单沿物体长度测量。
2. Calculating Moments | 力矩的计算
When the force is not perpendicular to the lever arm, the moment is calculated using the perpendicular component of the force or the perpendicular distance from the pivot to the line of action.
当力不与杠杆臂垂直时，需使用力的垂直分量或从转轴到力作用线的垂直距离来计算力矩。
M = F d sin θ
In this expression, θ is the angle between the force vector and the lever arm. For perpendicular forces, θ = 90° and sin 90° = 1, so the formula reduces to M = F × d.
式中 θ 是力矢量与杠杆臂之间的夹角。对于垂直力，θ = 90° 且 sin 90° = 1，因此公式简化为 M = F × d。
Worked examples often require resolving a force into components or extending the line of action to find the perpendicular distance.
典型例题通常需要分解力，或延长力作用线以求出垂直距离。
3. Principle of Moments | 力矩原理
For an object in rotational equilibrium, the sum of the clockwise moments about any pivot equals the sum of the anticlockwise moments about that same pivot.
对于处于转动平衡的物体，关于任意转轴，顺时针力矩之和等于逆时针力矩之和。
This statement is known as the principle of moments. It is extremely useful for calculating unknown forces or distances in balanced beams, bridges and levers.
这一表述称为力矩原理。它在计算平衡横梁、桥梁和杠杆中的未知力或距离时非常有用。
Σ clockwise moments = Σ anticlockwise moments
When applying this principle, always choose a clear pivot and take moments about that point. If a force acts through the pivot, its moment is zero because its perpendicular distance is zero.
应用该原理时，应始终明确选择转轴并对其取矩。如果某个力通过转轴，其力矩为零，因为垂直距离为零。
4. Couples and Torque | 力偶与转矩
A couple consists of two equal and opposite parallel forces whose lines of action do not coincide. A couple produces rotation without any resultant linear force.
力偶由两个大小相等、方向相反的平行力组成，且它们的作用线不重合。力偶产生转动，但不产生净线力。
The torque of a couple is the product of one of the forces and the perpendicular distance between their lines of action.
力偶的转矩等于其中一个力与两力作用线之间垂直距离的乘积。
T = F × d
Here, d is the perpendicular separation between the two forces, not the distance from a pivot. A couple has the same turning effect about any point in the plane.
这里 d 是两个力之间的垂直间距，而不是到某个转轴的距离。力偶在平面内任意点的转动效应相同。
5. Conditions for Equilibrium | 平衡条件
A body is in complete equilibrium when two independent conditions are satisfied simultaneously.
当物体同时满足两个独立条件时，才处于完全平衡状态。
First, the resultant force acting on the body must be zero in all directions. This prevents translational acceleration.
第一，作用在物体上的合力必须在所有方向上为零。这避免了平动加速度。
Second, the resultant moment about any point must be zero. This prevents rotational acceleration.
第二，关于任意点的合力矩必须为零。这避免了转动加速度。
ΣF = 0 and ΣM = 0
In CIE examination problems, you often need to combine resolving forces with taking moments to find unknown reaction forces on a supported beam.
在 CIE 考试题中，通常需要结合力的分解与取矩，来求解支承梁上的未知反作用力。
6. Centre of Gravity | 重心
The centre of gravity of an object is the point through which the entire weight of the object appears to act, regardless of the object’s orientation.
物体的重心是无论物体朝向如何，其全部重量似乎都通过该点作用的点。
For a uniform regular solid, the centre of gravity lies at its geometrical centre. For irregular objects, it can be found experimentally by suspending the object from different points.
对于均匀规则物体，重心位于其几何中心。对于不规则物体，可以通过从不同点悬挂物体的实验方法确定重心。
When calculating moments due to weight, the weight can be represented as a single downward force acting at the centre of gravity.
在计算由重力产生的力矩时，可以将重量表示为作用在重心处的单个向下力。
7. Stability and Toppling | 稳定性与倾倒
An object is stable if a small displacement produces a restoring moment that returns it to its original position.
如果物体轻微位移后会产生使其回到原位的恢复力矩，则该物体是稳定的。
An object topples when the vertical line through its centre of gravity falls outside its base of support. At the point of toppling, the line of action of the weight passes through the edge of the base.
当通过重心的竖直线落在支撑面之外时，物体会倾倒。在倾倒瞬间，重力作用线正好通过支撑面边缘。
Objects with a low centre of gravity and a wide base are more stable because they can be tilted through a larger angle before toppling.
重心低、底面宽的物体更稳定，因为在倾倒前可以倾斜更大的角度。
This principle explains why racing cars are low and wide, and why tall buses require careful loading to keep their centre of gravity low.
这一原理解释了为什么赛车低而宽，以及为什么高公交车需要小心装载以保持重心较低。
8. Moments in Everyday Structures | 日常结构中的力矩
The turning effect of a force is used in levers, spanners, wheelbarrows, cranes and bridges to amplify forces or to maintain balance.
力的转动效应被应用于杠杆、扳手、手推车、起重机和桥梁中，以放大力量或保持平衡。
For example, a longer spanner increases the perpendicular distance d, so a smaller force can produce the same moment needed to loosen a tight bolt.
例如，较长的扳手增大了垂直距离 d，因此较小的力就能产生拧松紧螺栓所需的相同力矩。
In a crane, the counterweight is positioned to produce an anticlockwise moment that balances the clockwise moment of the lifted load, preventing the crane from tipping.
在起重机中，配重的位置产生逆时针力矩，以平衡所吊重物的顺时针力矩，从而防止起重机翻倒。
Understanding moments allows engineers to calculate safe loads and choose appropriate supports for structures.
理解力矩使工程师能够计算安全载荷，并为结构选择合适的支撑。
9. Experimental Determination of Unknown Weight | 利用力矩原理测量未知重量
A classic CIE practical task involves balancing a uniform metre rule at its centre of gravity, then using known weights and the principle of moments to determine an unknown mass.
一个典型的 CIE 实验任务是先将均匀米尺在其重心处支起，然后利用已知砝码和力矩原理测定未知质量。
Place a known mass m₁ at a measured distance d₁ on one side of the pivot, and suspend the unknown mass m₂ at a distance d₂ on the other side until the rule is horizontal and balanced.
在转轴一侧距其 d₁ 处放已知质量 m₁，另一侧距其 d₂ 处悬挂未知质量 m₂，直至米尺水平且平衡。
m₁ × g × d₁ = m₂ × g × d₂
Since g cancels, the unknown mass is given by m₂ = m₁ × d₁ ÷ d₂. Repeating for several distances improves accuracy.
由于 g 可约去，未知质量为 m₂ = m₁ × d₁ ÷ d₂。在多个距离下重复实验可提高准确性。
This experiment also highlights why the ruler must be horizontal: it ensures the distances are perpendicular to the forces.
该实验还说明了为什么米尺必须水平：它确保了距离与力垂直。
10. Common Misconceptions and Exam Tips | 常见误区与备考提示
A common error is using the distance from the pivot to the point of application of the force, even when the force is not perpendicular to the object.
一个常见错误是在力不垂直于物体时，仍使用从转轴到力的作用点的距离。
Always draw the line of action of the force and measure the perpendicular distance from the pivot to that line. If the force is at an angle, use the perpendicular component of the force.
务必画出力的作用线，并测量从转轴到该作用线的垂直距离。如果力有角度，则使用力的垂直分量。
Another common mistake is forgetting that a force through the pivot has no moment. Choosing the pivot at the point where an unknown force acts often eliminates that unknown from the moment equation.
另一个常见错误是忘记通过转轴的力不产生力矩。将转轴选在未知力作用点处，通常可以在力矩方程中消去该未知量。
In equilibrium problems, write down both the force balance and the moment balance. Marks are awarded for a clear pivot, correct distances, and a statement of the principle of moments.
在平衡问题中，应同时列出力的平衡和力矩平衡。清晰的转轴、正确的距离以及对力矩原理的表述都能得分。
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