📚 Centre of Gravity and Equilibrium Analysis | 重心的概念与平衡分析
The concept of the centre of gravity (CG) is fundamental to understanding how objects behave under the influence of gravity and how they maintain or lose balance. In this article, we will explore the definition of the centre of gravity, methods for determining it, and its crucial role in equilibrium analysis — a key topic in the CIE A-Level Physics syllabus.
重心的概念是理解物体在重力作用下如何表现以及如何保持或失去平衡的基础。本文将深入探讨重心的定义、确定方法及其在平衡分析中的关键作用——这是 CIE A-Level 物理课程大纲中的一个重要考点。
1. Definition of Centre of Gravity | 重心的定义
The centre of gravity of an object is defined as the single point through which the entire weight of the object may be considered to act, regardless of the object’s orientation. It is the average position of all the individual weight vectors of the particles that make up the object.
物体的重心被定义为这样一个点：无论物体朝向如何，物体的全部重量都可以被认为作用在该点上。它是构成物体的所有微粒的重量矢量的平均位置。
In a uniform gravitational field (which is a valid assumption for most terrestrial problems), the centre of gravity coincides with the centre of mass. The centre of mass depends only on the distribution of mass within the object, whereas the centre of gravity depends on the gravitational field. For A-Level purposes, these two points are treated as identical.
在均匀引力场中（对于大多数地面问题这是有效假设），重心与质心重合。质心仅取决于物体内部的质量分布，而重心取决于引力场。在 A-Level 考试中，这两个点被视为相同的。
Key distinctions:
- Centre of mass is a purely geometric property based on mass distribution | 质心是一个纯粹基于质量分布的几何属性
- Centre of gravity is the point where weight acts | 重心是重量作用点的位置
- In a uniform gravitational field, CG = CM | 在均匀引力场中，重心 = 质心
2. Finding the Centre of Gravity of Regular Objects | 规则物体重心的确定
For objects with uniform density and regular geometric shapes, the centre of gravity lies at the geometric centre of the object. This is because the mass is distributed symmetrically about that point.
对于密度均匀且具有规则几何形状的物体，重心位于物体的几何中心。这是因为质量关于该点对称分布。
- Rectangular lamina: CG at the intersection of the diagonals | 矩形薄板：重心在对角线交点处
- Circular disc: CG at the centre of the circle | 圆形薄板：重心在圆心处
- Triangular lamina: CG at the centroid, which is ⅔ of the median from a vertex | 三角形薄板：重心在质心处，即从中线顶点起 ⅔ 处
- Solid sphere: CG at the centre of the sphere | 实心球体：重心在球心处
- Uniform rod: CG at the midpoint | 均匀细杆：重心在中点处
3. Centre of Gravity of Composite Objects | 组合物体的重心
For objects made up of multiple regular shapes, the centre of gravity can be found by taking moments about a chosen axis. The method involves treating each component as a point mass concentrated at its own centre of gravity.
对于由多个规则形状组成的物体，可以通过选取某轴取力矩的方法来确定重心。该方法将每个组成部分视为一个质点，集中在各自的重心上。
Consider two masses m₁ and m₂ at positions x₁ and x₂. The centre of gravity x̄ is given by:
考虑位于 x₁ 和 x₂ 处的两个质量 m₁ 和 m₂。重心 x̄ 由以下公式给出：
x̄ = (m₁x₁ + m₂x₂) / (m₁ + m₂)
This principle extends to any number of masses and to two-dimensional problems by applying the formula independently along the x and y axes.
该原理可推广到任意数量的质量，并可通过沿 x 轴和 y 轴分别应用该公式来拓展到二维问题。
4. Experimental Determination of the Centre of Gravity | 重心测量的实验方法
The centre of gravity of an irregular lamina can be determined experimentally using the plumb-line method:
不规则薄板的重心可以通过铅垂线法在实验中测定：
- Suspend the lamina freely from a point near its edge | 将薄板从边缘附近的一点自由悬挂
- Hang a plumb line from the same point and draw the corresponding vertical line on the lamina | 在同一点悬挂铅垂线，并在薄板上画出相应的竖直线
- Repeat with a different suspension point | 更换悬挂点，重复上述步骤
- The intersection of the lines gives the centre of gravity | 各竖直线段的交点即为重心
This method works because, when an object hangs in equilibrium, the centre of gravity must lie directly below the point of suspension. The same principle formsthe basis of stability analysis in engineering and design.
该方法的原理是：当物体悬挂平衡时，重心必须位于悬挂点正下方。同样的原理也是工程和设计中稳定性分析的基础。
5. Conditions for Equilibrium | 平衡条件
For an object to be in complete equilibrium, two conditions must be satisfied simultaneously:
物体要处于完全平衡状态，必须同时满足两个条件：
First condition (translational equilibrium): The vector sum of all forces acting on the object must be zero. That is, ΣF = 0.
第一条件（平动平衡）：作用在物体上的所有力的矢量和必须为零，即 ΣF = 0。
ΣFₓ = 0 and ΣFᵧ = 0
Second condition (rotational equilibrium): The sum of all moments about any point must be zero. That is, ΣM = 0.
第二条件（转动平衡）：关于任意一点的所有力矩之和必须为零，即 ΣM = 0。
ΣM = 0 (clockwise moments = anticlockwise moments)
ΣM = 0（顺时针力矩 = 逆时针力矩）
6. Types of Equilibrium | 平衡的类型
Depending on how an object responds when displaced from equilibrium, we classify equilibrium into three types:
根据物体偏离平衡位置后的响应方式，平衡可分为三种类型：
| Type | 类型 | Response to displacement | 对位移的响应 | CG behaviour | 重心行为 |
| Stable | 稳定 | Returns to original position | 返回原位置 | CG rises and then lowers | 重心升高后降低 |
| Unstable | 不稳定 | Moves further away | 进一步远离 | CG falls | 重心降低 |
| Neutral | 随遇 | Stays in new position | 保持在新位置 | CG height unchanged | 重心高度不变 |
Classic examples include: a sphere at the bottom of a bowl (stable), a sphere on top of a dome (unstable), and a sphere on a flat surface (neutral).
经典实例包括：碗底的球（稳定）、圆顶上的球（不稳定）以及平面上的球（随遇平衡）。
7. Stability and the Centre of Gravity | 稳定性与重心
The stability of an object is determined by the position of its centre of gravity relative to its base of support. An object will topple when the vertical line through its centre of gravity falls outside the base of support.
物体的稳定性取决于其重心相对于支撑基底的位置。当通过重心的竖直垂线落在支撑基底之外时，物体将倾倒。
Factors affecting stability:
- Base area: A wider base increases stability | 基底面积：基底越宽，稳定性越高
- CG height: A lower centre of gravity increases stability | 重心高度：重心越低，稳定性越高
- CG position: A central CG within the base is more stable | 重心位置：重心位于基底中心区域更稳定
This explains why racing cars are designed with low centres of gravity and wide wheelbases, and why double-decker buses have their heavy machinery at the bottom.
这解释了为什么赛车设计得重心低、轴距宽，以及为什么双层巴士将重型机械装置安装在底部。
8. Solving Equilibrium Problems | 平衡问题的解题方法
A systematic approach to equilibrium problems is essential for exam success. Follow these steps:
处理平衡问题需要系统的方法，这对考试取得好成绩至关重要。请遵循以下步骤：
- Draw a clear free-body diagram showing all forces | 画出清晰受力分析图，标明所有力
- Resolve forces into horizontal and vertical components | 将力分解为水平和竖直分量
- Apply ΣFₓ = 0 and ΣFᵧ = 0 | 应用 ΣFₓ = 0 和 ΣFᵧ = 0
- Choose a suitable pivot point and apply ΣM = 0 | 选择合适的支点，应用 ΣM = 0
- Solve the simultaneous equations for unknown quantities | 解联立方程求未知量
A common exam-style example is a uniform ladder resting against a smooth wall. The key insight is that the ladder’s weight acts through its centre, which is at the midpoint, and that reactions at the wall and ground provide the necessary forces for equilibrium. Taking moments about the foot of the ladder eliminates the ground reaction from the moment equation, simplifying the analysis considerably.
一个常见的考试题型是梯子靠在光滑墙上的问题。关键思路在于梯子的重力作用于其重心（即梯子中点），而墙和地面的反作用力提供了平衡所需的力。以梯子底部为支点取力矩，可以从力矩方程中消去地面反作用力，从而大大简化分析过程。
9. Worked Example | 例题分析
Problem: A uniform beam AB of length 6.0 m and weight 200 N is supported at points C and D, located 1.0 m and 4.0 m from A respectively. A load of 500 N is placed at a point 2.5 m from A. Determine the support reactions at C and D.
题目：一根长为 6.0 m、重为 200 N 的均匀梁 AB，在距 A 端 1.0 m 的 C 点和距 A 端 4.0 m 的 D 点处受到支撑。一个 500 N 的负载放置在距 A 端 2.5 m 处。求 C 点和 D 点的支撑反力。
Solution:
解答：
Since the beam is uniform, its weight acts at the midpoint, which is 3.0 m from A. For a uniform beam, the centre of gravity is at the geometric centre.
由于梁是均匀的，其重力作用于中点，即距 A 端 3.0 m 处。对于均匀梁，重心位于几何中心。
Taking moments about C: Let R_D be the reaction at D. The weight of the beam (200 N) acts 2.0 m from C. The load (500 N) acts 1.5 m from C. The reaction at D is 3.0 m from C.
取 C 点的力矩：设 D 点的反力为 R_D。梁的重力（200 N）作用点距 C 点 2.0 m。负载（500 N）作用点距 C 点 1.5 m。D 点的反力距 C 点 3.0 m。
Clockwise moments: (200 × 2.0) + (500 × 1.5) = 400 + 750 = 1150 N·m
逆时针力矩：(200 × 2.0) + (500 × 1.5) = 400 + 750 = 1150 N·m
Anticlockwise moments: R_D × 3.0
顺时针力矩：R_D × 3.0
1150 = 3.0 × R_D → R_D = 383.3 N
Applying ΣFᵧ = 0: R_C + R_D = 200 + 500 = 700 N. Therefore R_C = 700 − 383.3 = 316.7 N.
应用 ΣFᵧ = 0：R_C + R_D = 200 + 500 = 700 N。因此 R_C = 700 − 383.3 = 316.7 N。
Answer: R_D = 383 N upward, R_C = 317 N upward (both to 3 significant figures)
答案：R_D = 383 N 向上，R_C = 317 N 向上（均保留三位有效数字）
10. Common Mistakes and Exam Tips | 常见错误与考试技巧
Common pitfalls that lose marks:
常见的失分陷阱：
- Forgetting to convert all distances to metres before calculation | 忘记在计算前将所有距离换算成米
- Incorrectly identifying where the weight of a non-uniform rod acts | 错误确定非均匀杆重力作用点的位置
- Omitting the weight of the object itself in force summation | 在力求和时遗漏物体自身的重力
- Choosing an inconvenient pivot point that does not simplify the equations | 选择不方便的支点，未能简化方程
- Mixing up clockwise and anticlockwise directions when taking moments | 取力矩时混淆顺时针和逆时针方向
Exam tips for success:
- State the principle of moments explicitly when applying it | 应用力矩原理时明确写出该原理
- Show all working clearly, even for simple arithmetic | 即使简单计算也要展示完整的解题过程
- Draw large, clear diagram with all forces labelled | 绘制大而清晰且标注所有力的示意图
- Check that your answers are physically reasonable (reactions should be positive and sum to the total weight) | 检查答案是否在物理上合理（反力应为正且总和等于总重量）
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