📚 A-Level Mathematics: Analysis of Toppling of Objects | A-Level 数学：物体倾覆问题分析
Toppling is a key application of moments and centre of mass in mechanics. In A-Level Mathematics, you may be asked to determine whether a rigid object will slide or topple when placed on a rough inclined plane, or when subjected to external forces.
倾覆是力学中力矩与质心应用的关键问题。在 A-Level 数学中，你常常需要判断一个刚体放在粗糙斜面上或受外力作用时，究竟会滑动还是会倾覆。
1. What Is Toppling? | 什么是倾覆？
Toppling occurs when a rigid body rotates about one of its edges or corners instead of translating. For a block on a slope, this usually means rotating about the lower edge of the base.
倾覆是指刚体绕其底面某个边或角发生转动，而不是整体平移。对于斜面上的物体，通常指绕底面的下边缘转动。
The critical condition involves the line of action of the weight. If this line passes outside the base of support, the object will rotate about the edge it is closest to.
临界条件与重力作用线有关。如果重力作用线落在支撑面之外，物体将绕其最近的边缘发生转动。
2. Centre of Mass and Stability | 质心与稳定性
For a uniform rectangle, the centre of mass is at the geometric centre. The stability of an object depends on the horizontal position of its centre of mass relative to the base edges.
对于均匀矩形物体，质心位于几何中心。物体的稳定性取决于质心相对于底面边缘的水平位置。
If the centre of mass lies above the base, the object is in stable equilibrium. When it moves outside the base, the weight creates a moment that topples the object.
当质心位于支撑面正上方时，物体处于稳定平衡。一旦质心移出支撑面，重力产生的力矩就会使物体倾覆。
3. The General Toppling Condition | 倾覆的一般条件
Consider a body resting on a horizontal plane. The object will topple if the vertical line through its centre of mass falls outside the base area.
考虑一个物体在水平面上。如果过质心的竖直垂线落在底面区域之外，物体就会倾覆。
For a uniform rectangular block of width w and height h, pushing it horizontally at its top requires a force F. Taking moments about the lower edge, the block is on the point of toppling when:
对于宽为 w、高为 h 的均匀矩形物体，在其顶部施加水平力 F。绕下边缘取矩，物体处于临界倾覆状态时：
F h = mg × (w / 2)
Thus the minimum force to topple it is F = mgw / (2h). Note that this assumes friction is large enough to prevent sliding.
因此使其倾覆的最小力为 F = mgw / (2h)。注意这里假设摩擦力足够大，能够阻止滑动。
4. Toppling on an Inclined Plane | 斜面上的倾覆
In many exam questions, a uniform rectangular block rests on a rough plane inclined at angle θ to the horizontal. The block has width a along the plane and height b perpendicular to the plane.
在许多考试题目中，一个均匀矩形物体静止在倾角为 θ 的粗糙斜面上。该物体沿斜面方向的宽度为 a，垂直于斜面的高度为 b。
When the plane is tilted, the weight line may move outside the lower edge of the block. The critical angle for toppling is found by aligning the centre of mass vertically above the lower edge.
当斜面倾斜时，重力作用线可能移出物体的下边缘。倾覆临界角通过使质心恰好位于下边缘正上方来求得。
tan θ_t = a / b
So a wider base or a lower height makes the object harder to topple.
因此底面越宽或高度越低，物体越不容易倾覆。
5. Deriving the Critical Angle | 推导临界角
Take the lower edge of the block as the pivot. In the tilted position, the centre of mass is displaced along the plane by a/2 and perpendicular to the plane by b/2 from that edge.
以物体下边缘为支点。在倾斜状态下，质心相对于该边缘沿斜面方向偏移 a/2，垂直于斜面方向偏移 b/2。
Resolving the position horizontally and vertically, the critical condition occurs when the horizontal displacement of the centre of mass from the pivot is zero:
将位移分解到水平和竖直方向，质心相对支点的水平位移为零时即为临界条件：
– (a/2) cos θ + (b/2) sin θ = 0
This simplifes to:
化简得：
tan θ = a / b
The same logic applies to any object if a is replaced by the horizontal distance from the pivot to the centre of mass, and b by its vertical height above the plane.
对任意物体，若将 a 换成支点到质心沿斜面的距离，b 换成质心到斜面的垂直高度，上述结论同样成立。
6. Sliding vs Toppling | 滑动与倾覆的比较
On a rough incline, the object may also slide down the plane. Sliding occurs when the component of weight down the plane exceeds the maximum friction force.
在粗糙斜面上，物体也可能沿斜面下滑。当下滑的重力分量超过最大静摩擦力时，物体开始滑动。
The critical angle for sliding is determined by the coefficient of friction μ:
滑动的临界角由摩擦系数 μ 决定：
tan θ_s = μ
To decide what happens first, compare θ_t and θ_s. If θ_s < θ_t, the object slides before it topples. If θ_t < θ_s, it topples before it slides.
要知道先发生哪种运动，需比较 θ_t 和 θ_s。若 θ_s < θ_t，物体先滑动；若 θ_t < θ_s，物体先倾覆。
7. Worked Example: Comparing the Two Angles | 示例：比较两个临界角
A uniform rectangular block has base width 0.4 m and height 0.6 m. It rests on a rough plane with μ = 0.5. Determine whether it slides or topples as the angle increases.
一个均匀矩形物体，底面宽度为 0.4 m，高度为 0.6 m，放在摩擦系数 μ = 0.5 的粗糙斜面上。随着角度增大，判断它先滑动还是先倾覆。
For sliding:
对于滑动：
tan θ_s = μ = 0.5 ⇒ θ_s = 26.6°
For toppling:
对于倾覆：
tan θ_t = a / b = 0.4 / 0.6 = 0.6667 ⇒ θ_t = 33.7°
Since θ_s < θ_t, the block slides before it can topple.
因为 θ_s < θ_t，物体在倾覆之前已经发生滑动。
8. What If Friction Is Large? | 如果摩擦力很大呢？
Suppose the same block has μ = 0.8. Then θ_s = arctan 0.8 = 38.7°.
假设同一物体 μ = 0.8。则 θ_s = arctan 0.8 = 38.7°。
Now θ_t = 33.7° < θ_s, so the block topples before it slides.
此时 θ_t = 33.7° < θ_s，因此物体会先倾覆。
| Case | Sliding angle | Toppling angle | Outcome |
| μ = 0.5 | 26.6° | 33.7° | Slides first |
| μ = 0.8 | 38.7° | 33.7° | Topples first |
9. Non-Uniform Objects | 非均匀物体的倾覆
If the object is not uniform, its centre of mass is not at the geometric centre. You may need to calculate its position using moments or integration.
如果物体不是均匀的，其质心不在几何中心。你可能需要用力矩或积分来计算质心位置。
Once the centre of mass is known, the critical toppling angle is still based on whether its vertical projection passes through the lower edge:
一旦知道质心位置，倾覆临界角仍取决于其竖直投影是否通过下边缘：
tan θ_t = x_G / y_G
where x_G is the horizontal distance from the lower edge to the centre of mass along the plane, and y_G is the vertical distance from the plane to the centre of mass.
其中 x_G 是下边缘到质心沿斜面的距离，y_G 是斜面到质心的垂直距离。
10. Exam Strategy | 考试应对策略
Follow these steps for toppling questions:
处理倾覆问题可遵循以下步骤：
- Draw a clear diagram showing the object, the pivot edge, and the centre of mass.
- 清晰作图，标出物体、支点边和质心位置。
- Write the sliding condition using F_max = μR.
- 写出滑动条件：F_max = μR。
- Take moments about the pivot edge to find the toppling condition.
- 绕支点边取矩，建立倾覆条件。
- Compare the two critical angles to decide which motion occurs first.
- 比较两个临界角，判断哪种运动先发生。
11. Common Mistakes | 常见错误
Students often forget that the normal reaction and friction may be concentrated at the pivot when toppling is imminent. You do not need to know their exact values in the limiting case if you take moments about the pivot.
同学们常常忘记，在即将倾覆时，支持力和摩擦力可以视为集中在支点处。如果绕支点取矩，不需要知道它们的具体数值。
Another common error is using the full width a instead of the half-width a/2 in the moment arm. Always use the horizontal distance from the pivot to the vertical line through the centre of mass.
另一个常见错误是把力臂误用为整个宽度 a 而不是半宽 a/2。务必使用从支点到质心竖直投影线的水平距离。
12. Summary | 总结
Toppling in A-Level Mathematics is controlled by the position of the centre of mass relative to the support base. On an inclined plane, compare the toppling angle tan θ_t = a/b with the sliding angle tan θ_s = μ.
A-Level 数学中的倾覆问题由质心相对于支撑面的位置决定。在斜面上，比较倾覆角 tan θ_t = a/b 与滑动角 tan θ_s = μ 的大小即可。
Understanding these principles will help you solve both calculation and explanation problems with confidence.
理解这些原理，你将能自信地解决计算题和解释题。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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