Moments and Equilibrium | 力矩与平衡考点精讲

📚 Moments and Equilibrium | 力矩与平衡考点精讲

In IGCSE OCR Mathematics, moments and equilibrium form a core part of the mechanics syllabus. Understanding how to calculate moments and apply the equilibrium conditions is essential for solving a range of practical problems, from balancing beams to determining unknown forces. This article covers every key point you need for exam success.

在 IGCSE OCR 数学中，力矩与平衡是力学部分的核心内容。掌握如何计算力矩并应用平衡条件，对于解决从平衡横梁到确定未知力的一系列实际问题至关重要。本文涵盖了考试成功所需的每个关键知识点。

1. What is a Moment? | 什么是力矩？

A moment measures the turning effect of a force about a pivot (also called a fulcrum). Whenever a force is applied at a distance from a pivot, it can cause rotation. The size of the moment depends on how large the force is and how far from the pivot its line of action passes.

力矩衡量一个力绕支点（也称为支点）产生的转动效果。每当力在离支点一定距离处施加时，就能引起转动。力矩的大小取决于力的大小以及力的作用线离支点的距离。

Moments can be clockwise or anticlockwise. In problem solving, it is usual to choose anticlockwise as positive and consistently apply this sign convention.

力矩可以是顺时针或逆时针方向。在解题时，通常选择逆时针方向为正，并始终使用这一符号约定。

2. Formula and Units | 公式与单位

The moment M about a pivot is given by the product of the force F and the perpendicular distance d_⊥ from the pivot to the line of action of the force.

关于某支点的力矩 M 等于力 F 与支点到力作用线的垂直距离 d_⊥ 的乘积。

M = F × d_⊥

The SI unit of moment is the newton metre (N m). The distance must be measured at right angles to the force; using the simple distance along a lever only works if the force is already perpendicular.

力矩的国际单位是牛顿·米 (N m)。距离必须沿与力垂直的方向量取；只有当力已经垂直于杠杆时，才可直接使用杠杆上的距离。

If a force acts at an angle θ to the lever, the perpendicular distance is d sinθ, so the moment becomes M = F d sinθ. Alternatively, you can resolve the force into a perpendicular component.

如果力与杠杆成 θ 角，则垂直距离为 d sinθ，因此力矩为 M = F d sinθ。或者你也可以将力分解为垂直分量。

3. The Principle of Moments | 力矩原理

For a system that is in rotational equilibrium (not turning), the total clockwise moment about any pivot equals the total anticlockwise moment about that same pivot.

对于处于转动平衡（不发生旋转）的系统，关于任一支点的总顺时针力矩等于关于同一点的总逆时针力矩。

Σ M_clockwise = Σ M_{anticlockwise}

This principle is sometimes called the law of the lever. It allows you to calculate unknown forces or distances when a beam or lever is balanced.

该原理有时也称为杠杆定律。当横梁或杠杆平衡时，可以利用它计算未知的力或距离。

4. Conditions for Equilibrium | 平衡条件

A rigid body is in static equilibrium when two conditions are satisfied simultaneously:

刚体同时满足以下两个条件时，便处于静力平衡：

1. The resultant force in any direction is zero – this ensures no translational motion.

1. 任意方向的合力为零——这保证了没有平动。

2. The resultant moment about any point is zero – this ensures no rotational motion.

2. 关于任意点的合力矩为零——这保证了没有转动。

In many IGCSE questions, you will use the moment condition to find an unknown reaction force at a support after using the force condition to find another unknown.

在许多 IGCSE 考题中，你会先利用合力条件求出某个未知力，再利用合力矩条件求出另一个未知力。

5. Applying Moments to Uniform Beams | 力矩在均匀横梁中的应用

A uniform beam has its weight acting exactly at its geometric centre. When the beam is pivoted at its midpoint, its own weight produces no net moment about that pivot and can often be ignored.

均匀横梁的重量恰好作用在其几何中心。当梁在其中点支起时，其自身重量对该支点不产生净力矩，通常可以忽略。

If the pivot is not at the centre, the weight of the beam must be included as a force acting downwards at the midpoint. Always take moments about a convenient pivot to eliminate one of the unknown forces.

如果支点不在中心，则必须将梁的重量视为作用在中点向下的一个力。始终选择一个便利的支点取力矩，以消去其中一个未知力。

Example: A uniform beam of length 4.0 m and weight 100 N is pivoted 1.0 m from its left end. To keep it horizontal, a vertical force is applied at the far right end. By taking moments about the pivot, the required force can be found using the beam weight and its distance from the pivot.

示例：一根长 4.0 m、重 100 N 的均匀横梁在距左端 1.0 m 处支起。为保持水平，需在最右端施加一个竖直力。通过对支点取力矩，可利用梁重及其离支点的距离求出所需力的大小。

6. Solving Equilibrium Problems Step by Step | 逐步解平衡问题

A systematic approach is vital:

系统性的解法至关重要：

Step 1: Draw a clear diagram, marking all forces, distances and the chosen pivot.

第一步：画一张清晰的图，标出所有力、距离和所选支点。

Step 2: Identify the direction (clockwise or anticlockwise) of each moment about the pivot.

第二步：确定每个力对支点的力矩方向（顺时针或逆时针）。

Step 3: Write an equation using the principle of moments: sum of clockwise moments = sum of anticlockwise moments.

第三步：根据力矩原理写出等式：顺时针力矩之和 = 逆时针力矩之和。

Step 4: Solve for the unknown quantity. Substitute values in consistent units (N and m).

第四步：求解未知量。代入数值时单位要统一（用 N 和 m）。

Step 5: If there are two unknowns, first apply the zero resultant force condition to relate them, then take moments.

第五步：若有两个未知量，可先利用合力为零的条件建立它们的关系，再取力矩。

7. Moments with Non-Uniform Objects | 非均匀物体的力矩

A non-uniform beam’s centre of mass is not at its geometric centre. You may be asked to find its position using moments. Usually, the beam is supported at one end and a known pivot, or balanced on a pivot with a known weight, allowing the centre of mass distance to be calculated.

非均匀梁的质心不在其几何中心。你可能需要利用力矩求出其位置。通常，梁一端被支撑并有一个已知支点，或在一个支点上与已知重量平衡，从而可以算出质心距离。

For example, a plumber’s rod of weight 80 N balances on a pivot when a 50 N load is hung 0.40 m from the left end. If the rod is 2.0 m long and the pivot is 0.60 m from the left end, taking moments about the pivot yields the unknown centre of mass position.

例如，一根重 80 N 的管子在一个支点上平衡，此时左端 0.40 m 处挂有 50 N 的重物。若管子长 2.0 m，支点距左端 0.60 m，对支点取力矩即可求出未知的质心位置。

8. Centre of Mass and Moments | 质心与力矩

The centre of mass of an object is the point at which its entire weight may be considered to act. For a symmetrical, uniform lamina, it is at the intersection of lines of symmetry.

物体的质心是可以认为其全部重量集中作用的点。对于对称、均匀的薄板，它位于对称线的交点。

Experimentally, the centre of mass of a plane lamina can be found by suspending it from different points. The vertical line from the pivot always passes through the centre of mass. When two such lines are drawn, their intersection locates the centre of mass.

实验上，平面薄板的质心可通过从不同点悬挂来找到。从支点向下的竖直线总经过质心。画出两条这样的线，其交点即为质心位置。

In calculations, if an unknown mass must be placed to balance a non-uniform object, taking moments about a pivot allows you to use the position of the centre of mass to find the required force or distance.

在计算中，若需放置一个未知质量来使非均匀物体平衡，则可对支点取力矩，利用质心位置求出所需的力或距离。

9. Common Mistakes to Avoid | 常见错误避免

Not using the perpendicular distance – many students simply use the distance along the bar even when the force is at an angle. Always draw the line of action and measure the shortest distance to the pivot.

未使用垂直距离——许多学生在力不垂直时仍直接使用沿杆的距离。务必将力的作用线画出，并量取支点到该线的最短距离。

Ignoring the weight of the beam – when the pivot is not at the centre of a uniform beam, the beam’s weight produces a moment that must be included.

忽略了梁的自重——当支点不在均匀梁的中心时，梁的重量会产生力矩，必须计入。

Direction sign errors – mixing clockwise and anticlockwise without a consistent sign convention leads to wrong equations. Decide early which direction is positive.

方向符号错误——没有采用一致的符号约定而混淆顺时针和逆时针会导致方程出错。一开始就要确定哪个方向为正值。

Unit mixing – forces in newtons and distances in centimetres will give incorrect moments. Always convert distances to metres unless the question explicitly asks for N cm.

单位混用——力用牛顿、距离用厘米无法得到正确的力矩。除非题目明确要求使用 N cm，否则一律将距离转换为米。

10. Worked Exam-style Question | 考试风格例题讲解

A uniform plank AB of length 5.0 m and weight 150 N rests on a pivot at its centre. A box of weight 400 N is placed 1.5 m to the left of the centre. A second load W is placed on the right side so that the plank balances horizontally. Calculate W when it is placed 1.2 m to the right of the centre.

一根长 5.0 m、重 150 N 的均匀木板在其中点支起。一个重 400 N 的箱子放在中心左侧 1.5 m 处。另一个载荷 W 放在右侧，使木板水平平衡。当 W 放在中心右侧 1.2 m 处时，计算 W 的大小。

Take moments about the centre pivot. The plank’s weight acts through the pivot, so its moment is

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Moments and Equilibrium | 力矩与平衡 考点精讲