Calculating Resultant Moments | 合力矩的计算方法

📚 Calculating Resultant Moments | 合力矩的计算方法

In A-Level Mathematics and Further Mathematics, the concept of a moment is fundamental to understanding the rotational effect of forces. This article provides a comprehensive guide to calculating resultant moments, covering the necessary definitions, sign conventions, and step-by-step problem-solving techniques.

在A-Level数学与进阶数学中，力矩是理解力之转动效应的核心概念。本文将系统讲解合力矩的计算方法，涵盖必要的定义、正负号约定以及逐步解题技巧。

1. Definition of a Moment | 力矩的定义

A moment (or torque) is the turning effect produced by a force acting at a distance from a pivot point. Mathematically, the magnitude of a moment about a point O is given by the product of the force F and the perpendicular distance d from the line of action of the force to point O.

力矩（或转矩）是力作用于距支点一定距离处所产生的转动效应。从数学上讲，力F对某点O的力矩大小等于力F与其作用线到O点垂直距离d的乘积。

M = F × d

Here, M is the moment in newton-metres (N·m), F is the force in newtons (N), and d is the perpendicular distance in metres (m). The distance d is measured along a line perpendicular to the force vector, not along the object itself.

其中，M为力矩，单位为牛顿·米（N·m）；F为力，单位为牛顿（N）；d为垂直距离，单位为米（m）。距离d沿垂直于力矢量的方向量取，而非沿物体本身量取。

2. Units of Moment | 力矩的单位

The SI unit of moment is the newton-metre (N·m). It is important to note that although N·m is dimensionally equivalent to the joule (J), the two are never used interchangeably in mechanics. The joule is reserved exclusively for energy and work.

力矩的国际单位制（SI）单位是牛顿·米（N·m）。需要特别注意的是，尽管N·m与焦耳（J）在量纲上等价，但在力学中二者绝不能混用。焦耳仅专用于能量与功。

When expressing moments, ensure that all distances are converted to metres and all forces to newtons before performing calculations. This avoids common dimensional errors in examination solutions.

在表达力矩时，务必在计算前将所有距离换算为米、所有力换算为牛顿，以免在考试解答中出现量纲错误。

3. Sign Convention | 正负号约定

Moments are vector quantities, but in two-dimensional problems they can be treated as scalars with a sign. A common convention is: clockwise moments are taken as negative and anticlockwise moments as positive. Alternatively, some textbooks adopt the opposite; the key is consistency throughout a solution.

力矩是矢量，但在二维问题中可视为带正负号的标量。常用约定为：顺时针力矩取负，逆时针力矩取正。有些教材采用相反约定，关键在于整个解题过程中保持一致。

ΣM = M₁ + M₂ + M₃ + …

The resultant moment about a point is the algebraic sum of all individual moments about that same point. Each term must carry its correct sign according to the chosen convention.

对某点的合力矩等于所有力对同一点各自力矩的代数和。每一项必须按照所选约定携带正确的正负号。

4. Calculating Moment of a Single Force | 单个力的力矩计算

To calculate the moment of a single force about a point O, first identify the perpendicular distance from O to the line of action of the force. This often requires resolving the force into components or extending the line of action geometrically.

要计算单个力对O点的力矩，首先确定从O点到力的作用线的垂直距离。这通常需要将力分解为分量，或从几何上延长力的作用线。

Suppose a force of 10 N acts at an angle of 30° above the horizontal, applied at a point 2 m horizontally from O. The perpendicular distance from O to the line of action is 2 × sin(30°) = 1 m. Hence the moment is M = 10 × 1 = 10 N·m.

设一大小为10 N的力沿水平方向上方30°角方向作用，作用点距O点水平距离2 m。从O到力作用线的垂直距离为2 × sin(30°) = 1 m，因此力矩为M = 10 × 1 = 10 N·m。

M = F × (r sin θ)

In general, if r is the distance from O to the point of application and θ is the angle between the force vector and the line connecting O to the point of application, then the moment is F × r sin θ.

一般情况下，若r为O到作用点的距离，θ为力矢量与O到作用点连线之间的夹角，则力矩为F × r sin θ。

5. Resultant Moment of Multiple Forces | 多个力的合力矩

When several forces act on a rigid body, the resultant moment about a point O is simply the algebraic sum of the individual moments. For forces F₁, F₂, …, Fₙ with corresponding perpendicular distances d₁, d₂, …, dₙ, the resultant moment is:

当多个力作用于刚体时，对O点的合力矩等于各分力矩的代数和。设力F₁、F₂、…、Fₙ对应的垂直距离分别为d₁、d₂、…、dₙ，则合力矩为：

M_R = F₁d₁ + F₂d₂ + … + Fₙdₙ

This expression includes positive terms for forces causing anticlockwise rotation and negative terms for those causing clockwise rotation (under the standard convention). Note that forces whose line of action passes through O contribute zero moment.

该表达式包含使物体逆时针旋转的正项和使物体顺时针旋转的负项（采用标准约定）。请注意，作用线通过O点的力的力矩为零。

6. Worked Example 1: Simple Beam | 实例1：简支梁

A horizontal beam of length 4 m is pivoted at its left end O. Forces of 20 N downward act at 1 m from O, 30 N upward act at 3 m from O, and 40 N downward act at 4 m from O. Calculate the resultant moment about O.

一长度为4 m的水平梁，左端O为支点。在距O点1 m处有一20 N向下的力，距O点3 m处有一30 N向上的力，距O点4 m处有一40 N向下的力。求这些力对O点的合力矩。

Using the convention that anticlockwise is positive: the 20 N downward force creates a clockwise moment (negative), the 30 N upward force creates an anticlockwise moment (positive), and the 40 N downward force creates a clockwise moment (negative).

采用逆时针为正的约定：20 N向下的力产生顺时针力矩（负），30 N向上的力产生逆时针力矩（正），40 N向下的力产生顺时针力矩（负）。

M_R = −(20 × 1) + (30 × 3) − (40 × 4) = −20 + 90 − 160 = −90 N·m

The negative sign indicates that the resultant moment is 90 N·m in the clockwise direction about O.

负号表示合力矩为90 N·m，方向为绕O点顺时针。

7. Forces Not Perpendicular to the Beam | 不垂直于梁的力

Real exam problems often involve forces acting at angles. In such cases, compute the perpendicular component of the force that is effective in creating rotation about the pivot.

实际考试题目中，力的作用方向往往与梁成一定角度。此时需计算能有效产生绕支点转动的垂直分量。

Consider a force F applied at an angle θ to the beam, at a distance L from the pivot. The perpendicular component is F sin θ, and the moment is F L sin θ. Alternatively, one may multiply the full force by the perpendicular distance from the pivot to the line of action.

考虑一个与梁成θ角的力F，作用点距支点距离为L。垂直分量为F sin θ，力矩为F L sin θ。也可以直接用整个力乘以支点到力作用线的垂直距离。

M = F L sin θ

This formula is valid regardless of whether the force points towards or away from the pivot, as long as θ is measured between the force direction and the beam direction.

无论力指向支点还是远离支点，只要θ是力方向与梁方向之间的夹角，该公式均成立。

8. Couples | 力偶

A couple consists of two equal and opposite forces acting along parallel lines of action but separated by a perpendicular distance d. The moment of a couple is the product of one of the forces and the perpendicular distance between their lines of action.

力偶由两个大小相等、方向相反且作用线平行的力构成，二者之间垂直距离为d。力偶矩等于其中一个力的大小乘以两力作用线之间的垂直距离。

C = F × d

For example, a steering wheel of radius 0.2 m with two equal and opposite 15 N forces applied tangentially produces a couple of C = 15 × 0.4 = 6 N·m. The moment of a couple is the same about any point in the plane.

例如，半径为0.2 m的方向盘，在切向施加两个大小相等、方向相反的15 N力，产生的力偶矩为C = 15 × 0.4 = 6 N·m。力偶矩在平面内任意一点取值均相同。

9. Distributed Loads | 分布载荷

In mechanics problems, loads are sometimes distributed uniformly along a beam rather than applied at single points. A uniform distributed load (UDL) of intensity w N/m over a length L is equivalent to a single concentrated force of magnitude wL acting at the midpoint of the loaded section.

在力学问题中，载荷有时沿梁均匀分布而非作用于单点。强度为w N/m、长度为L的均布载荷等效于一个大小为wL的集中力，作用于该载荷段的中点。

For a non-uniform load, the magnitude equals the area under the load diagram (length × average intensity), and its line of action passes through the centroid of that area. In the case of a triangular load, this is at the point one-third of the base length from the right-angled side.

对于非均匀载荷，其等效大小等于载荷图下的面积（长度 × 平均强度），其作用线通过该面积的形心。对于三角形载荷，形心位于距直角边三分之一底边长度处。

Once the equivalent concentrated forces are identified, apply the standard summation procedure to find the resultant moment about any pivot.

一旦确定了等效集中力，即可采用标准叠加方法计算对任意支点的合力矩。

10. Principle of Moments in Equilibrium | 力矩平衡原理

When a rigid body is in static equilibrium under the action of coplanar forces, the resultant moment about any point must be zero, provided the resultant force is also zero. This is known as the principle of moments:

当刚体在共面力系作用下处于静态平衡时，若合力也为零，则对任意一点的合力矩必须为零。这称为力矩平衡原理：

ΣM = 0

This principle is extensively used to determine unknown reaction forces at supports. By taking moments about one support, the reaction at the other support can be isolated and solved directly.

该原理广泛用于求解支承处的未知反力。通过取关于一个支点的力矩方程，可直接分离并解出另一支点的反力。

For a beam of length L with forces F₁ and F₂ at distances a and b from the left support, and reactions R₁ (left) and R₂ (right), taking moments about the left support gives:

对于长度L的梁，左支承反力R₁，右支承反力R₂，力F₁与F₂分别距左支承a与b，则对左支承取矩得：

R₂ × L = F₁ × a + F₂ × b

11. Moments in Three Dimensions | 三维问题中的力矩

In three-dimensional problems, the moment of a force about a point is represented as a vector. The moment vector M about point O due to a force F applied at position vector r is given by the cross product:

在三维问题中，力对点的矩以矢量表示。在位置矢量r处施加力F，对O点的力矩矢量为叉积：

M = r × F

The direction of M follows the right-hand screw rule, and its magnitude is |M| = |r||F| sin θ, where θ is the angle between r and F. In component form, if r = (x, y, z) and F = (Fₓ, Fᵧ, F_z), then:

M的方向遵循右手螺旋定则，其大小为|M| = |r||F| sin θ，其中θ为r与F之间的夹角。若r = (x, y, z)，F = (Fₓ, Fᵧ, F_z)，则分量形式为：

M = (yF_z − zFᵧ, zFₓ − xF_z, xFᵧ − yFₓ)

For A-Level purposes, three-dimensional moment questions are typically straightforward substitutions into this formula, but a clear diagram and careful sign tracking are essential.

就A-Level而言，三维力矩题通常只是代入公式计算，但清晰的图示和仔细的正负号追踪必不可少。

12. Common Exam Pitfalls | 常见考试误区

A common mistake is using the wrong perpendicular distance. Always measure the distance from the pivot to the line of action of the force, not to the point of application along the beam. This distinction frequently causes lost marks.

常见错误之一是使用了错误的垂直距离。务必量取从支点到力作用线的垂直距离，而不是沿梁到作用点的距离。这一区别经常导致失分。

Another frequent error is inconsistent sign convention. Students sometimes change the sign rule mid-solution. Decide on one convention at the start and apply it uniformly.

另一常见错误是正负号约定前后不一致。有些学生在解题中途改变符号规则。应在开始时确定一种约定并始终如一地执行。

Finally, do not forget that a force whose line of action passes through the pivot produces zero moment. Including such a force in the moment equation with a non-zero distance is a common oversight.

最后，不要忘记作用线通过支点的力产生零力矩。在力矩方程中为这样的力赋予非零距离是常见疏忽。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Calculating Resultant Moments | 合力矩的计算方法