📚 A-Level Mathematics: Moments – Basic Concepts and Calculations | A-Level数学：力矩的基本概念与计算
In mechanics, a moment is the turning effect produced by a force acting at a distance from a pivot. It is a central topic in the Mechanics section of A-Level Mathematics and is essential for solving problems about rods, beams, ladders and seesaws.
在力学中，力矩描述的是一个力绕某个支点所产生的转动效果。它是 A-Level 数学力学部分的核心内容，也是解决杆件、梁、梯子和跷跷板等问题的关键。
1. What Is a Moment? | 什么是力矩？
When a force is applied to an object, it can cause the object to rotate about a fixed point called a pivot or fulcrum. The size of this turning effect is called the moment of the force about that point.
当一个力作用在物体上时，它可能使物体绕着一个固定点（称为支点或转轴）转动。这个转动效果的大小就称为该力对这一点产生的力矩。
ng at a perpendicular distance d from the pivot, the moment is given by the product F × d.
对于大小为 F、到支点垂直距离为 d 的力，力矩等于 F × d。
Moment = F × d
The moment is measured in newton-metres (N m). A large force or a large distance can both produce a large moment.
力矩的单位是牛顿米（N m）。较大的力或较大的力臂都可以产生较大的力矩。
2. Perpendicular Distance and Line of Action | 垂直距离与力的作用线
The distance d used in the moment formula is the perpendicular distance from the pivot to the line of action of the force. This is not necessarily the straight-line distance from the pivot to the point where the force is applied.
力矩公式中的距离 d 是支点到力的作用线的垂直距离，而不是支点到力作用点的直线距离。
If the force is applied at an angle, you must first find the perpendicular component of the force or the perpendicular distance, otherwise the moment will be incorrect.
如果力是倾斜施加的，必须先求出力的垂直分量或垂直距离，否则计算出的力矩就是错误的。
Moment = F × d⊥ = F sin θ × r
Here θ is the angle between the force and the line joining the pivot to the point of application, and r is the straight-line distance from the pivot to the point of application.
其中 θ 是力的方向与支点到受力点连线之间的夹角，r 是支点到受力点的直线距离。
3. Clockwise and Anticlockwise Moments | 顺时针与逆时针力矩
A moment can turn an object either clockwise or anticlockwise. In calculations we choose one direction as positive and the other as negative. A common convention is to take anticlockwise as positive.
力矩可以使物体顺时针或逆时针转动。在计算中我们通常选择其中一个方向为正，另一个为负。常见的约定是取逆时针为正。
For example, if a force tends to rotate the object clockwise, its moment is written as negative; if it tends to rotate anticlockwise, it is positive.
例如，如果一个力使物体顺时针转动，则其力矩为负；如果使物体逆时针转动，则为正。
It is important to state your sign convention clearly in an exam answer.
在考试作答中，清楚地说明你的正方向约定非常重要。
4. Resultant Moment | 合力矩
When several forces act on a rigid body, the total turning effect about a point is the algebraic sum of all the individual moments. This is called the resultant moment.
当多个力作用在一个刚体上时，物体绕某点的总转动效果等于各个力矩的代数和，称为合力矩。
Resultant moment = ΣM = M₁ + M₂ + M₃ + ...
Remember to include signs: clockwise moments may be negative and anticlockwise moments positive, or vice versa.
注意带上正负号：顺时针力矩可以取负，逆时针力矩取正，也可以反过来。
5. The Principle of Moments | 力矩原理
For a body in rotational equilibrium, the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about the same point. This is the principle of moments.
对于处于转动平衡的物体，绕任意一点顺时针力矩之和等于绕同一点逆时针力矩之和。这就是力矩原理。
ΣM_clockwise = ΣM_anticlockwise
We can use this equation to find unknown forces or distances in equilibrium problems.
我们可以利用这个方程来求解平衡问题中的未知力或未知距离。
6. Conditions for Equilibrium | 平衡条件
A rigid body is in complete equilibrium when both the resultant force and the resultant moment are zero. That means there is no translational acceleration and no angular acceleration.
一个刚体处于完全平衡状态时，合外力为零且合力矩为零。也就是说没有平动加速度，也没有角加速度。
For equilibrium we need to solve two sets of equations:
处于平衡时，我们需要解两组方程：
ΣFx = 0 and ΣFy = 0: the resultant force in both directions is zero.
ΣFx = 0 和 ΣFy = 0：两个方向上的合力均为零。
ΣM = 0 about any chosen point: the resultant moment is zero.
ΣM = 0（绕任意选定点）：合力矩为零。
The point about which moments are taken can be chosen to simplify the algebra, such as at a support where an unknown reaction acts.
力矩的取矩点可以任意选择，以简化计算，例如选在未知支座反力所在的支撑点处。
7. Support Reactions and Levers | 支座反力与杠杆
In lever problems, a rod is supported by one or more supports. The supports exert upward reactions. By taking moments about a support, the reaction at that support is eliminated because its distance is zero.
在杠杆问题中，杆由一个或多个支点支撑，支点施加向上的反力。当我们对某一支点取矩时，该支点处的反力因为距离为零而被消去。
This technique is very useful for finding unknown reaction forces.
这种技巧对于求解未知反力非常有用。
For example, if a beam of length L rests on two supports, taking moments about the left support gives the reaction at the right support immediately.
例如，一根长度为 L 的梁放在两个支点上，对左支点取矩可以直接求出右支点的反力。
8. Moment of a Couple | 力偶的力矩
A couple consists of two equal and opposite forces acting at different points on a body, not along the same line. A couple produces rotation without any resultant force.
力偶是一对大小相等、方向相反但不共线的力，作用在物体的不同点上。力偶会产生转动，但不会产生合力。
The moment of a couple is the product of one of the forces and the perpendicular distance between their lines of action.
力偶矩等于其中一个力的大小与两力作用线之间垂直距离的乘积。
Moment of couple = F × d
This moment is the same about any point on the body, so it does not depend on where you choose the pivot.
力偶矩在物体上任意一点的值都相同，因此它与取矩点的位置无关。
9. Worked Example – A Seesaw | 例题一：跷跷板
A child of mass 30 kg sits 2 m to the left of the pivot. Another child of mass 45 kg sits 1.5 m to the right. Determine whether the seesaw is balanced.
一名质量为 30 kg 的儿童坐在支点左侧 2 m 处，另一名质量为 45 kg 的儿童坐在右侧 1.5 m 处。判断跷跷板是否平衡。
Take g = 9.8 m/s². The weights are:
取 g = 9.8 m/s²。两边的重力为：
W₁ = 30 × 9.8 = 294 N
W₂ = 45 × 9.8 = 441 N
Taking anticlockwise as positive, the resultant moment about the pivot is:
取逆时针为正，绕支点的合力矩为：
ΣM = (294 × 2) − (441 × 1.5) = 588 − 661.5 = −73.5 N m
The resultant moment is negative, so the seesaw rotates clockwise. It is not balanced.
合力矩为负，说明跷跷板顺时针转动，所以不平衡。
10. Worked Example – A Loaded Beam | 例题二：受载横梁
A uniform beam AB of length 6 m and weight 200 N is supported at A and B. A load of 500 N hangs at a point 2 m from A. Find the reactions at A and B.
一根均匀横梁 AB 长 6 m，重量 200 N，在 A 和 B 处受到支撑。一个 500 N 的负载挂在距 A 端 2 m 处。求 A 和 B 处的反力。
For a uniform beam, the weight acts at the midpoint, 3 m from A.
对于均匀横梁，其重量作用在中点，即距 A 端 3 m 处。
Take moments about A. The reaction at A has zero moment, so:
对 A 点取矩。A 点反力的力矩为零，因此：
RB × 6 = (500 × 2) + (200 × 3)
6R_B = 1000 + 600 = 1600
RB = 1600 / 6 = 266.7 N
Vertical equilibrium gives RA + RB = 500 + 200 = 700 N, so RA = 700 − 266.7 = 433.3 N.
竖直方向平衡可得 RA + RB = 500 + 200 = 700 N，所以 RA = 700 − 266.7 = 433.3 N。
Check by taking moments about B: RA × 6 = 500 × 4 + 200 × 3 = 2000 + 600 = 2600, so RA = 433.3 N. The result is consistent.
验算：对 B 点取矩：RA × 6 = 500
Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图