📚 Moments and Equilibrium | 力矩与平衡
A moment is the turning effect of a force about a pivot. It depends on both the magnitude of the force and the perpendicular distance from the pivot to the line of action of the force. Understanding moments is essential for analysing the equilibrium of rigid bodies in mechanics.
力矩是力对某个支点产生的转动效应。它既取决于力的大小，也取决于支点到力作用线的垂直距离。理解力矩对于分析刚体在力学中的平衡至关重要。
1. Definition of a Moment | 力矩的定义
The moment of a force about a point is defined as the product of the force and the perpendicular distance from the point to the line of action of the force.
力对某一点的力矩定义为该力与从该点到力作用线的垂直距离的乘积。
Moment = Force × Perpendicular distance
力矩 = 力 × 垂直距离
The SI unit of a moment is the newton metre (N m). Note that although this is dimensionally equivalent to the joule, moments are not a form of energy and are always expressed in N m, not joules.
力矩的国际单位是牛顿米（N m）。请注意，虽然其量纲与焦耳相同，但力矩不是能量的一种形式，始终用N m表示，不用焦耳。
2. Calculating Moments | 力矩的计算
To calculate a moment correctly, you must identify the perpendicular distance d from the pivot to the line of action of the force. If the force is not perpendicular, you can either resolve the force into a perpendicular component or find the perpendicular distance using trigonometry.
要正确计算力矩，必须确定从支点到力作用线的垂直距离 d。如果力不垂直，可以将力分解为一个垂直分量，或利用三角学求出垂直距离。
For a force F acting at an angle θ to a lever arm of length r, the perpendicular distance is r sin θ, so the moment = F r sin θ.
对于一个与长度为 r 的杠杆臂成 θ 角的力 F，垂直距离为 r sin θ，因此力矩 = F r sin θ。
| Scenario | Perpendicular distance |
| Force at 90° to a rod of length L | L |
| Force at angle θ to a rod of length L | L sin θ |
3. Direction of Moments: Clockwise and Anticlockwise | 力矩的方向：顺时针与逆时针
Moments can cause rotation in two possible directions: clockwise (CW) and anticlockwise (ACW). In equilibrium analysis, it is conventional to define one direction as positive and the other as negative.
力矩可以引起两个方向的转动：顺时针（CW）和逆时针（ACW）。在平衡分析中，通常规定一个方向为正，另一个方向为负。
A body in rotational equilibrium must have the sum of clockwise moments about any pivot equal to the sum of anticlockwise moments about that same pivot.
处于转动平衡的物体，关于任意支点的顺时针力矩之和必须等于关于同一支点的逆时针力矩之和。
When solving problems, clearly state your sign convention, e.g. ‘take anticlockwise moments as positive’.
解题时，要明确注明正负号规定，例如“规定逆时针力矩为正”。
4. The Principle of Moments and Equilibrium | 力矩原理与平衡条件
The principle of moments states that for an object in rotational equilibrium, the sum of clockwise moments about any point equals the sum of anticlockwise moments about that point.
力矩原理指出，对于一个处于转动平衡的物体，关于任意点的顺时针力矩之和等于关于该点的逆时针力矩之和。
Σ MCW = Σ MACW
This condition, together with the condition for translational equilibrium (resultant force = 0 in all directions), ensures that a rigid body remains at rest.
该条件与平动平衡条件（各方向的合力均为零）一起，确保刚体保持静止。
In many problems, you will take moments about a support or a pivot where an unknown force acts, so that this unknown force has zero moment and is eliminated from the equation.
在许多问题中，你会对未知力作用的支点或支座取矩，这样该未知力的力矩为零，从而从方程中消去。
5. Resultant Force and Resultant Moment | 合力与合力矩
For a rigid body to be in static equilibrium, two conditions must be satisfied: the resultant force in any direction must be zero, and the resultant moment about any point must be zero.
刚体要保持静力平衡，必须满足两个条件：任意方向的合力必须为零，且关于任意点的合力矩必须为零。
This gives up to three independent equations for coplanar forces: ΣFx = 0, ΣFy = 0, and ΣM = 0 about a chosen point.
对于共面力系，这可以给出最多三个独立的方程：ΣFx = 0，ΣFy = 0，以及关于某选定点的 ΣM = 0。
Choosing the pivot wisely – for instance at a point where two unknown forces intersect – simplifies the moment equation and reduces the number of unknowns.
明智地选择支点——例如选在两个未知力交点处取矩——可简化力矩方程，减少未知数数目。
6. Centre of Gravity and Centre of Mass | 重心与质心
The centre of gravity (COG) of an object is the point through which the entire weight of the object appears to act. For uniform gravitational fields, the centre of gravity coincides with the centre of mass.
物体的重心是物体全部重量似乎所作用的点。在均匀引力场中，重心与质心重合。
In moment calculations, the weight of an object can be taken to act at its centre of gravity. This simplifies problems where distributed weights are involved.
在力矩计算中，物体的重量可以视为作用在其重心上。这简化了涉及分布重量的问题。
The position of the centre of mass of a system of particles can be found by taking moments about a point, using the total weight times the distance to the centre of mass equals the sum of individual weight moments.
通过取矩可以求出一个质点系的质心位置，即总重量乘以重心到某点的距离等于各重量力矩之和。
7. Centre of Gravity of Uniform Objects | 均匀物体的重心
For a uniform rod, the centre of gravity is at its midpoint. For a uniform lamina such as a rectangle, triangle or circle, the centre of gravity lies at the geometric centre.
对于均匀杆，重心在其中点。对于均匀薄片，如矩形、三角形或圆形，重心位于其几何中心。
When composite shapes are made from uniform material, the overall centre of mass can be found by treating each part as a point mass at its own centre of mass and using moments.
当组合形状由均匀材料制成时，可通过将每一部分视为位于自身质心的质点，并利用力矩来求出整体质心。
A common exam scenario is a beam with an additional weight attached. The combined centre of gravity shifts towards the heavier side.
常见的考题场景是梁上附加一个重物，组合重心会向较重的一侧移动。
8. Tilting and Stability | 倾斜与稳定度
An object placed on a surface will tilt if the line of action of its weight falls outside its base of support. The object is stable when the weight vector passes through the base.
放在表面上的物体，如果其重力的作用线超出支撑基础，就会倾斜。当重力矢量穿过支撑基础时，物体是稳定的。
In tilting problems, at the point of tilting, the reaction force from the surface is concentrated at the pivot edge and the moment of the weight about that edge causes rotation.
在倾斜问题中，处于倾斜临界点时，来自表面的反作用力集中在支点边缘，重力对该边缘的力矩会导致转动。
To prevent tilting, the maximum moment that can be applied by external forces must not exceed the restoring moment provided by the weight about the pivot edge.
为防止倾斜，外力能施加的最大力矩不得超过重力绕支点边缘提供的恢复力矩。
9. Equilibrium of Non-Concurrent Forces | 非共点力的平衡
When forces acting on a rigid body do not all intersect at a single point, they are non-concurrent. In such cases, both force and moment equilibrium must be applied.
当作用在刚体上的力不全都交汇于一点时，它们是非共点力。此时，必须同时应用力的平衡和力矩的平衡。
A typical problem involves a beam supported at two points with loads placed along its length. By taking moments about one support, the reaction at the other support can be found.
一个典型问题是梁在两点支撑，沿长度方向施加载荷。通过对其中一个支座取矩，可以求出另一支座的反力。
Always resolve forces into horizontal and vertical components, and write a moment equation about a point where forces intersect if possible, to simplify algebra.
始终将力分解为水平和竖直分量，并在可能的情况下对力交汇点写出力矩方程，以简化代数运算。
10. Worked Example: Ladder Problem | 典型例题：梯子问题
A uniform ladder of length L and weight W rests against a smooth vertical wall, with its foot on rough horizontal ground. The ladder makes an angle θ with the ground. Find the reaction forces and the minimum coefficient of friction to prevent slipping.
一个长度为 L、重量为 W 的均匀梯子斜靠在光滑竖直墙上，梯脚放在粗糙水平地面上。梯子与地面成 θ 角。求反作用力以及防止滑倒的最小摩擦系数。
The smooth wall exerts only a normal reaction Rwall. The ground exerts a normal reaction N and a friction force F. Taking moments about the foot of the ladder eliminates N and F, giving Rwall = (W/2) cot θ. Resolving forces horizontally and vertically then yields N = W and F = Rwall.
光滑墙只施加法向反力 Rwall。地面施加法向反力 N 和摩擦力 F。对梯脚取矩可消去 N 和 F，得到 Rwall = (W/2) cot θ。然后水平与竖直方向分解力得出 N = W，F = Rwall。
The friction must satisfy F ≤ μ N, so the minimum μ is (1/2) cot θ. This shows that a steeper ladder (larger θ) requires a smaller coefficient of friction.
摩擦力必须满足 F ≤ μ N，因此最小 μ 为 (1/2) cot θ。这表明梯子越陡（θ越大），所需的摩擦系数越小。
11. Worked Example: Combined Moments and Equilibrium | 典型例题：力矩与平衡的综合
A non-uniform plank of length 4 m and weight 200 N rests horizontally on two supports at its ends. Its centre of gravity is 1.5 m from the left end. A weight of 300 N is placed 1 m from the right end. Find the reactions at the supports.
一根长4 m、重200 N的非均匀木板水平放置在两端支座上。其重心距左端1.5 m。在距右端1 m处放置一个300 N的重物。求支座反力。
Let RL and RR be the reactions at the left and right supports. Taking moments about the left support: (200 N)(1.5 m) + (300 N)(3 m) = RR(4 m). Solving gives RR = 300 N. Then using ΣFy = 0: RL + 300 = 200 + 300, so RL = 200 N.
设左、右支座反力为 RL 和 RR。对左支座取矩：(200 N)(1.5 m) + (300 N)(3 m) = RR(4 m)。解得 RR = 300 N。然后利用 ΣFy = 0：RL + 300 = 200 + 300，所以 RL = 200 N。
Always check that the sum of vertical forces is zero and that the moment about a different point also yields zero to verify your solution.
始终检查竖直方向合力为零，并对另一点取矩验证结果为零，以确认解答正确。
12. Exam Tips and Common Mistakes | 考试技巧与常见错误
Always draw a clear, labelled diagram showing all forces and their distances from the chosen pivot. Mark the perpendicular distance, not just the line of the rod.
始终画一个清楚、标注完整的受力图，显示所有力及其到所选支点的距离。标记垂直距离，而不只是杆的方向。
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Do not forget to include the weight of the object itself acting at its centre of gravity. It is easy to miss when focusing on applied loads.
不要忘记包含作用在物体自身重心上的重量。当注意力集中在施加的载荷上时，很容易漏掉它。
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When a support is a roller or smooth contact, the reaction is perpendicular to the surface – resolve correctly.
当支座是滚轮或光滑接触时，反力垂直于表面——要正确分解。
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Be consistent with units: convert centimetres to metres if moments are expressed in N m, but keep consistent throughout.
单位要一致：如果力矩用 N m 表达，则将厘米转换为米，但整体保持一致。
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If a problem asks for the ‘maximum’ or ‘minimum’ value of a force before tilting or slipping, consider the limiting equilibrium condition.
如果题目要求倾斜或滑动前的“最大”或“最小”力值，要考虑极限平衡条件。
Marking schemes often award marks for the correct moment equation even if arithmetic is wrong, so write the equation clearly before substituting numbers.
评分标准通常会给正确的力矩方程分数，即使算术错误，所以先写清楚方程再代入数值。
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