📚 Modelling with Statics | 静力学建模
Statics is the branch of mechanics that deals with objects in equilibrium, whether they are at rest or moving with constant velocity. In Edexcel A-Level Mechanics, modelling with statics brings together forces, moments, friction and equilibrium conditions. A good model strips away unnecessary detail while keeping the essential features needed to predict whether a particle or rigid body will remain in balance.
静力学是力学中研究物体平衡状态的分支，平衡既包括静止也包括匀速直线运动。在 Edexcel A-Level 力学中，静力学建模将力、力矩、摩擦和平衡条件整合在一起。一个好的模型会剔除不必要的细节，但保留预测质点或刚体是否能保持平衡所必需的关键特征。
1. The Role of Statics in Mechanics | 静力学在力学中的作用
In statics, we assume a system is in equilibrium, so the net force and net moment acting on it are both zero. This principle underpins the design of bridges, shelves, ladders, cranes and many other everyday structures. By modelling real objects as particles or rigid bodies, we can calculate unknown forces and decide whether a structure or object is safe.
在静力学中，我们假设系统处于平衡状态，因此作用在其上的合外力与合力矩均为零。这一原理是桥梁、架子、梯子、起重机以及许多日常结构设计的基础。通过将真实物体建模为质点或刚体，我们可以计算未知力并判断结构或物体是否安全。
Statics contrasts with dynamics, where objects accelerate and Newton’s second law gives F = ma. In statics problems, acceleration is zero, so the equations of motion become simpler, but the geometry of forces and moments is often more demanding. Careful resolution of forces and correct choice of moment points are therefore central skills.
静力学与动力学不同，动力学中的物体有加速度，需要应用牛顿第二定律 F = ma。静力学问题中加速度为零，所以运动方程更简单，但力和力矩的几何关系往往更难处理。因此，准确分解力和正确选择取矩点是核心技能。
2. Modelling Assumptions in Statics | 静力学中的建模假设
Before solving any statics problem, you must identify the modelling assumptions from the wording of the question. Words such as ‘light’, ‘smooth’, ‘inextensible’, ‘uniform’ and ‘particle’ tell you which forces can be ignored, where weight acts, and whether rotation needs to be considered.
在解决任何静力学问题之前，必须从题干中识别建模假设。诸如 ‘light’（轻质）、’smooth’（光滑）、’inextensible’（不可伸长）、’uniform’（均匀）和 ‘particle’（质点）等词语会告诉你哪些力可以忽略、重力作用在哪里以及是否需要考虑转动。
| Assumption | 中文含义 | Modelling consequence |
|---|---|---|
| Particle | 质点 | All forces act at a single point; moments and rotation are ignored. |
| Light string or rod | 轻绳或轻杆 | Its mass is negligible, so its weight is ignored. |
| Inextensible string | 不可伸长的绳 | Both connected bodies move with the same speed and acceleration. |
| Smooth surface | 光滑表面 | No friction acts along the contact surface. |
| Rough surface | 粗糙表面 | Friction may act, up to a maximum value of μR. |
| Uniform body | 均匀物体 | Its weight acts at the geometric centre. |
| Rigid body | 刚体 | The body does not bend or deform; moments must be considered. |
3. Forces as Vectors | 力作为矢量
A force is a vector: it has both magnitude and direction. In statics, the common forces are weight, normal reaction, tension and friction. Weight, usually written as mg, acts vertically downwards through the centre of mass. The normal reaction acts perpendicular to a contact surface, while tension acts along a string or rod, pulling away from the object. Friction acts along a surface and opposes relative motion or the tendency to move.
力是矢量：它既有大小又有方向。静力学中常见的力包括重力、法向反作用力、张力和摩擦力。重力通常写作 mg，竖直向下作用在质心处。法向反作用力垂直于接触面，张力沿绳或杆方向并拉离物体。摩擦力沿接触面方向，阻碍相对运动或运动趋势。
When representing forces on a diagram, draw arrows from the point of application in the correct direction and label each force clearly. Accurate free-body diagrams are the foundation of successful statics solutions, because a missing or incorrect force can invalidate all subsequent equations.
在受力图上表示力时，应从作用点沿正确方向画出箭头，并清楚标注每个力。准确的自体受力图是成功求解静力学问题的基础，因为遗漏或画错任何一个力都可能使后续所有方程失效。
4. Resolving Forces and Components | 力的分解与分量
To apply equilibrium conditions in two dimensions, resolve each force into perpendicular components. If a force P acts at an angle θ to the horizontal, its horizontal component is P cos θ and its vertical component is P sin θ. This process lets you convert a two-dimensional vector problem into two separate one-dimensional equations.
为了在二维情况下应用平衡条件，需要将每个力分解为相互垂直的分量。如果力 P 与水平方向成 θ 角，则其水平分量为 P cos θ，竖直分量为 P sin θ。这一过程可以将二维矢量问题转化为两个独立的一维方程。
Horizontal component = P cos θ, Vertical component = P sin θ
On a slope inclined at an angle α to the horizontal, it is often easier to resolve the weight parallel and perpendicular to the plane. The component down the slope is mg sin α, and the component perpendicular to the slope is mg cos α. This choice of axes often simplifies friction and normal reaction calculations.
在倾角为 α 的斜面上，通常更方便将重力沿斜面方向和垂直斜面方向分解。沿斜面向下的分量为 mg sin α，垂直斜面的分量为 mg cos α。这样选择坐标轴通常能简化摩擦力和法向反作用力的计算。
Parallel to slope = mg sin α, Perpendicular to slope = mg cos α
5. Equilibrium of a Particle | 质点的平衡
For a particle in equilibrium, the vector sum of all forces acting on it is zero. In component form, this means the sum of horizontal components is zero and the sum of vertical components is zero. These two scalar equations can be solved simultaneously to find unknown forces or angles.
对于处于平衡状态的质点，作用在其上的所有力的矢量和为零。用分量形式表示，就是所有水平分量之和为零，所有竖直分量之和也为零。可以联立这两个标量方程来求解未知力或未知角度。
ΣFx = 0 and ΣFy = 0
Because a particle has negligible size, all forces act through the same point, so there is no turning effect to consider. This means moments are not needed for particle equilibrium problems. However, as soon as an object is modelled as a rigid body, you must also consider rotational equilibrium.
由于质点的大小可以忽略，所有力都通过同一点作用，因此不需要考虑转动效应。这意味着质点平衡问题不需要使用力矩。然而，一旦将物体建模为刚体，就必须同时考虑转动平衡。
6. The Triangle and Polygon of Forces | 力的三角形与多边形
If three coplanar forces act on a particle in equilibrium, their vectors can be arranged head-to-tail to form a closed triangle. This geometric representation is especially useful when the angle between forces is known, because you can then apply the sine rule or cosine rule to find unknown magnitudes.
如果三个共面力作用在处于平衡状态的质点上，它们的矢量可以首尾相接构成一个闭合三角形。当已知力与力之间的夹角时，这种几何表示特别有用，因为可以利用正弦定理或余弦定理求解未知力的大小。
For more than three forces, the vectors form a closed polygon. This is a direct consequence of zero resultant force. Although component resolution is often more systematic, the polygon method can provide quick visual checks and is sometimes required by examination questions that ask for scale drawings or geometric solutions.
对于三个以上的力，矢量会构成闭合多边形。这是合力为零的直接结果。虽然分量分解法通常更系统化，但多边形法可以提供快速的直观检验，有时考试中会要求用比例图或几何方法求解。
7. Friction and the Coefficient of Friction | 摩擦与摩擦系数
Friction acts between two rough surfaces to oppose sliding. For two surfaces in contact, the frictional force F can take any value up to a maximum value of μR, where R is the normal reaction and μ is the coefficient of friction. The inequality F ≤ μR holds whenever an object is not moving or about to move.
摩擦力作用在两个粗糙表面之间以阻碍滑动。对于两个接触表面，摩擦力 F 可以取从零到最大值 μR 之间的任意值，其中 R 为法向反作用力，μ 为摩擦系数。当物体没有运动或尚未即将运动时，始终满足不等式 F ≤ μR。
F ≤ μR
When an object is exactly at the point of sliding, the friction has reached its maximum possible value. This special case is called limiting equilibrium and is written as F = μR. In many problems, this equation provides the extra relationship needed to find μ, the angle of a slope, or a critical applied force.
当物体恰好处于即将滑动的临界状态时，摩擦力达到最大可能值。这种特殊情况称为极限平衡，写作 F = μR。在许多问题中，这个方程提供了额外的关系，用于求解 μ、斜面倾角或临界作用力。
At limiting equilibrium: F = μR
8. Limiting Equilibrium and Static Problems | 极限平衡与静力问题
Many examination questions state that a body is in ‘limiting equilibrium’ or is ‘on the point of sliding’. These phrases are signals that you should use F = μR, not the inequality. If the body is not on the point of sliding, friction is an unknown force that must be found from the equilibrium equations and then checked against the condition F ≤ μR.
许多考题会说明物体处于 ‘极限平衡’ 或 ‘即将滑动’ 状态。这些表述提示你应该使用 F = μR，而不是不等式。如果物体并非即将滑动，摩擦力就是一个未知力，需要通过平衡方程求出，然后再用 F ≤ μR 检验是否合理。
In static problems involving a rough inclined plane, a typical strategy is to resolve perpendicular to the plane to find R, then resolve parallel to the plane to find F. With limiting equilibrium, the equation F = μR then connects these two results and allows μ or the critical angle to be determined.
在涉及粗糙斜面的静力问题中，典型策略是先垂直斜面分解求出 R，再沿斜面分解求出 F。在极限平衡条件下，方程 F = μR 将这两个结果联系起来，从而求出 μ 或临界角度。
9. Moments of a Force | 力的力矩
The moment of a force about a point measures its turning effect. It is calculated as the product of the magnitude of the force and the perpendicular distance from the point to the line of action of the force. The SI unit of moment is the newton metre, written N m.
力对某点的力矩衡量该力的转动效果。它等于力的大小乘以从该点到力的作用线的垂直距离。力矩的国际单位是牛·米，写作 N m。
Moment = F × d
Moments can cause clockwise or anticlockwise rotation, so you must choose a sign convention and use it consistently. For equilibrium of a rigid body, the total clockwise moment about any point must equal the total anticlockwise moment about that same point. This is known as the principle of moments.
力矩可以使物体顺时针或逆时针转动，因此必须选择正负号约定并始终一致使用。对于刚
Published by TutorHao | A-Level Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导