A-Level Mathematics: Analysis of Forces on Static Rigid Bodies | A-Level 数学：静态刚体的受力分析

📚 A-Level Mathematics: Analysis of Forces on Static Rigid Bodies | A-Level 数学：静态刚体的受力分析

In A-Level Mechanics, a rigid body is an object with definite shape and size. Analysing a static rigid body means finding the forces and moments that keep it completely at rest under any given loading.

在 A-Level 力学中，刚体是具有确定形状和尺寸的物体。分析静态刚体，就是找出使它在外载作用下完全保持静止所需的力与力矩。

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1. Force as a Vector | 力是矢量

A force has both magnitude and direction, so it is a vector. The magnitude is measured in newtons. In two dimensions, a force can be resolved into perpendicular components using its angle with a reference axis.

力同时具有大小和方向，因此是一个矢量。力的大小单位是牛顿。在二维问题中，可以通过力与参考轴的夹角把力分解为互相垂直的分量。

If a force F acts at an angle θ to the x-axis, its components are:

若力 F 与 x 轴夹角为 θ，则它的分量为：

Fₓ = F cos θ, Fᵧ = F sin θ, F = √(Fₓ² + Fᵧ²)

For a body at rest, the vector sum of all external forces must be zero. However, for a rigid body this is not enough: the position where each force acts also matters, because forces at different points create different turning effects.

对于静止物体，所有外力的矢量和必须为零。但对刚体来说，这还不够：每个力的作用点也很重要，因为力作用在不同位置上会产生不同的转动效果。

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2. Moment of a Force | 力的力矩

The moment of a force about a point is a measure of its turning effect. It depends on the magnitude of the force and the perpendicular distance from the point to the line of action of the force.

力对某一点的力矩衡量的是该力的转动效果。它取决于力的大小以及从该点到力的作用线的垂直距离。

M = F d

Here d is the perpendicular distance, and the unit is N m. If a force is not perpendicular to the object, the perpendicular component must be used.

其中 d 是垂直距离，单位是 N m。如果力不垂直于物体，就必须使用垂直方向上的分力。

In practice, for a force F applied at an angle θ to a rod of length r, the moment about one end is:

实际上，若长为 r 的杆端部受与杆成 θ 角的力 F 作用，则对另一端的力矩为：

M = F r sin θ

The sign of a moment is conventional: a clockwise moment can be called positive, or a counter-clockwise moment can be called positive. The important rule is to be consistent throughout the calculation.

力矩的正负是人为约定的：逆时针方向为正或顺时针方向为正都可以。重要的是在整个计算中保持一致。

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3. The Principle of Moments | 力矩原理

For a rigid body in equilibrium, the total clockwise moment about any point must equal the total counter-clockwise moment about that same point.

对于处于平衡状态的刚体，对任意一点取矩，顺时针方向的合力矩必然等于逆时针方向的合力矩。

∑M(clockwise) = ∑M(counter-clockwise)

Equivalently, if we assign signs to moments, the algebraic sum of all moments about any point is zero.

等价地说，如果为力矩规定正负号，则对任意点取矩，所有力矩的代数和为零。

The principle of moments is especially useful when a body is subject to concentrated loads, such as the weight of a rod, a tension in a cable, or a reaction at a support.

当物体受集中载荷作用时，例如杆的重力、缆绳的张力和支承反力，力矩原理尤其有用。

To simplify working, choose a point where one or more unknown forces act. A force that passes through the chosen point does not create a moment about that point.

为了简化计算，应选择某一个或多个未知力作用点为矩心。经过矩心的力不会对该点产生力矩。

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4. Equilibrium Conditions for a Rigid Body | 刚体的静态平衡条件

A static rigid body must satisfy three separate conditions:

静态刚体必须满足三个独立条件：

• No resultant horizontal force: ∑Fₓ = 0
• No resultant vertical force: ∑Fᵧ = 0
• No resultant moment about any point: ∑M = 0

• 水平方向的合力为零：∑Fₓ = 0
• 垂直方向的合力为零：∑Fᵧ = 0
• 对任意一点的合力矩为零：∑M = 0

These equations are the foundation of every static rigid-body problem. They can be solved to find unknown reactions, tensions, friction forces, or positions.

这三个方程是求解一切刚体静态问题的基石。利用它们可以求出未知反力、张力、摩擦力或作用位置。

Moment equilibrium can be taken about any convenient point, but the other two equations are always based on perpendicular components.

力矩平衡可以选任意方便的点，但另外两个方程始终基于互相垂直方向上的分力。

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5. Free-Body Diagrams | 受力图

Before writing any equations, always isolate the body and draw a free-body diagram. This diagram must include every external force acting on the body, with labels, directions, and the points where the forces act.

在列方程之前，必须先隔离物体并绘制受力图。受力图必须包括作用在物体上的所有外力，并标出大小、方向和力的作用点。

For a uniform rod, the weight acts at the midpoint of the rod. For a rectangular lamina, the weight acts at the centre of the lamina. This point is called the centre of mass.

对匀质杆，重力作用于杆的中点。对矩形薄板，重力作用于薄板的中心。这一点称为质心。

Internal forces are not drawn on a free-body diagram. For example, normal forces inside a rod or stress within a material must be ignored; only forces from supports, gravity, friction, and applied loads are included.

内力不画在受力图上。例如，杆内部的正压力和材料内部的应力都应忽略；只画来自支承体、重力、摩擦力和外载荷的力。

A good diagram should also show all relevant lengths and angles. Without these dimensions, moment equations cannot be written correctly.

好的受力图还应当标出所有相关的长度和角度。没有这些尺寸，就无法正确地写出力矩方程。

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6. Support Reactions and Smooth Contacts | 支承反力与光滑接触

Different supports allow different reaction forces. In A-Level mechanics, the most common supports are smooth contacts, rollers, and hinges.

不同类型的支承会产生不同的反力。在 A-Level 力学中，最常见的支承是光滑接触面、滚子和铰链。

Support 支承	Reaction 反力
Smooth contact 光滑接触	One force perpendicular to the surface 一个垂直于接触面的力
Roller 滚子	One force normal to the supporting surface 一个垂直于支承面的力
Hinge or pin 铰链或销钉	Two perpendicular components, often X and Y 两个互相垂直的分量，通常写成 X 和 Y

For a smooth wall or smooth floor, there is no friction. The contact force acts only along the normal direction.

对于光滑墙面或光滑地面，没有摩擦力。接触力只沿法线方向作用。

For a hinge connected to a support, the reaction has an unknown direction. It is normally written as two components: a horizontal component X and a vertical component Y.

对于连接在支座上的铰链，反力方向未知。通常将其写成两个分量：水平分量 X 和垂直分量 Y。

When a light string or cable supports a rod, the tension acts along the string, pulling away from the rod.

当轻绳或缆绳支承杆时，张力沿绳的方向作用，并且拉离杆。

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7. Friction and Limiting Equilibrium | 摩擦力与极限平衡

Friction acts along a rough surface and opposes the tendency of a body to slide. In static problems, the magnitude of friction can be any value from zero up to a maximum value.

摩擦力沿粗糙表面作用，并阻碍物体滑动的趋势。在静态问题中，摩擦力的大小可以是零到最大值之间的任意值。

F ≤ μR

Here F is the friction force, R is the normal reaction force, and μ is the coefficient of friction between the two surfaces.

其中 F 是摩擦力，R 是法向反力，μ 是两个接触面之间的摩擦系数。

When the body is on the point of slipping, friction is at its maximum value and the condition becomes:

当物体处于即将滑动的临界状态时，摩擦力达到最大值，此时条件变为：

F = μR

This is called limiting equilibrium. In exam questions, the phrase “limiting friction” or “on the point of slipping” usually signals that the equality must be used.

这种状态称为极限平衡。在考题中，出现“最大静摩擦力”或“即将滑动”等短语时，通常表示必须使用等式。

If the question only asks for the minimum required coefficient, use the inequality F ≤ μR to find μ ≥ F/R.

如果题目只要求最小摩擦系数，则使用不等式 F ≤ μR 来求 μ ≥ F/R。

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8. Ladder Problems | 梯子类问题

The classic static rigid-body problem is a uniform ladder resting against a smooth vertical wall on rough horizontal ground.

经典的刚体静态问题是：一把匀质梯子靠在光滑的竖直墙上，梯脚放在粗糙的水平地面上。

Let the ladder have length L, weight W, and make an angle θ with the ground. A smooth wall exerts a normal reaction N_w at the top of the ladder. The rough ground exerts an upward normal reaction R and a horizontal friction force F at the foot.

设梯子的长度为 L，重量为 W，并与地面成 θ 角。光滑墙在梯子顶端施加法向反力 N_w。粗糙地面在梯脚处施加竖直向上的法向反力 R 和水平摩擦力 F。

Resolving vertically gives:

竖直方向受力平衡得：

R = W

Resolving horizontally gives:

水平方向受力平衡得：

F = N_w

Taking moments about the foot of the ladder eliminates both R and F from the equation. The wall reaction acts at a perpendicular distance L sin θ from the foot, while the weight acts at a distance (L/2) cos θ:

对梯脚取矩可以消去 R 和 F。墙反力到梯脚的垂直距离是 L sin θ，重力到梯脚的垂直距离是 (L/2) cos θ：

N_w L sin θ = W (L/2) cos θ

Therefore:

因此：

N_w = (W/2) cot θ

This is the horizontal push of the wall on the ladder, and it is equal to the friction force F at the foot. If the floor is rough enough, the minimum coefficient of friction is:

这就是墙对梯子的水平推力，并且等于梯脚处的摩擦力 F。若地面足够粗糙，所需最小摩擦系数为：

μ = F/R = (W/2) cot θ ÷ W = (1/2) cot θ

As θ decreases, cot θ increases, so a ladder placed at a smaller angle to the ground requires more friction.

当 θ 减小时，cot θ 增大，因此梯子与地面的夹角越小，所需要的摩擦力就越大。

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9. Hinged Rods and Beams | 铰接杆与梁

A rod hinged to a support and held in equilibrium by a cable is another important exam situation. Taking moments about the hinge gives the tension directly, because the hinge reaction passes through that point.

一端由铰链固定、另一端由缆绳拉住的杆是另一种重要题型。对铰链取矩可以直接求出张力，因为铰链反力经过该点而不产生力矩。

Consider a uniform rod AB of length 4 m and weight 60 N, hinged at A and held horizontal by a light cable attached at B. The cable makes an angle 30° with the rod.

考虑一根长 4 m、重 60 N 的匀质杆 AB，A 端铰接，B 端由轻缆绳拉住并保持水平。缆绳与杆成 30° 角。

Take moments about A. If the tension in the cable is T, its vertical component is T sin 30°, and it acts at distance 4 m from A:

对 A 点取矩。若缆绳张力为 T，其垂直分量为 T sin 30°，作用在距 A 点 4 m 处：

T sin 30° × 4 = 60 × 2

Since sin 30° = 0.5:

因为 sin 30° = 0.5：

T × 0.5 × 4 = 120 ⇒ T = 60 N

Now resolve horizontally. The cable has horizontal component T cos 30° = 30√3 N. The hinge reaction component Aₓ must balance this force:

现在水平方向受力平衡。缆绳水平分量为 T cos 30° = 30√3 N。铰链反力水平分量 Aₓ 必须与之平衡：

Aₓ = 30√3 N

Resolving vertically:

竖直方向受力平衡：

Aᵧ + T sin 30° = 60 ⇒ Aᵧ + 30 = 60 ⇒ Aᵧ = 30 N

The magnitude of the hinge reaction is:

铰链反力的大小为：

A = √(Aₓ² + Aᵧ²) = √((30√3)² + 30²) = √(2700 + 900) = 60 N

The angle of the hinge reaction with the horizontal satisfies tan α = Aᵧ/Aₓ = 1/√3, so α = 30°.

铰链反力与水平方向的夹角满足 tan α = Aᵧ/Aₓ = 1/√3，因此 α = 30°。

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