A-Level Maths: Equilibrium of a Static Particle | A-Level 数学：静态质点的平衡条件

📚 A-Level Maths: Equilibrium of a Static Particle | A-Level 数学：静态质点的平衡条件

In mechanics, a particle is said to be in static equilibrium when it is at rest and remains at rest.

在力学中，当质点保持静止且始终静止时，称其处于静态平衡。

This is a central topic in the Mechanics section of A-Level Mathematics. You will be required to resolve forces, draw free-body diagrams, and apply the condition ΣF = 0 systematically.

这是 A-Level 数学力学部分的核心主题。你将被要求分解力、绘制受力分析图，并系统地应用 ΣF = 0 的平衡条件。

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1. Conditions for Equilibrium | 平衡条件

For a static particle, the vector sum of all external forces must be zero. In symbols:

对于静态质点，所有外力的矢量和必须为零。用符号表示为：

ΣF = 0

Since force is a vector, this is equivalent to the sum of components in any two perpendicular directions being zero.

由于力是矢量，这等价于任意两个互相垂直方向上的分量之和为零。

For convenience, we usually choose horizontal (x) and vertical (y) directions. The scalar conditions are:

为方便起见，我们通常选择水平（x）和竖直（y）方向。标量条件为：

方向 / Direction	平衡条件 / Equilibrium condition
水平 / Horizontal	ΣFₓ = 0
竖直 / Vertical	ΣFᵧ = 0

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2. Free-Body Diagrams | 受力分析图

Before solving any equilibrium problem, draw a free-body diagram (FBD) of the particle. This diagram isolates the particle and shows every force acting on it as an arrow.

在解决任何平衡问题之前，先画出质点的受力分析图（FBD）。该图将质点孤立出来，并将每个作用力用箭头表示出来。

Key forces to identify are weight (W = mg), normal reaction (R or N), tension (T), and friction (F).

需要识别的主要力有：重力（W = mg）、法向反作用力（R 或 N）、张力（T）和摩擦力（F）。

Make sure you label the forces clearly and indicate angles accurately.

确保清晰地标注各力，并准确标明角度。

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3. Resolving Forces | 力的分解

In many problems, forces act at angles. To apply the equilibrium equations, resolve each force into horizontal and vertical components using trigonometry.

在许多问题中，力以一定角度作用。为了应用平衡方程，需要使用三角函数将每个力分解为水平和竖直分量。

Fₓ = F cosθ, Fᵧ = F sinθ

Here, θ is the angle the force makes with the horizontal axis.

这里，θ 是力与水平轴之间的夹角。

For example, a tension T inclined at 30° above the horizontal has components T cos30° horizontally and T sin30° vertically.

例如，与水平方向成 30° 角的张力 T，其水平分量为 T cos30°，竖直分量为 T sin30°。

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4. Worked Example 1: Horizontal Surface | 示例 1：水平面

A particle of weight 20 N rests on a smooth horizontal surface. A horizontal force of 8 N acts to the right. Find the force needed to keep the particle in equilibrium, and the normal reaction from the surface.

一个重力为 20 N 的质点静止在光滑水平面上。一个大小为 8 N 的水平力向右作用。求保持质点平衡所需的力，以及表面对质点的法向反作用力。

Solution: For vertical equilibrium, the normal reaction R must balance the weight, so:

解：对于竖直方向的平衡，法向反作用力 R 必须平衡重力，因此：

R = 20 N

For horizontal equilibrium, the applied force must be balanced by an equal and opposite force. Hence the required force is 8 N to the left.

对于水平方向的平衡，施加的力必须由等大反向的力平衡。因此所需的力为 8 N，方向向左。

Thus, the particle remains at rest when a leftward force of 8 N is applied.

因此，当施加一个向左的 8 N 的力时，质点保持静止。

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5. Worked Example 2: Inclined Plane | 示例 2：斜面

A particle of mass 2 kg rests on a smooth plane inclined at 30° to the horizontal. Find the normal reaction and the force F required, parallel to the plane, to keep the particle in equilibrium.

一个质量为 2 kg 的质点静止在倾角为 30° 的光滑斜面上。求法向反作用力以及为使质点保持平衡而需要的平行于斜面的力 F。

Take g = 9.8 m/s². The weight is W = 2g = 19.6 N.

取 g = 9.8 m/s²。重力 W = 2g = 19.6 N。

Resolving perpendicular to the plane:

垂直于斜面方向分解：

R = W cos30° = 19.6 × 0.866 = 16.97 N

Resolving parallel to the plane:

平行于斜面方向分解：

F = W sin30° = 19.6 × 0.5 = 9.8 N

Thus the normal reaction is approximately 17 N, and an upward force of 9.8 N parallel to the plane is needed to keep the particle at rest.

因此，法向反作用力约为 17 N，需要沿斜面向上施加一个大小为 9.8 N 的力才能使质点保持静止。

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6. Worked Example 3: Suspended Particle | 示例 3：悬挂质点

A particle of mass 3 kg is suspended from two light strings that make angles of 40° and 50° with the horizontal. Find the tensions T₁ and T₂ in the strings.

一个质量为 3 kg 的质点由两根轻绳悬挂，两绳与水平方向的夹角分别为 40° 和 50°。求两根绳中的张力 T₁ 和 T₂。

Resolve horizontally: the horizontal components of the tensions must cancel.

水平方向分解：两个张力的水平分量必须相互抵消。

T₁ cos40° = T₂ cos50°

Resolve vertically: the sum of the vertical components balances the weight 3g = 29.4 N.

竖直方向分解：竖直分量之和平衡重力 3g = 29.4 N。

T₁ sin40° + T₂ sin50° = 29.4

Using cos40° ≈ 0.766, cos50° ≈ 0.643, sin40° ≈ 0.643, sin50° ≈ 0.766:

利用 cos40° ≈ 0.766，cos50° ≈ 0.643，sin40° ≈ 0.643，sin50° ≈ 0.766：

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