Analysis of Friction Forces Acting on a Stationary Point Mass | 静止质点所受摩擦力分析

📚 Analysis of Friction Forces Acting on a Stationary Point Mass | 静止质点所受摩擦力分析

In A-Level Mechanics, the analysis of a stationary point mass under friction is one of the most frequently tested topics in the Statics component. Understanding how friction behaves when a body is at rest is essential, not merely for solving textbook exercises, but also for building a robust foundation for dynamics, energy and momentum problems. This article provides a systematic treatment of friction acting on a stationary particle, covering key principles, derivations and exam-style techniques.

在A-Level力学中，静止质点所受摩擦力的分析是静力学部分最常考查的内容之一。理解物体静止时摩擦力的行为方式，不仅对解决教科书习题至关重要，更是学习动力学、能量与动量问题的坚实基础。本文将系统地讲解静止颗粒所受摩擦力的相关原理、推导过程及应试技巧。

1. The Nature of Friction on a Stationary Body | 静止物体上摩擦力的性质

When a point mass rests on a rough surface, friction is a reaction force that arises from the microscopic roughness of the two contacting surfaces. The crucial point is that friction is not a fixed value; it adjusts itself to oppose any tendency of motion. If no force is applied, friction is zero. If a small horizontal force is applied, friction automatically matches that force in the opposite direction, provided the body remains at rest.

当质点静止于粗糙表面上时，摩擦力是由两个接触表面的微观粗糙度所产生的反作用力。关键之处在于：摩擦力并非固定值，它会自行调节以阻碍任何运动趋势。若无外力作用，摩擦力为零；若施加较小水平力，摩擦力会自动在反方向与该力等值，前提是物体保持静止。

This self-adjusting behaviour is the defining characteristic of static friction. Unlike kinetic friction, which has a roughly constant magnitude, static friction can take any value between zero and a well-defined upper limit. The upper limit is called the limiting friction.

这种自调节行为是静摩擦力的典型特征。与动摩擦力大小大致恒定不同，静摩擦力可以在零到某一明确上限之间取任意值，该上限即称为极限摩擦力。

2. Normal Reaction Force and the Rough Surface Model | 法向反作用力与粗糙面模型

Friction and the normal reaction are intimately connected. The normal reaction force R is the perpendicular contact force exerted by a surface on the body. For a point mass of mass m resting on a horizontal plane, with no other vertical forces acting, the surface must support the full weight, so

摩擦力与法向反作用力密切相关。法向反作用力 R 是表面对物体施加的垂直于接触面的力。对于放置在水平面上的质量为 m 的质点，若无其他竖直方向外力作用，表面必须完全支撑物体的重量，因此

R = mg

where g is the acceleration due to gravity. The symbol R is used conventionally in A-Level Mathematics; some syllabuses also denote it as N. The normal reaction is essential because the maximum possible friction is proportional to R: the rougher the surface or the heavier the body, the greater the friction available.

其中 g 为重力加速度。符号 R 是A-Level数学中的惯用记号，部分考纲也以 N 表示。法向反作用力至关重要，因为最大静摩擦力与 R 成正比：表面越粗糙或物体越重，可获得的最大摩擦力就越大。

When additional forces act on the particle, such as a vertical push or a component of an inclined force, the value of R changes. This is a common source of error, as students often assume R equals mg in every situation. Only on a horizontal plane with no vertical component of applied force does R equal mg exactly.

当质点受到额外力（如竖直推力或斜向力的分力）作用时，R 的值会发生改变。这是常见的错误来源——学生往往在任何情况下都默认 R = mg。只有在水平面上且施力无竖直分量时，R 才严格等于 mg。

3. Coefficient of Friction and Limiting Friction | 摩擦系数与极限摩擦力

The coefficient of friction, denoted by the Greek letter μ (mu), is a dimensionless quantity that measures the roughness of the two surfaces in contact. For two given surfaces, μ is treated as a constant, independent of the area of contact. The limiting friction F_max is then given by

摩擦系数用希腊字母 μ 表示，是无量纲量，用于度量两个接触表面的粗糙程度。对于给定的两个表面，μ 被视为常数，与接触面积无关。极限摩擦力 F_max 由下式给出：

F_max = μR

This relation is known as the law of friction and is a key formula in every A-Level mechanics syllabus. Throughout this article, μ is taken to be positive, and it is usually stated in the question as a given constant. All surfaces in A-Level problems are treated as ‘rough’ only when μ is provided; otherwise the surface is smooth and friction is taken as zero.

该关系式称为摩擦定律，是每个A-Level力学考纲中的核心公式。在本文中，μ 为正值，通常在题目中作为已知常数给出。只有当题目给出 μ 时，表面才被视为”粗糙”；否则视为光滑表面，摩擦力取为零。

It is vital to keep in mind that the actual static friction F is not always equal to μR. In a wide range of problems the particle is at rest with a friction force well below its limiting value. The equality only holds at the very point where motion is about to start—the limiting case.

必须牢记：实际静摩擦力 F 并不总是等于 μR。在许多问题中，质点处于静止状态，摩擦力远低于极限值。只有当物体即将开始运动的临界时刻，等式才成立。

4. The Equilibrium Condition | 平衡条件

A particle at rest on a rough surface is in equilibrium, which means that both the resultant force and the resultant moment (for a body with size) must be zero. For a point mass, only the resultant force condition matters, and it separates into the horizontal and vertical components:

静止于粗糙表面上的质点处于平衡状态，这意味着合力为零（对于有一定尺寸的物体还要求合力矩为零）。对质点而言，只需考虑合力条件，将其分解为水平与竖直两个方向：

ΣF_x = 0, ΣF_y = 0

In addition, the friction force must not exceed its limiting value. Thus, for a body at rest, the mathematical condition is

此外，摩擦力不得超出其极限值。因此，对静止物体，其数学条件为：

F ≤ μR

This inequality is fundamental. It tells us whether a given configuration is possible. When solving problems, we often first assume equilibrium and solve for the unknown friction F using Newton’s first law; we then verify that F ≤ μR. If the condition fails, the particle cannot remain at rest.

这一不等式是基础性的。它告诉我们某一受力配置是否能够维持静止。解题时，我们通常先假设平衡，利用牛顿第一定律解出未知摩擦力 F，再验证是否满足 F ≤ μR。若条件不满足，则质点无法保持静止。

Situation	Friction Value	Condition for Rest
No applied force	F = 0	Always at rest
Force applied, body still at rest	0 < F < μR	F balances applied force
On the point of motion	F = μR	Equality holds exactly

5. Stationary Mass on a Horizontal Plane | 水平面上的静止质点

Consider a particle of mass m resting on a rough horizontal plane, with a horizontal force P applied to it. Resolving vertically, the normal reaction is R = mg. Resolving horizontally, equilibrium requires

考虑一个质量为 m 的质点静止于粗糙水平面上，受到水平力 P 的作用。竖直方向分解得法向反作用力 R = mg；水平方向分解得平衡条件为：

F = P

For the particle to remain at rest, we require P ≤ μmg. The largest force that can be applied without moving the particle is therefore

为使质点保持静止，需满足 P ≤ μmg。因此，不使物体移动所能施加的最大力为：

P_max = μmg

This is a very common exam question. Note carefully that if P is less than this maximum, friction simply takes the value P and the particle stays at rest; if P exceeds the maximum, friction cannot grow further, and the particle accelerates in the direction of P. In the latter case, the problem transitions from statics to dynamics.

这是非常常见的考题。注意：若 P 小于该最大值，摩擦力恰好等于 P，质点保持静止；若 P 超过最大值，摩擦力无法继续增大，质点将在 P 方向上加速运动。此时问题从静力学过渡到动力学。

A useful extension occurs when a force is applied at an angle to the horizontal, either pushing downward or pulling upward. This alters the normal reaction. If a force P is applied at angle α above the horizontal, the vertical component P sin α reduces the normal reaction, so R = mg − P sin α. Conversely, if the force pushes downward at an angle α below the horizontal, R = mg + P sin α.

一个重要的扩展情形是力与水平方向成一定角度的情况——无论是斜向下推还是斜向上拉，都会改变法向反作用力。若力 P 与水平方向成 α 角向上，其竖直分量 P sin α 会减小法向反作用力，即 R = mg − P sin α；反之，若力斜向下推，则 R = mg + P sin α。

6. Stationary Mass on an Inclined Plane | 斜面上的静止质点

When a particle rests on a rough plane inclined at an angle θ to the horizontal, the weight must be resolved into components parallel and perpendicular to the plane. These components are mg sin θ (down the plane) and mg cos θ (into the plane), respectively.

当质点静止于与水平方向成 θ 角的粗糙斜面上时，需要将重力分解为沿斜面方向的分量 mg sin θ（沿斜面向下）和垂直斜面方向的分量 mg cos θ（压入斜面）。

Resolving perpendicular to the plane gives the normal reaction

沿垂直于斜面方向分解得法向反作用力为：

R = mg cos θ

Since the particle tends to slide down the plane, friction acts up the plane. At rest, resolving parallel to the plane gives

由于质点有沿斜面向下滑的趋势，摩擦力沿斜面向上。静止时，沿斜面方向分解得：

F = mg sin θ

The particle remains at rest if this required friction does not exceed the limiting value:

若所需摩擦力不超过极限值，质点即保持静止：

mg sin θ ≤ μmg cos θ ⇒ tan θ ≤ μ

This elegant result shows that the ability to rest on an inclined plane depends only on the angle of inclination and the coefficient of friction, not on the mass of the particle. For a given surface, there is a critical angle θ_crit = arctan μ. If θ exceeds this angle, the particle slides.

这一简洁结论表明：质点能否在斜面上保持静止，仅取决于斜面倾角与摩擦系数，而与质点质量无关。对给定表面，存在临界角 θ_crit = arctan μ。若倾角超过此临界角，质点将下滑。

7. External Force Applied at an Angle on an Inclined Plane | 斜面上斜向施力分析

Consider a particle on an inclined plane with an additional force P acting up the plane. In this case, resolving parallel to the plane gives F = mg sin θ − P. If P is less than mg sin θ, friction acts up the plane; if P is greater, the particle would tend to move up, so friction acts down the plane. The direction of friction is not arbitrary—it always opposes the direction of impending motion.

考虑斜面上的质点额外受到一个沿斜面向上的力 P。此时沿斜面方向分解得 F = mg sin θ − P。若 P 小于 mg sin θ，摩擦力沿斜面向上；若 P 大于 mg sin θ，质点将有向上运动的趋势，摩擦力方向变为沿斜面向下。摩擦力的方向并非随意，而是始终阻碍即将发生的运动方向。

This illustrates an important two-case structure in friction problems. The magnitude of F is the absolute value of the net driving force, and the condition for rest is that this magnitude is no greater than μR. Students must carefully determine the direction of friction before writing any equilibrium equation.

这体现了摩擦问题中常见的双情形结构。摩擦力 F 的大小等于净驱动力的绝对值，静止条件为这个值不大于 μR。学生必须先仔细判断摩擦力的方向，再写出平衡方程。

8. Systematic Problem-Solving Strategy | 系统解题策略

To solve any friction problem involving a stationary point mass, follow this structured approach, which mirrors the method expected in A-Level examinations.

要解决任何涉及静止质点的摩擦力问题，可遵循以下结构化步骤，这与A-Level考试所要求的解题方法完全一致。

Step 1: Draw a clear free-body diagram. Show all forces: weight, normal reaction, applied forces and friction. Mark the direction of friction opposing the tendency to move.

第一步：画出清晰的受力分析图。标出所有力：重力、法向反作用力、外力和摩擦力。摩擦力的方向应阻碍运动趋势。
Step 2: Resolve forces perpendicular to the surface. Use this to determine the normal reaction R in terms of known quantities.

第二步：沿垂直于表面方向分解力，由此确定法向反作用力 R 与已知量之间的关系。
Step 3: Resolve forces parallel to the surface. Assuming equilibrium, set the net force parallel to the surface equal to zero, and solve for the required friction F.

第三步：沿平行于表面方向分解力。假设平衡，令平行方向合力为零，解出所需摩擦力 F。
Step 4: Apply the friction inequality. Verify that F ≤ μR. If the inequality holds, the body is at rest; otherwise, equilibrium is impossible.

第四步：应用摩擦不等式。验证是否满足 F ≤ μR。若成立，物体保持静止；否则无法平衡。
Step 5: For limit questions (‘find the maximum force’), set F = μR as the boundary condition and solve the resulting equation.

第五步：对于求极值的问题（如”求最大力”），令 F = μR 作为临界条件，解所得方程即可。

9. Worked Example 1: Horizontal Plane | 典型例题一：水平面

Example. A particle of mass 5 kg rests on a rough horizontal plane. The coefficient of friction is 0.4. Find the maximum horizontal force that can be applied to the particle without causing it to move. Take g = 9.8 m s⁻².

例题。质量为 5 kg 的质点静止于粗糙水平面上，摩擦系数为 0.4。求不使质点移动所能施加的最大水平力。取 g = 9.8 m s⁻²。

Solution. Resolving vertically, since there is no vertical component of the applied force:

解答。竖直方向分解，由于施加的力无竖直分量：

R = mg = 5 × 9.8 = 49 N

The limiting friction is

极限摩擦力为：

F_max = μR = 0.4 × 49 = 19.6 N

Therefore the maximum horizontal force is 19.6 N. If the applied force is less than this, the particle remains at rest and friction takes exactly the value of the applied force.

因此最大水平力为 19.6 N。若施加的力小于此值，质点保持静止，摩擦力恰等于所施加的力。

10. Worked Example 2: Inclined Plane | 典型例题二：斜面

Example. A particle of mass 2 kg rests on a plane inclined at 30° to the horizontal. The coefficient of friction between the particle and the plane is 0.6. Determine whether the particle remains at rest, and find the friction force acting on it. Take g = 9.8 m s⁻².

例题。质量为 2 kg 的质点静止于与水平方向成 30° 角的斜面上，质点与斜面间的摩擦系数为 0.6。试判断质点能否保持静止，并求其所受摩擦力。取 g = 9.8 m s⁻²。

Solution. First, resolve perpendicular to the plane:

解答。首先沿垂直于斜面方向分解：

R = mg cos 30° = 2 × 9.8 × (√3 ⁄ 2) ≈ 16.97 N

Resolving parallel to the plane, the component of weight down the plane is

沿斜面方向分解，重力沿斜面向下的分量为：

mg sin 30° = 2 × 9.8 × 0.5 = 9.8 N

The maximum available friction is

可提供的最大摩擦力为：

μR = 0.6 × 16.97 ≈ 10.18 N

Since 9.8 N < 10.18 N, the particle remains at rest. The actual friction equals the down-the-plane weight component:

因为 9.8 N < 10.18 N，质点保持静止。实际摩擦力等于重力沿斜面的分量：

F = 9.8 N (acting up the plane) | F = 9.8 N（方向沿斜面向上）

Notice that F is below the limiting value; the particle is not on the point of sliding. The margin is 10.18 − 9.8 = 0.38 N.

注意 F 低于极限值，质点并未处于即将滑动的临界状态。裕量为 10.18 − 9.8 = 0.38 N。

11. Exam Tips and Common Misconceptions | 考试技巧与常见误区

The following points summarise the most frequent pitfalls encountered by A-Level candidates when dealing with friction on a stationary particle, along with the corresponding correct reasoning.

以下要点总结了A-Level考生在处理静止质点摩擦力问题时最常见的误区及对应的正确思路。

Misconception: ‘Friction always equals μR.’ Correction: Friction equals μR only at the limiting point. In equilibrium problems, first solve for F from force balance, then check F ≤ μR.

误区：“摩擦力始终等于 μR。”纠正：只有在临界状态下摩擦力才等于 μR。在平衡问题中，应先通过力平衡解出 F，再验证 F ≤ μR。
Misconception: ‘Normal reaction always equals mg.’ Correction: R depends on all forces with components perpendicular to the surface. Always resolve perpendicular to the plane first.

误区：“法向反作用力始终等于 mg。”纠正：R 取决于所有在垂直于表面方向上有分量的力。务必先沿垂直于平面的方向进行分解。
Misconception: ‘Friction always opposes the applied force.’ Correction: Friction opposes the direction of potential motion, which is determined by the net effect of all other forces parallel to the surface.

误区：“摩擦力总是与施加的力方向相反。”纠正：摩擦力阻碍的是潜在运动方向，该方向由所有平行于表面方向的其他力的合力决定。
Misconception: ‘If F < μR, we can increase F indefinitely.' Correction: F is a reaction determined by the other forces; it cannot be chosen arbitrarily. The inequality F ≤ μR is a restriction, not a formula.

误区：“若 F < μR，就可以无限增大 F。"纠正：F 是由其他力决定的反应力，不可随意选取。不等式 F ≤ μR 是一种限制条件，而非计算公式。

Finally, always check whether the direction of friction is correctly drawn before writing equations. A free-body diagram with incorrectly placed friction will produce a wrong answer despite correct algebra. Practice resolving forces on inclined planes until the decomposition of weight into mg sin θ and mg cos θ becomes automatic.

最后，在写方程之前务必检查摩擦力的方向是否绘制正确。若受力分析图中摩擦力方向标错，即使代数运算完全正确，答案也是错误的。反复练习斜面问题中的力分解，直至将重力分解为 mg sin θ 和 mg cos θ 成为本能反应。

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Analysis of Friction Forces Acting on a Stationary Point Mass | 静止质点所受摩擦力分析