A-Level Maths: Forces on an Object on an Inclined Plane | A-Level 数学：斜面上的物体受力分析

📚 A-Level Maths: Forces on an Object on an Inclined Plane | A-Level 数学：斜面上的物体受力分析

When an object rests or moves on a sloping surface, its weight does not act perpendicular to the surface. Instead, it must be resolved into two components: one perpendicular to the plane, and one parallel to the plane. This is one of the most tested topics in A-Level Mechanics.

当一个物体在斜面上静止或运动时，它的重力并不垂直于斜面。我们必须将重力分解为两个分量：一个垂直于斜面，另一个平行于斜面。这是 A-Level 力学中最常考的考点之一。

1. Setting Up the Problem | 建立问题模型

Consider a particle of mass m kg on a plane inclined at an angle θ to the horizontal. The weight acts vertically downward with magnitude mg. We choose axes parallel and perpendicular to the plane, because motion, if any, occurs along the plane.

考虑一个质量为 m kg 的质点位于与水平方向成 θ 角的斜面上。重力竖直向下，大小为 mg。我们选择平行于斜面和垂直于斜面的坐标轴，因为如果有运动，运动方向是沿斜面的。

The weight mg makes an angle θ with the perpendicular to the plane. Therefore:

重力 mg 与斜面法线方向之间的夹角为 θ。因此：

Component perpendicular to plane: mg cos θ
Component parallel to plane: mg sin θ

The component mg sin θ pulls the object down the plane. The component mg cos θ presses the object into the plane.

分量 mg sin θ 将物体沿斜面向下拉；分量 mg cos θ 将物体压向斜面。

2. Normal Reaction Force | 法向支持力

The normal reaction R (or N) acts perpendicular to the plane, away from the surface. If there is no acceleration perpendicular to the plane, the forces in that direction balance:

法向支持力 R（或 N）垂直于斜面、背离表面方向。若物体在垂直于斜面方向无加速度，则该方向上的力平衡：

R = mg cos θ

Note that R is not equal to mg unless the plane is horizontal (θ = 0°). A common mistake is to write R = mg out of habit. On an inclined plane, the normal reaction is always smaller than the full weight.

注意：除非斜面水平（θ = 0°），否则 R 并不等于 mg。一个常见错误是习惯性地写成 R = mg。在斜面上，法向支持力总是小于物体的总重力。

3. Friction on an Inclined Plane | 斜面上的摩擦力

Friction acts parallel to the plane, opposing the direction of motion (or the direction in which the object would tend to move if the plane were smooth).

摩擦力平行于斜面作用，阻碍物体的运动方向（或阻碍物体在光滑斜面上将会运动的方向）。

For static friction, the maximum value is given by:

对于静摩擦力，其最大值由下式给出：

F_max = μR

where μ is the coefficient of friction and R is the normal reaction. If the object is moving, the kinetic friction is usually taken as F = μR as well, at A-Level.

其中 μ 为摩擦系数，R 为法向支持力。若物体在运动，A-Level 中通常也取动摩擦力 F = μR。

4. Equilibrium on the Plane | 在斜面上的平衡

An object is in equilibrium if the resultant force on it is zero. On an inclined plane, this requires both:

物体所受合力为零时处于平衡状态。在斜面上，这要求以下两个方向同时满足：

Perpendicular to the plane: R = mg cos θ

垂直于斜面：R = mg cos θ
Parallel to the plane: F = mg sin θ (if the object is on the point of slipping down)

平行于斜面：F = mg sin θ（若物体刚好处于下滑临界状态）

If the object is at rest on a rough plane, friction must be exactly equal to mg sin θ, but cannot exceed μR. Hence the condition for the object to stay at rest is:

若物体在粗糙斜面上静止，摩擦力必须恰好等于 mg sin θ，但不能超过 μR。因此物体保持静止的条件为：

mg sin θ ≤ μ mg cos θ → tan θ ≤ μ

This result is independent of mass. It is a classic exam fact: a particle remains at rest on a rough plane as long as the angle of inclination is less than or equal to the angle of friction, where tan α = μ.

这一结果与质量无关。这是一个经典考点：只要斜面的倾角小于或等于摩擦角 α（其中 tan α = μ），物体就能在粗糙斜面上保持静止。

5. Acceleration Down a Smooth Plane | 光滑斜面上的下滑加速度

If the plane is smooth (no friction), the only force along the plane is mg sin θ. By Newton’s Second Law:

若斜面光滑（无摩擦），沿斜面方向唯一的力是 mg sin θ。根据牛顿第二定律：

mg sin θ = ma → a = g sin θ

Thus, on a smooth plane, every object slides down with the same acceleration, regardless of its mass.

因此，在光滑斜面上，任何物体下滑的加速度都相同，与质量无关。

6. Acceleration Down a Rough Plane | 粗糙斜面上的下滑加速度

If the plane is rough and the object is sliding down, friction acts up the plane. The resultant force down the plane is:

若斜面粗糙且物体向下滑动，摩擦力沿斜面向上。沿斜面向下的合力为：

mg sin θ − F = ma

Using F = μR = μmg cos θ:

代入 F = μR = μmg cos θ：

a = g (sin θ − μ cos θ)

This acceleration is positive only if sin θ > μ cos θ, i.e. tan θ > μ. This matches the equilibrium condition derived earlier: if tan θ ≤ μ, the object will not slide on its own.

只有当 sin θ > μ cos θ，即 tan θ > μ 时，加速度才为正。这与前面导出的平衡条件一致：若 tan θ ≤ μ，物体不会自行下滑。

7. Worked Example 1: Particle at Rest | 示例 1：静止质点

A particle of mass 5 kg rests on a rough plane inclined at 20° to the horizontal. The coefficient of friction is 0.4. Show that the particle remains at rest and find the friction force.

一个质量为 5 kg 的质点静止在倾角为 20° 的粗糙斜面上，摩擦系数为 0.4。证明该质点保持静止，并求出摩擦力。

First check the equilibrium condition:

首先检验平衡条件：

tan 20° ≈ 0.3640, μ = 0.4

Since tan 20° ≤ 0.4, the particle stays at rest.

因为 tan 20° ≤ 0.4，所以质点保持静止。

The friction force is exactly the component of weight down the plane:

摩擦力恰好等于重力沿斜面的分量：

F = mg sin θ = 5 × 9.8 × sin 20° ≈ 16.76 N

The normal reaction is R = mg cos 20° ≈ 46.05 N. The maximum possible friction is μR ≈ 18.42 N, which is greater than 16.76 N, so indeed the particle does not slip.

法向支持力 R = mg cos 20° ≈ 46.05 N。最大静摩擦力 μR ≈ 18.42 N，大于 16.76 N，因此质点确实不会滑动。

8. Worked Example 2: Particle Sliding Down | 示例 2：质点下滑

A block of mass 2 kg slides down a rough plane inclined at 30° to the horizontal. The coefficient of kinetic friction is 0.25. Find the acceleration of the block.

一个质量为 2 kg 的物块沿与水平方向成 30° 的粗糙斜面下滑，动摩擦系数为 0.25。求物块的加速度。

Resolve perpendicular to the plane:

沿垂直于斜面方向分解：

R = mg cos 30° = 2 × 9.8 × cos 30° ≈ 16.97 N

Friction F = μR ≈ 0.25 × 16.97 ≈ 4.24 N.

摩擦力 F = μR ≈ 0.25 × 16.97 ≈ 4.24 N。

Apply Newton’s Second Law down the plane:

沿斜面方向应用牛顿第二定律：

mg sin 30° − F = ma

2 × 9.8 × 0.5 − 4.24 = 2a

9.8 − 4.24 = 2a → a = 2.78 m s⁻²

Check using the formula: a = g(sin θ − μ cos θ) = 9.8(sin 30° − 0.25 cos 30°) ≈ 9.8(0.5 − 0.2165) ≈ 2.78 m s⁻². The two methods agree.

用公式验证：a = g(sin θ − μ cos θ) = 9.8(sin 30° − 0.25 cos 30°) ≈ 9.8(0.5 − 0.2165) ≈ 2.78 m s⁻²。两种方法结果一致。

9. Additional Applied Forces | 附加外力

Sometimes a force P is applied to the object, either up or down the plane. In such problems, resolve all forces along and perpendicular to the plane, and be careful with the direction of friction.

有时物体还受到一个外力 P 的作用，方向可能沿斜面向上或向下。解决这类问题时，将所有力沿斜面方向和垂直斜面方向分解，并注意摩擦力的方向。

For example, if a force P pulls the object up the plane at constant speed, friction acts down the plane. Then:

例如，若力 P 拉着物体沿斜面匀速上升，摩擦力沿斜面向下。此时：

P = mg sin θ + μmg cos θ

If the object is on the point of moving up but is still at rest, F = μR acts down the plane and this is the maximum static friction. If the object is on the point of slipping down, friction acts up the plane.

若物体处于即将向上运动的临界状态但仍静止，此时摩擦力 F = μR 沿斜面向下，且为最大静摩擦力。若物体处于即将下滑的临界状态，则摩擦力沿斜面向上。

10. Changing the Direction of Friction | 摩擦力的方向判断

Friction always opposes the motion or the tendency to move. On an inclined plane:

摩擦力总是阻碍运动或运动趋势。在斜面上：

If the object slides down, friction acts up the plane.

若物体向下滑动，摩擦力沿斜面向上。
If the object slides up, friction acts down the plane.

若物体向上滑动，摩擦力沿斜面向下。
If the object is at rest but on the point of slipping down, friction acts up the plane and F = μR.

若物体静止但处于下滑临界状态，摩擦力沿斜面向上且 F = μR。
If the object is at rest but on the point of moving up, friction acts down the plane and F = μR.

若物体静止但处于上滑临界状态，摩擦力沿斜面向下且 F = μR。

Drawing a clear free-body diagram is essential. Label every force: weight mg, normal reaction R, friction F, and any applied force P. Then resolve along the chosen axes.

画出清晰的受力分析图至关重要。标出每个力：重力 mg、法向支持力 R、摩擦力 F 以及任何外力 P。然后沿所选坐标轴分解。

11. Common Mistakes | 常见错误

Writing R = mg instead of R = mg cos θ. Always resolve perpendicular to the plane.

写成 R = mg 而不是 R = mg cos θ。一定要沿垂直于斜面方向分解。
Using the wrong friction direction. Friction opposes motion or tendency to move; draw it correctly on the diagram.

摩擦力方向判断错误。摩擦力阻碍运动或运动趋势；在图中正确标出。
Forgetting that when an object is at rest, friction is not necessarily μR; it is whatever value balances the other forces, up to μR.

忘记物体静止时摩擦力不一定等于 μR；它等于平衡其他力所需的数值，但不超过 μR。
Mixing up sin and cos. The component of weight parallel to the plane is mg sin θ; perpendicular is mg cos θ. Remember: when θ = 0, sin θ = 0 (no component down the plane) and cos θ = 1 (full weight perpendicular).

混淆 sin 和 cos。重力沿斜面的分量为 mg sin θ；垂直于斜面的分量为 mg cos θ。记住：当 θ = 0 时，sin θ = 0（沿斜面方向无分量），cos θ = 1（重力完全垂直于平面）。

12. Summary | 总结

For any inclined plane problem, follow these steps:

对于任何斜面问题，按以下步骤操作：

Draw a free-body diagram showing all forces.

画出受力分析图，标出所有力。
Choose axes parallel and perpendicular to the plane.

选择平行于斜面和垂直于斜面的坐标轴。
Resolve the weight into mg sin θ (down the plane) and mg cos θ (into the plane).

将重力分解为 mg sin θ（沿斜面向下）和 mg cos θ（垂直压向斜面）。
Write equations for forces perpendicular and parallel to the plane.

分别列出垂直于斜面和平行于斜面的力方程。
Use F ≤ μR for static cases, F = μR for limiting or kinetic cases, and apply F = ma where there is acceleration.

静止时用 F ≤ μR，临界或运动时用 F = μR；若有加速度，则应用 F = ma。

Mastering the inclined plane is central to A-Level Mechanics. With a clear diagram and careful resolution of forces, you can handle any question involving slopes, rough surfaces, friction, and applied forces.

掌握斜面上的受力分析是 A-Level 力学的核心。凭借清晰的受力图和仔细的力分解，你就能应对任何涉及斜面、粗糙表面、摩擦力和外力的题目。

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A-Level Maths: Forces on an Object on an Inclined Plane | A-Level 数学：斜面上的物体受力分析