Inclined Planes | 斜面

📚 Inclined Planes | 斜面

In Edexcel A-Level Mechanics, inclined planes appear whenever a particle or block rests on a surface at an angle to the horizontal. You need to resolve forces parallel and perpendicular to the plane, apply F = ma, and use the friction model F ≤ μR. This article covers the core methods, common question types, and exam tips.

在 Edexcel A-Level 力学中，只要物体或物块静止在倾斜表面上，就会出现斜面问题。你需要沿斜面平行方向和垂直方向分解力，应用 F = ma，并使用摩擦模型 F ≤ μR。本文涵盖核心方法、常见题型与应试技巧。

1. Resolving the Weight Component | 分解重力分量

The weight mg always acts vertically downwards. On an inclined plane at angle θ to the horizontal, replace mg by two components: mg sin θ down the slope and mg cos θ perpendicular to the plane.

重力 mg 始终竖直向下。在倾角为 θ 的斜面上，将 mg 分解为两个分量：沿斜面向下的 mg sin θ 和垂直于斜面的 mg cos θ。

The angle between the vertical and the normal to the plane is equal to the plane’s angle θ. This is why the components are mg sin θ parallel to the slope and mg cos θ perpendicular to it.

竖直方向与斜面法线之间的夹角等于斜面倾角 θ。因此沿斜面方向的分量是 mg sin θ，垂直斜面的分量是 mg cos θ。

Weight components: mg sin θ down slope | mg cos θ into plane

2. Normal Reaction and Friction | 法向反力与摩擦力

The normal reaction R is the force from the plane on the object. It balances the component of weight perpendicular to the plane when there is no other perpendicular force: R = mg cos θ.

法向反力 R 是斜面对物体的作用力。当没有其他垂直斜面的力时，它平衡重力的垂直分量：R = mg cos θ。

Friction F acts along the plane to oppose slipping. If the object is in limiting equilibrium or moving, F = μR; otherwise F can be smaller than μR and must be found from equilibrium.

摩擦力 F 沿斜面方向作用，阻碍滑动。若物体处于极限平衡或运动状态，F = μR；否则 F 可以小于 μR，必须根据平衡条件求出。

3. Equilibrium on an Inclined Plane | 斜面上的平衡

For an object at rest on a rough inclined plane, the resultant force is zero. Resolving parallel to the plane gives F = mg sin θ, provided no other parallel forces act.

对于静止在粗糙斜面上的物体，合外力为零。若没有其他平行于斜面的力，沿斜面方向分解可得 F = mg sin θ。

Resolving perpendicular to the plane gives R = mg cos θ. The friction is not necessarily μR; it is just enough to prevent sliding.

垂直斜面方向分解得到 R = mg cos θ。此时摩擦力不一定是 μR；它的大小刚好足以阻止滑动。

For equilibrium, friction must satisfy F ≤ μR. Substituting gives tan θ ≤ μ, so the object remains at rest if the slope angle is small enough.

平衡时，摩擦力必须满足 F ≤ μR。代入可得 tan θ ≤ μ，因此只要斜面倾角足够小，物体就能保持静止。

4. Newton’s Second Law Along the Slope | 沿斜面的牛顿第二定律

When an object accelerates along a smooth inclined plane, the resultant force down the slope is mg sin θ. From F = ma, the acceleration is a = g sin θ.

当物体在光滑斜面上加速运动时，沿斜面向下的合力为 mg sin θ。由 F = ma 可得加速度 a = g sin θ。

The acceleration is independent of the mass. A 1 kg block and a 10 kg block released from rest on the same smooth plane accelerate at exactly the same rate.

加速度与质量无关。在同一光滑斜面上，从静止释放 1 kg 和 10 kg 的物块，两者的加速度完全相同。

a = g sin θ (smooth plane)

5. Friction and the Coefficient of Friction | 摩擦与摩擦系数

On a rough plane, if the block moves down the slope, friction acts up the slope. The resultant force is mg sin θ − μR. Since R = mg cos θ, F = ma gives a = g(sin θ − μ cos θ).

在粗糙斜面上，如果物块向下滑动，摩擦力沿斜面向上。合力为 mg sin θ − μR。由于 R = mg cos θ，由 F = ma 可得 a = g(sin θ − μ cos θ)。

If the block is projected up the rough plane, both the weight component mg sin θ and friction μR act down the slope. The acceleration magnitude is a = g(sin θ + μ cos θ), directed down the plane.

若物块沿粗糙斜面向上抛出，则重力分量 mg sin θ 和摩擦力 μR 都沿斜面向下。加速度大小为 a = g(sin θ + μ cos θ)，方向沿斜面向下。

Moving down: a = g(sin θ − μ cos θ)

Moving up: a = g(sin θ + μ cos θ)

6. Angles of Friction and Limiting Equilibrium | 摩擦角与极限平衡

Limiting equilibrium occurs when the object is just about to slide. At this point F = μR and the forces are still in equilibrium. The maximum angle for equilibrium on a rough plane is called the angle of friction, given by tan θ = μ.

极限平衡发生在物体刚好将要滑动的瞬间。此时 F = μR，但力系仍处于平衡。粗糙斜面上能保持平衡的最大倾角称为摩擦角，满足 tan θ = μ。

At any smaller angle, friction is less than μR. At any larger angle, the object accelerates down the plane because mg sin θ exceeds μ mg cos θ.

任何较小的倾角下，摩擦力都小于 μR；任何较大的倾角下，由于 mg sin θ 超过 μ mg cos θ，物体会沿斜面加速下滑。

Limiting equilibrium on rough plane: tan θ = μ

7. Acceleration Down and Up the Plane | 沿斜面下滑与上滑的加速度

For an object released from rest on a rough plane, the acceleration is a = g(sin θ − μ cos θ). If sin θ > μ cos θ, the object accelerates downward; if sin θ < μ cos θ, it stays at rest.

物体在粗糙斜面上从静止释放时，加速度为 a = g(sin θ − μ cos θ)。若 sin θ > μ cos θ，物体向下加速；若 sin θ < μ cos θ，物体保持静止。

For an object given an initial velocity up the plane, the acceleration during upward motion is a = −g(sin θ + μ cos θ) if taking up the slope as positive. This larger deceleration is why an object stops more quickly going up than coming down.

若物体获得沿斜面向上的初速度，则上滑过程中的加速度为 a = −g(sin θ + μ cos θ)（取沿斜面向上为正）。这种更大的减速说明为什么物体上滑时比下滑时更快停下。

To find the distance travelled up the plane, use v² = u² + 2as with the upward acceleration being negative. The time can be found from v = u + at.

求物块沿斜面上滑的距离时，使用 v² = u² + 2as，其中向上加速度为负；时间可用 v = u + at 求出。

8. Connected Particles on Inclined Planes | 斜面上的连接体

Many Edexcel questions combine an inclined plane with a particle hanging vertically over a pulley. Draw a clear diagram and treat each particle separately, writing F = ma in the direction of motion for each.

许多 Edexcel 题目将斜面与通过滑轮竖直悬挂的物体结合起来。画出清晰的力图，分别对每个物体在运动方向上应用 F = ma。

For a mass m on a smooth plane connected by a light inextensible string to a mass M hanging freely, assume the string is taut and the acceleration is the same for both particles. The tension T is also the same on both sides of a smooth pulley.

对于光滑斜面上的质量 m 通过轻质不可伸长细绳与自由悬挂的质量 M 相连的情况，假设细绳张紧，两个物体的加速度相同；光滑滑轮两侧的张力 T 也相同。

Write the equation for the hanging mass: Mg − T = Ma. For the mass on the plane: T − mg sin θ = ma. Adding eliminates T and gives the common acceleration.

对悬挂物写出方程：Mg − T = Ma。对斜面上的物块写出：T − mg sin θ = ma。两式相加可消去 T，得到共同加速度。

Common acceleration: a = (Mg − mg sin θ) ÷ (M + m)

9. Worked Example: Block Held by a String | 例题：绳子拉住的物块

A 4 kg block rests on a rough plane inclined at 30° to the horizontal. It is held in equilibrium by a string parallel to the slope. The coefficient of friction is 0.5. Find the possible values of tension T.

一个 4 kg 的物块静止在倾角为 30° 的粗糙斜面上，被一根平行于斜面的绳子拉住。摩擦系数为 0.5。求张力 T 的可能取值范围。

Weight component down slope = 4g sin 30° = 2g. Normal reaction R = 4g cos 30° = 2√3 g. Maximum friction = μR = 0.5 × 2√3 g = √3 g.

重力沿斜面向下分量 = 4g sin 30° = 2g。法向反力 R = 4g cos 30° = 2√3 g。最大摩擦力 = μR = 0.5 × 2√3 g = √3 g。

When the block is about to slip down, T + F = 2g with F = √3 g, so T_min = 2g − √3 g. When the block is about to slip up, T = 2g + F_max, so T_max = 2g + √3 g.

当物块即将向下滑动时，T + F = 2g 且 F = √3 g，因此 T_min = 2g − √3 g。当物块即将向上滑动时，T = 2g + F_max，因此 T_max = 2g + √3 g。

Taking g = 9.8 m s⁻² gives T_min ≈ 2.62 N and T_max ≈ 36.58 N. The string tension may be any value between these limits.

取 g = 9.8 m s⁻²，得到 T_min ≈ 2.62 N，T_max ≈ 36.58 N。绳子的张力可以是这两个界限之间的任意值。

10. Common Mistakes and Exam Tips | 常见错误与应试技巧

Always draw a force diagram showing weight, normal reaction, friction, tension, and any applied force. Do not draw ‘mg sin θ’ as an extra force; it is a component of weight.

始终画出受力图，标出重力、法向反力、摩擦力、张力和任何外力。不要把 mg sin θ 当成额外的力；它只是重力的分量。

Match the direction of friction to the direction of actual or potential motion. If friction is drawn in the wrong direction, the acceleration equation will be incorrect.

摩擦力的方向必须与实际运动或可能运动的方向相反。如果摩擦力方向画错，加速度方程就会出错。

Use consistent positive directions for connected particles. If the hanging mass moves down, the block on the plane moves up; write the equations so the same acceleration a appears with the same sign.

对连接体要使用一致的正方向。如果悬挂物向下运动，斜面上的物块就向上运动；写方程时要让同一个加速度 a 以相同符号出现。

Draw a clear force diagram before writing any equations. | 写任何方程之前先画出清晰的受力图。
Check friction direction against possible motion. | 根据可能的运动方向检查摩擦方向。
Do not forget R = mg cos θ when friction is involved. | 涉及摩擦时不要忘记 R = mg cos θ。
In limiting equilibrium problems, test both slipping down and slipping up. | 在极限平衡问题中，要分别检验向下滑动和向上滑动。

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Inclined Planes | 斜面