A-Level Mathematics: Resolution of Forces | 力的分解方法

📚 A-Level Mathematics: Resolution of Forces | 力的分解方法

In A-Level Mechanics, forces are vector quantities. They have both magnitude and direction, so a single force acting at an angle can be replaced by two perpendicular forces that together have exactly the same effect. This process is called resolving a force, and it is one of the most important techniques in the whole of the mechanics syllabus.

在 A-Level 力学中，力是矢量。力既有大小又有方向，因此一个倾斜作用的力可以用两个互相垂直的分力来替代，这两个分力合起来与原力效果完全相同。这个过程称为力的分解，是整个力学考纲中最重要的技巧之一。

1. Why Resolve Forces? | 为什么要分解力？

Forces are often not aligned with the directions we care about. For example, a box pulled by a rope at an angle has a horizontal effect that moves it along the ground and a vertical effect that reduces the normal reaction from the ground. Without resolution, we cannot apply Newton’s second law separately in the horizontal and vertical directions.

力通常并不沿着我们关心的方向作用。例如，用绳子斜拉一个箱子时，拉力既有水平方向的效果使箱子沿地面移动，又有竖直方向的效果从而减小地面对箱子的支持力。如果不进行力的分解，就不能分别在水平方向和竖直方向应用牛顿第二定律。

Resolving a force means finding its components along chosen perpendicular axes. The original force is then replaced by two forces acting along those axes, and all the mathematics becomes much simpler.

分解一个力，就是求出它在选定互相垂直的坐标轴上的分力。原力被两个沿轴方向的分力替代后，所有计算都会变得简单得多。

2. The Basic Formula: Fₓ = F cos θ, Fᵧ = F sin θ | 基本公式：Fₓ = F cos θ, Fᵧ = F sin θ

Suppose a force of magnitude F makes an angle θ above the positive x-axis. The horizontal component and vertical component are given by the two basic formulae:

设一个大小为 F 的力与 x 轴正方向成 θ 角，则其水平分力和竖直分力由以下两个基本公式给出：

Fₓ = F cos θ

Fᵧ = F sin θ

The component adjacent to the angle always uses cosine, while the component opposite the angle uses sine. This is exactly the same as finding the adjacent and opposite sides of a right-angled triangle.

与角度相邻的分量使用余弦，与角度相对的分量使用正弦。这完全等同于在直角三角形中求邻边和对边。

If the angle is given with respect to the y-axis instead, simply swap the formulae: the component along the y-axis becomes F cos θ, and the component along the x-axis becomes F sin θ.

如果已知角是与 y 轴的夹角，只需交换公式：沿 y 轴的分量为 F cos θ，沿 x 轴的分量为 F sin θ。

3. Choosing the Axes | 选择坐标轴

You can choose any two perpendicular directions as axes, but a wise choice makes the algebra much shorter. The usual strategy is to choose one axis along the direction of motion or the direction in which the acceleration is known, and the other axis perpendicular to it.

你可以选择任意两个互相垂直的方向作为坐标轴，但聪明的选择能大大简化代数运算。常用策略是：一个轴沿着运动方向或已知加速度的方向，另一个轴与之垂直。

On a horizontal surface, use horizontal and vertical axes.
On an inclined plane, use axes parallel and perpendicular to the plane.
For a particle moving in a circle, use radial and tangential axes.

在水平面上，使用水平轴和竖直轴。
在斜面上，使用平行于斜面和垂直于斜面的轴。
对圆周运动的质点，使用径向和切向轴。

If a force is already along an axis, do not resolve it. Only forces that are at an angle to an axis need to be broken into components.

如果某个力已经沿某一坐标轴方向，就不要再去分解它。只有与坐标轴成角度的力才需要分解。

4. Components of Weight on an Inclined Plane | 斜面上的重力分量

One of the most common situations is a particle of mass m resting on a plane inclined at angle θ to the horizontal. The weight mg acts vertically downwards, which is not along either chosen axis.

最常见的情形之一是质量为 m 的质点静止在倾角为 θ 的斜面上。重力 mg 竖直向下，并不沿任何选定坐标轴。

When the axes are taken parallel and perpendicular to the plane, the weight must be resolved. The component perpendicular to the plane is mg cos θ, and the component down the plane is mg sin θ.

当坐标轴取为平行于斜面和垂直于斜面时，必须分解重力。垂直于斜面的分量为 mg cos θ，沿斜面向下的分量为 mg sin θ。

Component perpendicular to plane: mg cos θ

Component down the plane: mg sin θ

Notice that the angle θ of the plane appears in both components, but in opposite positions compared with the horizontal-force case. A common memory aid is: the weight component along the plane is always mg sin θ.

注意斜面的倾角 θ 在两个分量中都会出现，但位置与水平力的情况相反。一个常用的记忆法是：重力沿斜面的分量始终为 mg sin θ。

5. Equilibrium and Resultant | 平衡与合力

When a particle is in equilibrium, the resultant force on it is zero. This means that the sum of all components in any direction must be zero. Usually we write two equations:

当质点处于平衡状态时，它所受的合力为零。这意味着任意方向上的所有分力之和必须为零。通常我们写出两个方程：

ΣFₓ = 0

ΣFᵧ = 0

If the particle is accelerating, Newton’s second law is used instead. Along the direction of motion we write ΣF = ma, and perpendicular to the motion the acceleration is usually zero, so ΣF = 0.

如果质点正在加速，则应使用牛顿第二定律。沿运动方向写 ΣF = ma，垂直于运动方向通常加速度为零，因此 ΣF = 0。

The resultant of two or more forces can also be found by first resolving each force into components, adding the components, and then using Pythagoras’ theorem to find the magnitude:

求两个或多个力的合力，也可以先分解每个力，再分别相加各方向分量，最后用勾股定理求出合力大小：

R = √(Rₓ² + Rᵧ²)

The direction of the resultant is given by tan α = Rᵧ / Rₓ, where α is the angle from the x-axis.

合力方向由 tan α = Rᵧ / Rₓ 给出，其中 α 是与 x 轴的夹角。

6. Connected Particles and Tension | 连接体与张力

When two particles are connected by a light inextensible string over a smooth pulley, the tension is the same throughout the string. To solve such problems, draw a separate force diagram for each particle and resolve forces along the direction of motion for each.

当两个质点通过跨过光滑滑轮的轻绳连接时，绳中张力处处相等。解决这类问题时，需要对每个质点分别画受力图，并沿各自的运动方向分解力。

If one particle hangs vertically and the other rests on a rough horizontal table, the hanging particle’s weight provides the driving force, while friction opposes the motion of the particle on the table. Both particles have the same magnitude of acceleration because the string is inextensible.

如果一个质点竖直悬挂，另一个质点在粗糙水平桌面上滑动，则悬挂质点的重力提供驱动力，而摩擦力阻碍桌面上质点的运动。由于绳不可伸长，两个质点的加速度大小相等。

For a particle on a smooth inclined plane connected to a hanging particle, the component of weight down the plane, mg sin θ, may act with or against the motion. You must define a positive direction and then resolve every force along that direction.

对于斜面上与悬挂质点相连的光滑质点，重力沿斜面的分量 mg sin θ 可能与运动同向或反向。你必须先确定正方向，然后沿该方向分解所有力。

7. Friction and Normal Reaction | 摩擦力与法向反力

The normal reaction R acts perpendicular to the contact surface. Friction F acts along the surface, opposing motion or attempted motion. The maximum value of friction is given by:

法向反力 R 垂直于接触面。摩擦力 F 沿接触面作用，阻碍运动或运动趋势。最大摩擦力由下式给出：

F_max = μR

where μ is the coefficient of friction. If a particle is on the point of sliding, the friction is at its maximum value. If it is moving, the friction usually has this same maximum value, and the direction is opposite to the velocity.

其中 μ 是摩擦系数。如果质点处于即将滑动的临界状态，摩擦力达到最大值。如果质点正在运动，摩擦力通常也取这个最大值，且方向与速度方向相反。

Resolving forces perpendicular to the surface is essential to find R. On a horizontal surface, R is usually equal to the total downward component of all forces. On an inclined plane, R is usually equal to mg cos θ minus any other component perpendicular to the plane.

沿垂直于接触面的方向分解力是求 R 的关键。在水平面上，R 通常等于所有力向下的竖直分量之和。在斜面上，R 通常等于 mg cos θ 减去其他垂直于平面的分量。

Never assume R = mg unless the only vertical forces are the weight and the normal reaction. Any extra vertical component, such as a pulling force at an angle, changes R and therefore changes the friction.

除非竖直方向只有重力和法向反力，否则绝不能假设 R = mg。任何额外的竖直分量，例如斜向的拉力，都会改变 R，从而改变摩擦力。

8. Worked Example: Particle on a Rough Inclined Plane | 例题：粗糙斜面上的质点

A particle of mass 2 kg is held at rest on a rough plane inclined at 30° to the horizontal. The coefficient of friction is 0.4. Find the minimum horizontal force P required to prevent the particle from sliding down the plane.

一个质量为 2 kg 的质点静止在倾角为 30° 的粗糙斜面上，摩擦系数为 0.4。求阻止质点沿斜面下滑所需的最小水平力 P。

Take axes parallel and perpendicular to the plane. The weight has components mg sin 30° down the plane and mg cos 30° into the plane.

取平行于斜面和垂直于斜面的坐标轴。重力的分量为沿斜面向下的 mg sin 30° 和垂直于斜面向里的 mg cos 30°。

mg = 2 × 9.8 = 19.6 N

mg sin 30° = 19.6 × 0.5 = 9.8 N

mg cos 30° = 19.6 × 0.866 = 16.97 N

The horizontal force P must be resolved parallel and perpendicular to the plane. If P acts horizontally into the plane, its component up the plane is P cos 30° and its component perpendicular to the plane is P sin 30°.

水平力 P 必须沿平行和垂直于斜面两个方向分解。若 P 水平压向斜面，则沿斜面向上的分量为 P cos 30°，垂直于斜面的分量为 P sin 30°。

Perpendicular to the plane, there is no acceleration, so:

垂直于斜面方向没有加速度，因此：

R = mg cos 30° + P sin 30°

At the point when the particle is just about to slide down, friction acts up the plane at its maximum value μR. For equilibrium parallel to the plane:

当质点刚好要下滑时，摩擦力沿斜面向上并达到最大值 μR。沿斜面方向的平衡方程为：

P cos 30° + μR = mg sin 30°

Substitute R:

代入 R：

P cos 30° + 0.4(16.97 + 0.5P) = 9.8

0.866P + 6.788 + 0.2P = 9.8

1.066P = 3.012

P = 2.83 N

So the minimum horizontal force needed is about 2.83 newtons.

因此所需最小水平力约为 2.83 牛顿。

9. Common Mistakes | 常见错误

Many students lose marks in mechanics not because they cannot solve the equations, but because they resolve forces incorrectly. Here are the most common errors and how to avoid them.

许多学生在力学中失分，不是因为不会解方程，而是因为力的分解出错。以下是最常见的错误及避免方法。

Using sine instead of cosine: always check which component is adjacent to the given angle.
Forgetting to resolve a force that is at an angle to the axes.
Assuming R = mg on an inclined plane or when extra vertical forces exist.
Drawing the weight component down the plane as mg cos θ instead of mg sin θ.
Choosing inconsistent axes for different particles in a connected system.
Ignoring the direction of friction when the motion is about to change direction.

把正弦和余弦用反：始终检查哪个分量与已知角相邻。
忘记分解与坐标轴成角度的力。
在斜面上或存在额外竖直力时假设 R = mg。
把重力沿斜面分量误写成 mg cos θ，而不是 mg sin θ。
在连接体问题中对不同质点选择不一致的坐标轴。
当运动方向即将改变时忽视摩擦力的方向。

A useful habit is to label every angle clearly on your force diagram. If an angle is between the force and the x-axis, the x-component is F cos θ. If the angle is between the force and the y-axis, the y-component is F cos θ.

一个有用的习惯是在受力图中清楚标出每个角度。如果角在力与 x 轴之间，则 x 分量为 F cos θ。如果角在力与 y 轴之间，则 y 分量为 F cos θ。

10. Practice Strategies for Exams | 备考策略

To master resolution of forces, practise drawing force diagrams before doing any algebra. A correct diagram is worth half the marks. Then resolve every force systematically along the two chosen axes, and write down the two component equations.

要熟练掌握力的分解，应先在动手计算之前练习画受力图。正确的受力图等于拿到一半分数。然后系统地沿两个选定坐标轴分解所有力，并写出两个分量方程。

Work through past paper questions involving inclined planes, connected particles, and friction. Pay special attention to questions where the direction of friction is not obvious, such as when an external force is just strong enough to move a particle up or down the plane.

认真做涉及斜面、连接体和摩擦力的历年真题。特别注意那些摩擦力方向不明确的问题，例如外力刚好能使质点沿斜面向上或向下运动的情况。

Finally, check your answers using units and common sense. A force component cannot be larger than the original force unless you have made a mistake. If P comes out negative, the direction you assumed was wrong.

最后，用单位和常识检查答案。分力不可能大于原力，除非你算错了。如果求出的 P 为负值，说明你假设的方向有误。

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A-Level Mathematics: Resolution of Forces | 力的分解方法