📚 Horizontal and Vertical Components | 水平与竖直分量
In Edexcel A Level Mathematics, and especially in the mechanics sections of Papers 3 (Statistics and Mechanics), many vector quantities such as displacement, velocity, acceleration and force are most usefully analysed by splitting them into horizontal and vertical components. This process, called resolving, simplifies problems involving motion, equilibrium and Newton's laws, because the horizontal and vertical directions are perpendicular and therefore completely independent of one another.
在 Edexcel A Level 数学中，尤其是在第三套试卷（统计与力学）的力学部分，位移、速度、加速度和力等许多矢量量，通过将其分解为水平与竖直分量来分析最为有用。这个过程称为分解（resolving），它简化了涉及运动、平衡和牛顿定律的问题，因为水平方向与竖直方向互相垂直，因此彼此完全独立。
By resolving a vector into two independent perpendicular directions, you turn a single two-dimensional problem into two much simpler one-dimensional problems. This is the core idea behind most mechanics questions involving forces, equilibrium, Newton's laws and projectile motion, and it is the single technique that appears most often across the whole mechanics paper.
通过将矢量分解到两个独立的垂直方向，你可以把一个二维问题转化为两个更简单的一维问题。这是大多数涉及力、平衡、牛顿定律和抛体运动的力学题目的核心思想，也是整张力学试卷中出现频率最高的技巧。
This article covers the definition of components, the trigonometric rules for resolving, worked examples on forces, equilibrium, Newton's second law, projectiles and inclined planes, plus the most common exam mistakes and a set of practice questions with full answers.
本文将涵盖分量的定义、分解所用的三角法则、力与平衡与牛顿第二定律与抛体与斜面的例题，以及最常见的考试错误和一组带完整答案的练习题。
1. What Are Horizontal and Vertical Components? | 什么是水平分量和竖直分量？
A vector quantity has both magnitude and direction. In mechanics the most common vector quantities are displacement, velocity, acceleration, force, momentum and impulse. To work with such quantities, we often split a vector into two perpendicular parts: a horizontal component acting along the x-axis, and a vertical component acting along the y-axis.
矢量量既有大小又有方向。力学中最常见的矢量量是位移、速度、加速度、力、动量和冲量。为了处理这些量，我们通常把矢量分解成两个互相垂直的部分：沿 x 轴作用的水平分量，和沿 y 轴作用的竖直分量。
If a vector has magnitude r and makes an angle θ with the positive horizontal direction, its two components are found with a right-angled triangle. The horizontal component is adjacent to θ, so it uses cosine; the vertical component is opposite to θ, so it uses sine.
如果一个矢量的大小为 r，并与正水平方向成 θ 角，那么它的两个分量可以通过一个直角三角形求出。水平分量与 θ 相邻，所以用余弦；竖直分量与 θ 相对，所以用正弦。
Fₓ = r cos θ Fy = r sin θ
The magnitude and direction can always be recovered from the components by reversing the process, using Pythagoras' theorem and the tangent ratio.
只要反向操作，用勾股定理和正切比，就总能由分量重新求回大小和方向。
r = √Fₓ² + Fy² tan θ = Fy ÷ Fₓ
nly holds when θ is measured from the horizontal. If the angle is given from the vertical instead, the sine and cosine swap roles. Checking this before you start is the fastest way to avoid losing method marks.
一个关键细节：只有当 θ 是从水平方向量度时，Fₓ = r cos θ 才成立。如果题目给的是与竖直方向的夹角，正弦和余弦的角色就会互换。动笔前先检查这一点，是避免丢分最快的方法。
2. Why Perpendicular Directions Are Independent | 为什么垂直方向彼此独立
The reason resolving works so well is that perpendicular directions do not interact. A force that acts purely horizontally has zero vertical component, so it cannot change the vertical motion of an object at all. Equally, a vertical force cannot change horizontal motion. Mathematically this is a direct consequence of the components being at 90° to each other.
分解之所以如此有效，原因是垂直方向之间互不干涉。一个纯粹水平作用的力，其竖直分量为零，因此完全不能改变物体的竖直运动。同样，一个竖直方向的力也不能改变水平运动。从数学上看，这是两个分量互相成 90° 的直接结果。
Because of this independence, you can write down the equations for the horizontal direction and the equations for the vertical direction separately, and solve them one at a time. Each direction behaves like a simple one-dimensional problem from earlier in the course.
正因为这种独立性，你可以分别写出水平方向的方程和竖直方向的方程，然后一个一个求解。每个方向都像课程前面学过的一维简单问题一样处理。
chanics. For equilibrium you require the horizontal forces to balance and the vertical forces to balance. For Newton' ce in each direction. For projectiles, horizontal acceleration is zero while vertical acceleration is g downwards.
这个思想贯穿整个力学。平衡要求水平方向的力平衡、竖直方向的力也平衡。牛顿第二定律要应用两次 ΣF = ma，每个方向各一次。抛体运动中，水平加速度为零，而竖直加速度为 g 向下。
A useful check: if a problem gives you more unknowns than equations, you have probably forgotten to use one of the two directions. Almost every A Level mechanics question is solved with exactly two component equations.
一个实用的检查方法：如果题目给出的未知量比方程多，你很可能漏用了两个方向中的一个。几乎所有 A Level 力学题目都是用恰好两个分量方程解出的。
3. Resolving with Trigonometry: Signs and Special Angles | 用三角分解：正负号与特殊角
When resolving forces or velocities that point in different quadrants, you must decide on a positive direction before you start. The standard convention is right and up are positive, so a force pointing left or downwards contributes a negative component.
当分解指向不同象限的力或速度时，必须先约定正方向。标准约定是向右和向上为正，因此指向左或向下的力贡献负的分量。
Forces given as an angle measured anticlockwise from the positive x-axis are easiest: the horizontal component is r cos θ and the vertical component is r sin θ, with the signs handled automatically by the cosine and sine values of θ. For angles described in words, such as '30° above the horizontal pointing to the left', you must insert the minus sign yourself.
如果力是用从正 x 轴逆时针量度的角度给出的，那是最方便的：水平分量是 r cos θ，竖直分量是 r sin θ，正负号由 θ 的余弦和正弦值自动决定。对于用文字描述的角度，例如"向左上方与水平成 30°"，就需要你自己加上负号。
Exact values for common angles appear constantly in exam questions, so they should be memorised in both degrees and as surds.
常见角的精确值在考题中反复出现，所以既要记住角度形式，也要记住根式形式。
| θ | 0° | 30° | 45° | 60° | 90° |
| sin θ | 0 | ½ | √2/2 | √3/2 | 1 |
| cos θ | 1 | √3/2 | √2/2 | ½ | 0 |
| tan θ | 0 | √3/3 | 1 | √3 | undefined |
ematics, and keep full calculator accuracy until the final answer, rounding only at the end.
除非题目另有说明，Edexcel A Level 数学中取 g = 9.8 m s⁻²，并且在最终答案之前保留计算器的完整精度，只在最后一步取近似。
4. Worked Example: Resolving a Single Force | 例题：分解一个力
Example 1. A force of magnitude 50 N acts at 35° above the horizontal, pointing to the right. Find its horizontal and vertical components.
例 1。一个大小为 50 N 的力，在水平方向上方 35° 向右作用。求它的水平分量和竖直分量。
Fₓ = 50 cos 35° = 40.96 N Fy = 50 sin 35° = 28.68 N
The force therefore acts like a 41.0 N push to the right together with a 28.7 N pull upwards. Note that both components are positive because the force points up and to the right.
因此这个力相当于一个 41.0 N 向右的推力，加上一个 28.7 N 向上的拉力。注意两个分量都是正的，因为力指向右上方。
Example 2. A rope tension of 20 N acts at 25° to the vertical. Find the horizontal and vertical components.
例 2。一根绳的张力为 20 N，与竖直方向成 25°。求水平分量和竖直分量。
Horizontal = 20 sin 25° = 8.45 N Vertical = 20 cos 25° = 18.13 N
This is the case where the angle is measured from the vertical, so the sine and cosine swap. If you used 20 cos 25° for the horizontal component you would get 18.1 N, which is almost the whole tension, and the final answer would be badly wrong.
这就是角度从竖直方向量度的情形，正弦和余弦互换。如果你用 20 cos 25° 当作水平分量，会得到 18.1 N，几乎是整根绳子张力的全部，最终答案会严重错误。
5. Resolving Several Forces and Finding the Resultant | 分解多个力求合力
When several forces act on a single particle, resolve each force into components, add all the horizontal components to get the total horizontal force, and add all the vertical components separately. Then combine the two totals to obtain the resultant.
当多个力作用在同一个质点上时，先把每个力分解，再把所有水平分量相加得到合力在水平方向的总体分量，竖直分量同样单独相加。然后把这两个总量合成，得到最终的合力。
The method in four steps:
四步方法：
- Choose and state a positive direction, usually right and up. | 选定并写明正方向，通常取向右、向上为正。
- Resolve every force: list all Fₓ and all Fy with correct signs. | 分解每一个力：列出所有 Fₓ 和 Fy，注意正负号。
- Sum: ΣFₓ and ΣFy. | 求和：ΣFₓ 与 ΣFy。
- Combine with r = √ΣFₓ² + ΣFy² and tan θ = ΣFy ÷ ΣFₓ. | 用 r = √ΣFₓ² + ΣFy² 与 tan θ = ΣFy ÷ ΣFₓ 合成。
= 9 N at 20° below the positive x-axis. Find the magnitude and direction of the resultant.
例 3。两个力作用在一个质点上：F₁ = 12 N，在正 x 轴上方 40°；F₂ = 9 N，在正 x 轴下方 20°。求合力的大小和方向。
ΣFₓ = 12 cos 40° + 9 cos 20° = 9.19 + 8.46 = 17.65 N
ΣFy = 12 sin 40° − 9 sin 20° = 7.71 − 3.08 = 4.63 N
r = √17.65² + 4.63² = 18.2 N θ = tan⁻¹(4.63 ÷ 17.65) = 14.7°
The resultant is 18.2 N acting at 14.7° above the positive x-axis. Because ΣFy is positive, the resultant tilts upwards, and because ΣFₓ is positive it points to the right.
合力为 18.2 N，在正 x 轴上方 14.7°。由于 ΣFy 为正，合力向上倾斜；由于 ΣFₓ 为正，它指向右方。
6. Equilibrium Problems: ΣFₓ = 0 and ΣFy = 0 | 平衡问题：ΣFₓ = 0 与 ΣFy = 0
A particle is in equilibrium when the resultant force on it is zero. Since the two directions are independent, this single vector condition becomes two separate scalar equations.
当作用在质点上的合力为零时，该质点处于平衡状态。由于两个方向互相独立，这个单一的矢量条件就变成了两个独立的标量方程。
ΣFₓ = 0 and ΣFy = 0
If a particle moves with constant velocity, it is also in equilibrium, so the same two equations apply. Only when there is acceleration do the right-hand sides become non-zero.
如果质点以恒定速度运动，它同样处于平衡状态，因此这两个方程仍然适用。只有当存在加速度时，等式右边才不为零。
Example 4. A particle of weight 20 N hangs in equilibrium from two light inextensible strings. The first string makes 35° with the vertical and the second makes 50° with the vertical, on opposite sides. Find the tension in each string.
例 4。一个重 20 N 的质点由两根轻的不可伸长细绳悬吊并保持平衡。第一根绳与竖直方向成 35°，第二根在另一侧与竖直方向成 50°。求每根绳的张力。
in 35° = T₂ sin 50°, so T₁ = 1.336 T₂. Resolving vertically: T₁ cos 35° + T₂ cos 50° = 20.
水平方向分解：T₁ sin 35° = T₂ sin 50°，所以 T₁ = 1.336 T₂。竖直方向分解：T₁ cos 35° + T₂ cos 50° = 20。
T₂ = 11.5 N T₁ = 15.4 N
A very common exam variant replaces the second string with a smooth peg or a horizontal force. The method is identical: always resolve both ways, even when one direction looks trivial.
考试中很常见的变式是把第二根绳
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