📚 Vector Resolution | 矢量的分量分解
In A-Level Physics (CIE), vectors are quantities that have both magnitude and direction. Displacement, velocity, acceleration, force and momentum are all vectors. When two or more vectors act in different directions, we cannot simply add their magnitudes. Instead, we use a technique called vector resolution: we split each vector into two perpendicular components, usually horizontal and vertical, add the components separately, and then recombine them to find the resultant. This method turns a geometry problem into a simple algebra problem.
在CIE A-Level物理中，矢量是既有大小又有方向的物理量。位移、速度、加速度、力和动量都是矢量。当两个或多个矢量沿不同方向作用时，我们不能简单地将大小相加。这时需要使用矢量分解的技巧：将每个矢量分解为两个相互垂直的分量（通常为水平和竖直分量），分别相加后再合成，从而得到合力。这一方法将几何问题转化为简单的代数问题。
1. What Is Vector Resolution? | 什么是矢量分解？
Any vector F can be represented as the sum of two perpendicular vectors called its components. If we draw the vector as an arrow, the two components form the adjacent and opposite sides of a right-angled triangle, with the original vector as the hypotenuse. The process of finding these components is called resolution.
任何一个矢量F都可以表示为两个相互垂直的矢量之和，这两个矢量称为它的分量。若将矢量画成箭头，两个分量恰好构成直角三角形的两条直角边，而原矢量则为斜边。求取这些分量的过程就称为分解。
The most common choice is to resolve into horizontal (x) and vertical (y) components. This choice is natural because the x- and y-axes are perpendicular and independent in Cartesian coordinates. However, we may also resolve along any other pair of perpendicular directions, such as parallel and normal to an inclined plane, if that is more convenient for the problem.
最常见的选择是分解为水平（x）分量和竖直（y）分量。这一选择很自然，因为在笛卡尔坐标系中，x轴与y轴相互垂直且彼此独立。当然，我们也可以沿其他任意一对互相垂直的方向进行分解，例如沿斜面的平行方向和法向方向，只要对解题更方便即可。
2. Why Resolve Vectors? | 为什么要分解矢量？
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To add vectors that are not collinear: vectors at different angles cannot be added as scalars. Resolving all vectors into the same two directions lets us add the x-components together and the y-components together.
用于合成不在同一直线上的矢量：方向不同的矢量不能按标量直接相加。将所有矢量分解到相同的两个方向上，就可以把x分量相加、y分量相加。
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To analyse forces on an object: when a force acts at an angle, only its component along a given direction actually contributes to motion in that direction. For example, a force pulling a trolley at an angle has a smaller useful horizontal effect than the same force applied horizontally.
用于分析物体受力：当力沿某一角度作用时，只有沿特定方向的分量才对那个方向的运动有贡献。例如，斜拉小车时，拉力的水平分量小于同大小的力水平施加时的效果。
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To separate motion into independent directions: in projectile motion, the horizontal velocity is constant while the vertical velocity changes under gravity. Resolving the initial velocity separately in x and y makes the whole problem solvable.
用于将运动分解为独立方向：在抛体运动中，水平速度恒定，而竖直速度在重力作用下变化。将初速度分别分解到x和y方向，整个问题就可以求解了。
3. The Component Equations | 分量公式
Consider a vector F of magnitude F acting at an angle θ measured anticlockwise from the positive x-axis. From right-angled-triangle trigonometry, the horizontal component is the adjacent side and the vertical component is the opposite side:
考虑一个大小为F的矢量F，其方向与x轴正方向成夹角θ（逆时针测量）。根据直角三角形三角关系，水平分量为邻边，竖直分量为对边：
Fₓ = F cos θ
Fᵧ = F sin θ
The original magnitude can be recovered using Pythagoras’ theorem:
利用勾股定理可以还原原矢量的大小：
F = √(Fₓ² + Fᵧ²)
And the direction angle is found using the inverse tangent:
方向角则用反正切函数求出：
θ = tan⁻¹(Fᵧ / Fₓ)
A quick memory aid: if θ is the angle between the vector and the x-axis, the component along that axis uses cos, and the component perpendicular to that axis uses sin. If the angle is measured to the y-axis instead, the roles of sin and cos are swapped.
一个快速记忆技巧：若θ是矢量与x轴的夹角，则沿该轴的分量用cos，垂直该轴的分量用sin。若角度是相对于y轴给出的，则sin与cos的角色互换。
4. Worked Example 1: Resolving a Single Force | 例题1：分解单个力
A force of 50 N acts at an angle of 30° above the horizontal. Find its horizontal and vertical components.
一个大小为50 N的力沿水平面上方30°的方向作用。求它的水平分量和竖直分量。
Identify the known values: F = 50 N, θ = 30°. The x-component is:
已知条件：F = 50 N，θ = 30°。x分量为：
Fₓ = 50 cos 30° = 50 × 0.866 = 43.3 N
The y-component is:
y分量为：
Fᵧ = 50 sin 30° = 50 × 0.5 = 25.0 N
Check: √(43.3² + 25.0²) = √(1875 + 625) = √2500 = 50 N, which confirms the calculation. Always perform this check in calculations to catch mistakes such as swapping sin and cos.
检验：√(43.3² + 25.0²) = √(1875 + 625) = √2500 = 50 N，证实计算正确。在计算中应养成检验的习惯，这样可以发现如sin与cos互换之类的错误。
5. Resolving Multiple Vectors | 多个矢量的分解
To find the resultant of two or more vectors acting in different directions, follow these steps:
要求两个或多个方向不同矢量的合力，可按以下步骤进行：
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Resolve every vector into x- and y-components using Fₓ = F cos θ and Fᵧ = F sin θ.
用Fₓ = F cos θ和Fᵧ = F sin θ将每个矢量分解为x和y分量。
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Add all x-components together to get ΣFₓ, paying attention to signs. Add all y-components together to get ΣFᵧ.
将所有x分量相加得到ΣFₓ，注意正负号；将所有y分量相加得到ΣFᵧ。
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Combine the total components using Pythagoras to find the resultant magnitude: R = √(ΣFₓ² + ΣFᵧ²).
用勾股定理合成总分量，求出合力大小：R = √(ΣFₓ² + ΣFᵧ²)。
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Find the direction using θ = tan⁻¹(ΣFᵧ / ΣFₓ), then adjust the quadrant based on the signs of the components.
用θ = tan⁻¹(ΣFᵧ / ΣFₓ)求方向，再根据两个分量的符号判断所在象限并修正角度。
This component method always works, no matter how many vectors are involved. Since the x- and y-directions are perpendicular, the two components of each vector are independent and can be summed separately.
这种分量法无论涉及多少矢量都有效。因为x方向与y方向相互垂直，每个矢量的两个分量彼此独立，可以分别求和。
6. Worked Example 2: Resultant of Two Forces | 例题2：两个力的合力
Two forces act on a body: F₁ = 40 N at 60° to the positive x-axis, and F₂ = 30 N at 150° to the positive x-axis. Find the magnitude and direction of the resultant force.
两个力作用在同一物体上：F₁ = 40 N，方向与x轴正方向成60°；F₂ = 30 N，方向与x轴正方向成150°。求合力的大小和方向。
Step 1: Resolve each force into components.
第一步：分解每个力。
| Force | x-component / N | y-component / N |
| F₁ = 40 N at 60° | 40 cos 60° = 20.0 | 40 sin 60° = 34.6 |
| F₂ = 30 N at 150° | 30 cos 150° = −26.0 | 30 sin 150° = 15.0 |
Step 2: Sum the components.
第二步：分量求和。
ΣFₓ = 20.0 + (−26.0) = −6.0 N
ΣFᵧ = 34.6 + 15.0 = 49.6 N
Step 3: Find the magnitude of the resultant.
第三步：求合力大小。
R = √((−6.0)² + 49.6²) = √(36 + 2460) = √2496 = 50.0 N
Step 4: Find the direction. Since ΣFₓ is negative and ΣFᵧ is positive, the resultant lies in the second quadrant:
第四步：求方向。由于ΣFₓ为负、ΣFᵧ为正，合力位于第二象限：
θ = 180° − tan⁻¹(49.6 / 6.0) = 180° − 83.1° = 96.9°
The resultant is 50.0 N at 96.9° to the positive x-axis. The angle measured anticlockwise from the positive x-axis is reported because that is the CIE convention.
合力大小为50.0 N，方向为与x轴正方向成96.9°。按CIE的惯例，角度从x轴正方向逆时针测量并报告。
7. Applications on an Inclined Plane | 斜面中的应用
A common exam situation is a mass m sliding on a frictionless plane inclined at angle α to the horizontal. The weight W = mg acts vertically downwards. It is usually most convenient to resolve the weight into components parallel and perpendicular to the plane, not into horizontal and vertical components.
考试中常见的情形是：质量为m的物体在倾角为α（与水平面的夹角）的光滑斜面上滑动。重力W = mg竖直向下。通常最方便的做法是将重力分解为平行于斜面和垂直于斜面的两个分量，而不是分解为水平和竖直分量。
Using the geometry of the slope, the angle between the weight vector and the normal to the plane is α. Therefore:
根据斜面的几何关系，重力矢量与斜面法线之间的夹角为α。因此：
W_parallel = mg sin α (down the slope)
W_perpendicular = mg cos α (into the plane)
If the plane is frictionless, the net force down the slope is mg sin α, so the acceleration is:
若斜面光滑，沿斜面向下的合力为mg sin α，因此加速度为：
a = g sin α
The normal reaction force N balances the perpendicular component of the weight:
法向支持力N与重力的垂直分量平衡：
N = mg cos α
A useful check: when α = 0° (horizontal plane), mg sin 0° = 0 and mg cos 0° = mg, so N = mg, which matches a mass resting on a flat surface. When α = 90° (vertical plane), mg sin 90° = mg and mg cos 90° = 0, so the object falls freely with no normal contact.
有效的检验：当α = 0°（水平面）时，mg sin 0° = 0，mg cos 0° = mg，因此N = mg，与物体静止在水平面上相符。当α = 90°（竖直面）时，mg sin 90° = mg，mg cos 90° = 0，物体自由下落且无法向接触。
8. Sign Conventions and Quadrants | 符号约定与象限
Component signs are determined by the direction of each component on the Cartesian axes. A component pointing right or up is positive; a component pointing left or down is negative. When using θ = tan⁻¹(ΣFᵧ / ΣFₓ), the calculator gives an angle in the first or fourth quadrant only, so you must adjust it using the signs of the components.
分量的正负号由其在笛卡尔坐标轴上的方向决定：指向右或向上的分量为正；指向左或向下的分量为负。使用θ = tan⁻¹(ΣFᵧ / ΣFₓ)时，计算器只能给出第一或第四象限的角度，因此必须根据分量的符号进行修正。
| Quadrant | Sign of ΣFₓ | Sign of ΣFᵧ | Correct angle |
| First (0° to 90°) | + | + | θ as calculated |
| Second (90° to 180°) | 更多咨询请联系16621398022（同微信）
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