Addition of Forces (Vector Addition) | 力的合成法则

📚 Addition of Forces (Vector Addition) | 力的合成法则

Forces are vector quantities. When two or more forces act on a particle, the single force that has the same effect as all the original forces together is called the resultant force. The process of finding the resultant is known as addition of forces.

力是矢量。当两个或多个力作用在质点上时，能够产生与所有原力共同作用相同效果的单一力称为合力，求合力的过程称为力的合成。

1. Forces as Vectors | 力作为矢量

A vector quantity has both magnitude and direction. Force is a classic example: a 10 N force acting to the right is completely different from a 10 N force acting upward, even though their magnitudes are equal.

矢量既有大小又有方向。力是典型的矢量：向右的10 N力与向上的10 N力完全不同，尽管大小相同。

Because force is a vector, ordinary arithmetic does not always apply. Adding 5 N and 5 N can give any value between 0 N and 10 N, depending on the angle between the two forces.

因为力是矢量，普通的算术并不总是适用。将5 N和5 N相加时，结果可以是0 N到10 N之间的任何值，具体取决于两力之间的夹角。

Magnitude: a positive number with units (e.g. newtons)
Direction: a line of action and a sense (e.g. 30° above the horizontal)

大小：带单位的正数（如牛顿）
方向：作用线及其指向（如水平向上30°）

2. Scalars vs Vectors | 标量与矢量

Scalar quantities, such as mass, temperature and speed, have magnitude only. They obey normal algebraic addition. Vector quantities, such as force, velocity and displacement, obey the rules of vector addition.

标量只有大小，如质量、温度和速率，遵循普通代数加法。矢量如力、速度和位移，遵循矢量的加法规则。

Scalars 标量	Vectors 矢量
Mass 质量	Force 力
Energy 能量	Velocity 速度
Temperature 温度	Momentum 动量
Speed 速率	Acceleration 加速度

In A-Level examinations, you must clearly state whether a quantity is a vector or a scalar. CIE mark schemes frequently award a mark for writing “force is a vector” when explaining resultant motion.

在A-Level考试中，必须明确说明一个量是矢量还是标量。CIE评分方案中，解释合运动时写出“力是矢量”常可得一分。

3. Parallelogram Law | 平行四边形法则

The parallelogram law states that if two forces acting at a point are represented in magnitude and direction by the adjacent sides of a parallelogram, their resultant is represented by the diagonal drawn from the point.

平行四边形法则指出：如果作用于同一点的两个力，其大小和方向由平行四边形的相邻两边表示，则合力由从该点出发的对角线表示。

Consider two forces F₁ and F₂ with angle θ between them. The resultant R is given by:

考虑两个力F₁和F₂，夹角为θ。合力R为：

R = √(F₁² + F₂² + 2F₁F₂cos θ)

The direction of R makes an angle φ with F₁, where:

R与F₁之间的夹角φ满足：

tan φ = (F₂ sin θ) / (F₁ + F₂ cos θ)

Worked example: F₁ = 6 N, F₂ = 8 N, θ = 60°. Then R = √(36 + 64 + 48) = √148 ≈ 12.2 N. The direction is tan⁻¹(8 × 0.866 / (6 + 4)) = tan⁻¹(6.928/10) ≈ 34.7° to the 6 N force.

例题：F₁ = 6 N，F₂ = 8 N，θ = 60°。则R = √(36 + 64 + 48) = √148 ≈ 12.2 N。方向为tan⁻¹(8 × 0.866 / (6 + 4)) = tan⁻¹(6.928/10) ≈ 34.7°，即与6 N力夹角34.7°。

4. Triangle Law | 三角形法则

The triangle law is a special case of the parallelogram law. If two forces acting at a point are represented by two sides of a triangle taken in order, the third side taken in the reverse order represents the resultant.

三角形法则是平行四边形法则的特例。如果作用于同一点的两个力按顺序用三角形的两条边表示，则按相反顺序取第三边即为合力。

To draw the force triangle: place the tail of F₂ at the head of F₁. The resultant R is drawn from the tail of F₁ to the head of F₂. This is often easier for sketching and works for any two forces.

画力三角形时：将F₂的尾端放在F₁的头端。合力R从F₁的尾端画到F₂的头端。这种方法作图更方便，适用于任意两个力。

F₁ + F₂ = R (vector sum)

In CIE practical questions, students are often asked to draw a scale diagram using the triangle method. Use a sharp pencil and a protractor; state the scale clearly, e.g. 1 cm = 2 N.

在CIE实验题中，常要求用三角形法画比例图。要用削尖的铅笔和量角器，并清楚标明比例，如1 cm = 2 N。

5. Analytical Method | 解析法求合力

When forces are perpendicular, the resultant is easy to find using Pythagoras’ theorem. If F₁ is horizontal and F₂ is vertical, then:

当力相互垂直时，可用勾股定理方便地求合力。若F₁水平、F₂竖直，则：

R = √(F₁² + F₂²), tan θ = F₂ / F₁

This is the most common special case in A-Level mechanics. For example, a force of 3 N east plus 4 N north gives a resultant of 5 N at 53.1° north of east, because √(9 + 16) = 5 and tan⁻¹(4/3) ≈ 53.1°.

这是A-Level力学中最常见的特殊情况。例如，东向3 N加上北向4 N，合力为5 N，方向为东偏北约53.1°，因为√(9 + 16) = 5，tan⁻¹(4/3) ≈ 53.1°。

For non-perpendicular forces, the cosine rule may be applied directly to the force triangle. The formula is the same as the parallelogram formula, since the angle used is always the angle between the two forces.

对于不垂直的力，可直接对力三角形使用余弦定理。其结果与平行四边形公式相同，因为使用角永远是两力之间的夹角。

6. Resolving into Components | 正交分解法

Resolving a force means replacing it by two perpendicular components. This is the reverse of addition and is essential for solving equilibrium problems.

分解力是指用一个力的两个垂直分量代替它。这是合成的逆运算，对求解平衡问题至关重要。

If a force F makes an angle θ with the x-axis, then:

若力F与x轴夹角为θ，则：

Fₓ = F cos θ, Fᵧ = F sin θ

To add many forces analytically, resolve every force into x and y components, sum the components separately, and then combine:

用解析法合成多个力时，先将每个力分解为x和y分量，分别求和，再合成：

Rₓ = ΣFₓ, Rᵧ = ΣFᵧ, R = √(Rₓ² + Rᵧ²)

θ = tan⁻¹(Rᵧ / Rₓ)

This method avoids scale-drawing errors and is the preferred approach in CIE Paper 2 and Paper 4 calculations.

此法避免比例作图误差，是CIE Paper 2和Paper 4计算题的首选方法。

7. Adding Three or More Forces | 三个及以上力的合成

For three or more concurrent forces, the polygon rule applies: if forces are placed head-to-tail in order, the resultant is the side that closes the polygon, drawn from the tail of the first to the head of the last.

对于三个或更多共点力，使用多边形法则：将各力按顺序首尾相接，合力是从第一个力的尾端指向最后一个力的头端的封闭边。

Analytically, simply extend the component method. For forces F₁, F₂, F₃ at angles θ₁, θ₂, θ₃ to the x-axis:

解析法只需扩展分量法。对于与x轴夹角为θ₁、θ₂、θ₃的力F₁、F₂、F₃：

Rₓ = ΣFᵢ cos θᵢ , Rᵧ = ΣFᵢ sin θᵢ

Example: 10 N at 0°, 15 N at 90°, 20 N at 180°. Then Rₓ = 10 – 20 = -10 N, Rᵧ = 0 + 15 = 15 N. Hence R = √(100 + 225) = 18.0 N at tan⁻¹(15/-10) = 123.7° from the +x axis.

例：10 N沿0°，15 N沿90°，20 N沿180°。则Rₓ = 10 – 20 = -10 N，Rᵧ = 0 + 15 = 15 N。因此R = √(100 + 225) = 18.0 N，方向为从+x轴量起tan⁻¹(15/-10) = 123.7°。

8. Equilibrium and Closed Polygon | 平衡与闭合多边形

When several forces act on a particle and the particle remains at rest or moves with constant velocity, the forces are in equilibrium. The resultant of all forces is zero.

当多个力作用于质点，且质点保持静止或匀速运动时，各力处于平衡。此时所有力的合力为零。

For three forces in equilibrium, the force triangle closes exactly: the three vectors placed head-to-tail form a closed triangle. For more than three forces, they form a closed polygon.

三力平衡时，力三角形严格闭合：三个矢量首尾相接形成封闭三角形。三个以上力平衡时，形成封闭多边形。

ΣFₓ = 0 and ΣFᵧ = 0

This is the single most useful equation set in CIE mechanics. When you see “particle is at rest” or “moves with constant velocity”, immediately write the two equations above.

这是CIE力学中最有用的方程组。看到“质点静止”或“匀速运动”时，立即写出上述两个平衡方程。

9. Lami’s Theorem | 拉密定理

Lami’s theorem applies to three forces in equilibrium. If three forces F₁, F₂ and F₃ act at a point and are in equilibrium, then each force is proportional to the sine of the angle between the other two forces.

拉密定理适用于三力平衡。若三力F₁、F₂、F₃作用于一点且平衡，则每个力与另外两力夹角的正弦成正比。

F₁ / sin α = F₂ / sin β = F₃ / sin γ

where α is the angle between F₂ and F₃, β is the angle between F₃ and F₁, and γ is the angle between F₁ and F₂.

其中α是F₂与F₃之间的夹角，β是F₃与F₁之间的夹角，γ是F₁与F₂之间的夹角。

Note: In the diagram, α is opposite F₁. Students often mistakenly use the angle between F₂ and F₃ instead of its supplement. When forces pull outward from a point, the angles between the forces themselves are large; the opposite angles in the force triangle are the smaller angles.

注意：在图中，α与F₁相对。学生常误用F₂与F₃之间的夹角而不是其补角。当力从某点向外拉时，力与力之间的夹角较大；而力三角形中的对角是较小的角。

10. Common Mistakes and Exam Tips | 常见错误与应试提示

Mistake 1: Adding forces as scalars. Two forces of 5 N and 5 N can give 10 N only when they act in the same direction. If they act at 120°, the resultant is only 5 N.

错误一：把力当标量相加。两个5 N的力只有在同向时才等于10 N。若夹角为120°，合力只有5 N。

Mistake 2: Using the wrong angle in the cosine formula. The formula R² = F₁² + F₂² + 2F₁F₂cos θ uses the angle between the two forces when placed tail-to-tail, not the angle between them in the head-to-tail diagram.

错误二：余弦公式中用错角度。公式R² = F₁² + F₂² + 2F₁F₂cos θ中的θ是两力尾对尾时的夹角，不是首尾相连图中的夹角。

Mistake 3: Forgetting the direction of the resultant. Many students calculate R correctly but fail to state its direction, losing a mark in CIE exams.

错误三：忘记合力的方向。许多学生正确算出了R，但没有说明方向，在CIE考试中丢分。

Tip 1: Always draw a clear labelled sketch before starting vector algebra.

提示一：进行矢量代数运算前，先画出清晰标注的草图。

Tip 2: In equilibrium problems, resolve all forces along two perpendicular axes and set each sum to zero.

提示二：在平衡问题中，将所有力沿两个垂直轴分解，并令每轴合力为零。

Tip 3: When using a scale drawing, measure at least two significant figures to obtain a reliable answer. State your scale explicitly.

提示三：用比例作图时，测量至少保留两位有效数字，并明确写出比例尺。

11. Summary | 总结

The addition of forces is a core skill in A-Level Physics. You must know how to use the parallelogram law, the triangle law, the polygon method, and the component method. You must also understand the special case of equilibrium where the resultant is zero.

力的合成是A-Level物理的核心技能。必须掌握平行四边形法则、三角形法则、多边形法和正交分解法，还要理解合力为零的平衡特例。

In the CIE examination, force addition appears in nearly every mechanics question, from simple weight-and-reaction problems to inclined-plane dilemmas and connected particles. Mastering the component method will give you a reliable tool for all of these.

在CIE考试中，力的合成几乎出现在每道力学题中，从简单的重力与支持力问题到斜面问题和连接体问题。掌握分量法能为你解决所有这些问题提供可靠工具。

Finally, always remember: a vector answer is not complete unless it has both a magnitude and a direction. Write R = 12 N at 30° to the horizontal, not just R = 12 N.

最后，请记住：矢量答案必须包含大小和方向才算完整。要写“R = 12 N，与水平方向成30°”，而不要只写“R = 12 N”。

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Addition of Forces (Vector Addition) | 力的合成法则