Forces as Vectors | 力的矢量分析

📚 Forces as Vectors | 力的矢量分析

At AS level, forces are not just numbers; they are vector quantities with both magnitude and direction. In Edexcel Physics, you must be able to add forces, resolve them into components, and use vector methods to decide whether an object is in equilibrium.

在 AS 阶段，力不仅仅是数字，而是既有大小又有方向的矢量。在爱德思物理中，你必须能够合成力、将力分解为分量，并用矢量方法判断物体是否处于平衡状态。

1. Scalars and Vectors in Mechanics | 力学中的标量与矢量

A scalar quantity has magnitude only. Examples in mechanics include distance, speed, mass, energy, work and time.

标量只有大小。力学中常见的标量包括距离、速率、质量、能量、功和时间。

A vector quantity has both magnitude and direction. Examples include displacement, velocity, acceleration, force, momentum and weight.

矢量既有大小又有方向。常见的矢量包括位移、速度、加速度、力、动量和重力。

Force is a vector because the direction in which a force acts changes its total effect on an object. Pushing a box east is not the same as pushing it north.

力是矢量，因为力的作用方向会改变它对物体的总效果。向东推箱子和向北推箱子效果不同。

In vector diagrams, forces are drawn as arrows. The length of the arrow is proportional to the magnitude of the force, and the arrowhead gives the direction.

在矢量图中，力用箭头表示。箭头的长度与力的大小成正比，箭头指向表示力的方向。

2. Representing a Force as a Vector | 用矢量表示力

A force arrow must be drawn to a clear scale, for example 1 cm = 10 N. This lets you convert a measured length back into a force magnitude.

力的箭头必须按明确的比例尺绘制，例如 1 cm = 10 N。这样你可以将测量的长度换算回力的大小。

Direction should be stated from a reference line, such as 30° above the horizontal, or 20° clockwise from north. Without a reference direction, a vector description is incomplete.

方向必须相对于参考线给出，例如与水平线成 30° 向上，或从北顺时针转 20°。没有参考方向，矢量描述就是不完整的。

When labelling a force, write the symbol and magnitude, for example F₁ = 50 N at 30° to the horizontal. This tells you everything needed to redraw or resolve the force.

标注力时，写出符号和大小，例如 F₁ = 50 N，与水平方向成 30°。这提供了重新绘制或分解该力所需的全部信息。

Arrow length is calculated by length = magnitude ÷ scale. If the scale is 1 cm = 10 N, a 50 N force is drawn as 5 cm.

箭头长度由长度 = 力的大小 ÷ 比例尺计算。如果比例尺为 1 cm = 10 N，则 50 N 的力画成 5 cm。

3. Adding Forces: The Tip-to-Tail Method | 力的相加：首尾相接法

To add two or more forces, draw the first force vector, then draw the second vector starting at the arrowhead of the first. This is called the tip-to-tail method.

要合成两个或多个力，先画出第一个力矢量，再从第一个矢量的箭头端开始画第二个矢量。这称为首尾相接法。

The resultant force R is drawn from the tail of the first vector to the arrowhead of the last vector. It represents the combined effect of all the forces.

合力 R 从第一个矢量的尾端指向最后一个矢量的箭头端。它表示所有力的总效果。

Vector addition is commutative, so the order in which forces are drawn does not change the resultant.

矢量加法满足交换律，因此画力的顺序不会改变合力。

In a scale drawing, measure the length of R with a ruler and its direction with a protractor to find the magnitude and direction of the resultant.

在比例图中，用尺子测量 R 的长度，并用量角器测量其方向，即可得到合力的大小和方向。

4. The Parallelogram Rule | 平行四边形法则

If two forces act at the same point, draw both vectors from the same origin and complete a parallelogram. The resultant is the diagonal from the origin.

如果两个力作用在同一点，从同一原点画出两个矢量，并作出平行四边形。合力就是从原点引出的对角线。

When the angle between two forces F₁ and F₂ is θ, the magnitude of the resultant is given by the cosine rule:

当两个力 F₁ 和 F₂ 之间的夹角为 θ 时，合力的大小由余弦定理给出：

R = √(F₁² + F₂² + 2 F₁ F₂ cos θ)

The direction of the resultant can be found using the sine rule. If R makes an angle α with force F₁, then sin α / F₂ = sin θ / R.

合力的方向可用正弦定理求得。如果 R 与力 F₁ 的夹角为 α，则 sin α / F₂ = sin θ / R。

For perpendicular forces, θ = 90° and cos 90° = 0, so the formula simplifies to:

对于相互垂直的力，θ = 90° 且 cos 90° = 0，因此公式简化为：

R = √(F₁² + F₂²)

The parallelogram rule and tip-to-tail method give the same resultant; choose the method that is easier to draw accurately in the exam.

平行四边形法则和首尾相接法给出相同的合力；考试时选择更容易准确绘制的方法。

5. Resolving Forces into Components | 力的正交分解

Any force F acting at an angle θ to the x-axis can be split into a horizontal component F cos θ and a vertical component F sin θ.

任何与 x 轴成 θ 角的力 F 都可以分解为水平分量 F cos θ 和竖直分量 F sin θ。

The horizontal component is found by multiplying the force by cos θ, and the vertical component by sin θ. The components are perpendicular and independent of each other.

水平分量等于力乘以 cos θ，竖直分量等于力乘以 sin θ。两个分量相互垂直，并且彼此独立。

Choose axes that make the problem simple. For a slope, the most useful axes are usually parallel to the slope and perpendicular to the slope, not horizontal and vertical.

选择能让问题简化的坐标轴。对于斜面，最常用的坐标轴通常是沿斜面方向和垂直斜面方向，而不是水平和竖直方向。

Resolution is the reverse of vector addition. Two perpendicular components can replace one force without changing the overall physical effect.

分解是矢量合成的逆运算。两个相互垂直的分量可以替代一个力，而不改变整体物理效果。

For equilibrium, the sum of the horizontal components must be zero and the sum of the vertical components must be zero.

平衡时，水平分量的代数和必须为零，竖直分量的代数和也必须为零。

6. Resultant Force and Equilibrium | 合力与平衡

The resultant force is the single force that represents the combined effect of all forces acting on a body. If the resultant is zero, the forces are balanced.

合力是代表作用在物体上所有力的总效果的单个力。如果合力为零，则这些力相互平衡。

When the resultant force is zero, an object remains at rest or continues to move with constant velocity. This is Newton’s first law of motion.

当合力为零时，物体保持静止或继续以恒定速度运动。这就是牛顿第一运动定律。

The equilibrium condition is ΣF = 0. In component form, this means the sum of the x-components is zero and the sum of the y-components is zero.

平衡条件为 ΣF = 0。用分量形式表示，就是 x 方向分量的代数和为零，y 方向分量的代数和为零。

If three forces are in equilibrium, they form a closed triangle when drawn tip-to-tail. This closed triangle can be used to find an unknown force by scale drawing or trigonometry.

如果三个力平衡，首尾相接时会形成一个闭合三角形。这个闭合三角形可以用来通过比例图或三角学求未知力。

For more than three forces, a closed polygon is needed. Equilibrium means the vectors return to the starting point.

对于三个以上的力，则需要形成闭合多边形。平衡意味着各矢量首尾相接后回到起点。

7. Free-Body Diagrams | 受力分析图

A free-body diagram shows all the forces acting on a single object. It isolates the object from its surroundings so that only the forces on that object are drawn.

受力分析图显示作用在单个物体上的所有力。它将物体从周围环境中隔离出来，只画出作用在该物体上的力。

Common forces to include are weight W = mg acting vertically downwards, the normal reaction N perpendicular to a contact surface, tension T along a rope or string, and friction F opposing motion or potential motion.

常见的力包括重力 W = mg 竖直向下、法向反作用力 N 垂直于接触面、绳或线中的张力 T 沿绳的方向、摩擦力 F 与运动或运动趋势方向相反。

Do not include forces that the object exerts on other bodies. For example, if a book rests on a table, the free-body diagram of the book does not include the downwards force the book exerts on the table.

不要包含该物体施加给其他物体的力。例如，如果一本书放在桌面上，书的受力分析图中不应包含书对桌面向下的压力。

Before resolving forces, draw the coordinate axes on the free-body diagram and mark all relevant angles. This reduces mistakes when writing sin and cos components.

在分解力之前，在受力分析图上画出坐标轴并标出所有相关角度。这可以减少写正弦和余弦分量时的错误。

A well-drawn free-body diagram is often worth marks in exams, even if the final calculation goes wrong. Always label every force arrow clearly.

画得清晰的受力分析图在考试中通常能得分，即使最终计算有误。务必清楚地标注每个力箭头。

8. Worked Example: Two Tugboats | 例题：两艘拖船

Problem: Two tugboats pull a ship. Tug A exerts 40 kN at 20° to the forward direction. Tug B exerts 30 kN at 35° on the other side of the forward direction. Find the resultant force on the ship.

例题：两艘拖船拉一艘船。拖船 A 施加 40 kN，方向与前进方向成 20°。拖船 B 在前进方向另一侧施加 30 kN，与前进方向成 35°。求作用在船上的合力。

Resolve each tugboat force into forward and sideways components. For tug A, forward component = 40 cos 20° = 37.6 kN and sideways component = 40 sin 20° = 13.7 kN.

将每艘拖船的力分解为前进分量和侧向分量。对于拖船 A，前进分量 = 40 cos 20° = 37.6 kN，侧向分量 = 40 sin 20° = 13.7 kN。

For tug B, forward component = 30 cos 35° = 24.6 kN and sideways component = 30 sin 35° = 17.

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Forces as Vectors | 力的矢量分析