Newton's Laws of Motion | 牛顿运动定律

📚 Newton’s Laws of Motion | 牛顿运动定律

Newton’s laws of motion form the foundation of classical mechanics and are central to the CIE A-Level Physics syllabus. They describe the relationship between forces acting on an object and its motion, enabling us to analyse everything from simple equilibrium to complex systems of connected bodies and pulleys. Mastering these laws is essential for solving a wide range of exam problems.

牛顿运动定律构成了经典力学的基础，也是 CIE A-Level 物理大纲的核心内容。它们描述了作用在物体上的力与其运动之间的关系，使我们能够分析从简单平衡到复杂的连接体和滑轮系统等各类问题。掌握这些定律对于解决考试中广泛的问题至关重要。

1. Newton’s First Law – Inertia | 牛顿第一定律 – 惯性

Newton’s first law states: An object will remain at rest or move with constant velocity in a straight line unless acted upon by a resultant external force. This tendency to resist changes in motion is called inertia. Inertia is directly proportional to the mass of the object. The larger the mass, the more difficult it is to change its state of motion.

牛顿第一定律指出：物体将保持静止或沿直线作匀速运动，除非受到合外力的作用。这种抵抗运动状态变化的倾向称为惯性。惯性的大小与物体的质量成正比。质量越大，改变其运动状态就越困难。

If the resultant force on an object is zero, the object is in equilibrium. It may be at rest or moving at constant velocity. The first law also provides the concept of an inertial reference frame, which is essential for understanding relative motion.

如果物体所受的合外力为零，物体处于平衡状态，可能静止也可能以恒定速度运动。第一定律还提供了惯性参考系的概念，这对理解相对运动至关重要。

2. Newton’s Second Law – F = ma | 牛顿第二定律 – F = ma

Newton’s second law states that the rate of change of momentum of an object is directly proportional to the resultant force acting on it, and the change takes place in the direction of that force. For constant mass, this simplifies to the famous equation:

牛顿第二定律指出：物体动量的变化率与作用在其上的合外力成正比，并且变化沿力的方向发生。在质量不变的情况下，这简化为著名的方程：

F = m a

where F is the net force (N), m is the mass (kg) and a is the acceleration (m s⁻²). Both force and acceleration are vectors, so the direction of acceleration is always the same as the direction of the resultant force.

其中 F 为合外力（N），m 为质量（kg），a 为加速度（m s⁻²）。力和加速度都是矢量，因此加速度的方向始终与合外力的方向相同。

3. Net Force and Free-Body Diagrams | 合外力与受力分析图

A free-body diagram is a simplified sketch showing all the forces acting on a single body. The vectors represent forces such as weight (mg), normal reaction (R), tension (T) and friction (f). The net force, or resultant force, is the vector sum of all these forces. Only when the net force is non-zero does the object accelerate according to F = ma.

受力分析图是一个简化的示意图，显示作用在单个物体上的所有力。这些力矢量包括重力（mg）、法向反作用力（R）、张力（T）和摩擦力（f）等。合外力是所有力的矢量和。只有当合外力不为零时，物体才会按照 F = ma 加速。

Always draw a clear free-body diagram before applying the second law. Isolate the body of interest and show every force with its correct line of action. This will help you avoid sign errors when resolving forces along chosen axes.

在应用第二定律之前，一定要画出清晰的受力分析图。隔离感兴趣的物体，并准确画出每个力及其作用线。这将有助于在沿选定坐标轴分解力时避免正负号错误。

4. Resolution of Forces and Components | 力的分解与分量

When a force acts at an angle to the direction of motion, it must be resolved into perpendicular components, usually parallel and perpendicular to the surface or the direction of acceleration. For a force F at an angle θ to the horizontal, the horizontal component is F cos θ and the vertical component is F sin θ.

当一个力与运动方向成一定角度时，必须将其分解为相互垂直的分量，通常沿平面或加速度方向及其垂直方向。对于与水平方向成 θ 角的力 F，水平分量为 F cos θ，竖直分量为 F sin θ。

This technique is crucial for inclined plane problems. The weight mg is resolved into a component down the slope, mg sin θ, and a component perpendicular to the slope, mg cos θ. The net force along the slope then determines the acceleration.

这种分解技巧在斜面问题中至关重要。重力 mg 可分解为沿斜面向下的分量 mg sin θ 和垂直于斜面的分量 mg cos θ。沿斜面的合外力决定了加速度的大小。

5. Connected Bodies and Tension | 连接体与张力

In problems involving connected bodies, such as two masses pulled by a string, we can use the whole-system approach to find the common acceleration. The total external force accelerates the total mass: F_total = (m₁ + m₂) a. Then, to find the tension in the string, isolate one mass and apply F = ma to that mass alone.

在涉及连接体的问题中，例如两个质量块通过绳子连接并一起被拉动，我们可以使用整体法求出共同的加速度。总外力使总质量加速：F_total = (m₁ + m₂) a。然后，为了求出绳中的张力，隔离其中一个物体，并单独对其应用 F = ma。

It is essential to remember that a light inextensible string transmits the same tension throughout its length. Therefore, the magnitude of tension acting on each connected body is identical, though the directions are opposite along the string.

必须记住，轻质不可伸长的绳子会沿其长度传递相等的张力。因此，作用在每个连接体上的张力大小是相同的，尽管它们沿着绳子方向相反。

6. Smooth Inclined Planes | 光滑斜面问题

When an object slides down a smooth (frictionless) inclined plane, the only force causing acceleration down the slope is the component of weight parallel to the plane. Applying Newton’s second law along the slope gives:

当物体沿光滑（无摩擦）斜面下滑时，导致沿斜面向下加速的力只有重力沿斜面的分量。沿斜面应用牛顿第二定律可得：

mg sin θ = m a → a = g sin θ

The acceleration is independent of the mass of the object, similar to free fall but reduced by the factor sin θ. The normal reaction R equals mg cos θ and plays no part in the motion along the plane.

加速度与物体的质量无关，类似于自由落体，但减小了一个因子 sin θ。法向反作用力 R 等于 mg cos θ，不参与沿平面的运动。

7. Rough Inclined Planes and Friction | 粗糙斜面与摩擦力

If the inclined plane is rough, friction acts opposite to the direction of motion or the tendency to move. The friction force f is given by f = μ R, where μ is the coefficient of friction and R is the normal reaction. For a stationary object, static friction (μₛ) applies; once moving, kinetic friction (μₖ) is used.

如果斜面是粗糙的，摩擦力作用方向与运动方向或运动趋势相反。摩擦力 f 由 f = μ R 给出，其中 μ 为摩擦系数，R 为法向反作用力。对于静止的物体，采用静摩擦系数 μₛ；一旦开始运动，则采用动摩擦系数 μₖ。

The net force down the slope becomes mg sin θ − f. The object will only slide if mg sin θ exceeds the maximum static friction μₛ mg cos θ. For a body sliding down, a = g (sin θ − μₖ cos θ).

沿斜面向下的合外力变为 mg sin θ − f。只有当 mg sin θ 超过最大静摩擦力 μₛ mg cos θ 时，物体才会开始下滑。对于正在下滑的物体，加速度 a = g (sin θ − μₖ cos θ)。

8. Pulley Systems | 滑轮系统

A typical Atwood-type pulley problem involves two masses connected by a light string passing over a smooth, massless pulley. The heavier mass m₂ accelerates downwards while the lighter mass m₁ accelerates upwards, both with the same magnitude of acceleration a because the string is inextensible.

典型的阿特伍德式滑轮问题涉及到两个质量块，通过绕过光滑、无质量滑轮的轻绳相连。较重的质量 m₂ 加速向下，较轻的质量 m₁ 加速向上，由于绳子不可伸长，两者的加速度大小 a 相同。

Applying Newton’s second law to each mass and solving simultaneously gives the acceleration and tension:

分别对每个质量应用牛顿第二定律并联立求解，可得到加速度和张力：

a = (m₂ − m₁) g / (m₁ + m₂)

T = (2 m₁ m₂ g) / (m₁ + m₂)

These results assume that the pulley is light and frictionless, and the string is light and inextensible.

这些结果都假设滑轮是轻质且光滑的，绳子是轻质且不可伸长的。

9. Equilibrium and Statics | 平衡与静力学

An object is in translational equilibrium when the vector sum of all forces acting on it is zero: ΣF = 0. This means the object either remains at rest or moves with constant velocity in a straight line. In static problems, we often resolve forces into two perpendicular directions and set the sum of components to zero in each direction.

当作用在一个物体上的所有力的矢量和为零时，该物体处于平动平衡状态：ΣF = 0。这意味着物体要么保持静止，要么沿直线匀速运动。在静力学问题中，我们通常将力分解到两个互相垂直的方向上，并令每个方向上的分力之和为零。

For three non-parallel forces in equilibrium, they must be coplanar and their vector arrows form a closed triangle. This condition is known as the triangle of forces. Some exam questions may also use Lami’s theorem, which relates three forces in equilibrium to the angles between them.

对于三个平衡的非平行力，它们必须共面，且力矢量箭头构成一个闭合三角形。这一条件被称为力的三角形法则。有些考试题也可能用到拉密定理，它将处于平衡的三个力与它们之间的夹角联系起来。

10. Newton’s Third Law – Action-Reaction Pairs | 牛顿第三定律 – 作用力与反作用力

Newton’s third law states: If body A exerts a force on body B, then body B exerts an equal and opposite force on body A. These forces are of the same type, act on different bodies, and are always equal in magnitude and opposite in direction.

牛顿第三定律指出：如果物体 A 对物体 B 施加一个力，那么物体 B 会对物体 A 施加一个大小相等、方向相反的力。这两个力属于同种类型，作用在不同的物体上，并且总是在同一直线上大小相等、方向相反。

A classic example is the normal contact force: a book resting on a table exerts a downward force on the table (its weight acting on the table, plus any reaction), while the table exerts an equal upward normal force on the book. These two forces are an action-reaction pair. It is crucial to distinguish them from equilibrium forces, which act on the same body.

一个经典的例子是法向接触力：一本放在桌子上的书对桌子施加一个向下的力（书对桌子的压力），而桌子对书施加一个等大向上的法向力。这两个力是一对作用力与反作用力。关键是要将它们与平衡力区分开，平衡力是作用在同一个物体上的。

11. Impulse and the Second Law (F = Δp/Δt) | 冲量与第二定律的动量形式

Newton’s second law, in its most general form, is expressed in terms of momentum p = m v. The rate of change of momentum equals the resultant force:

牛顿第二定律最一般的形式是用动量 p = m v 来表达的。动量的变化率等于合外力：

F = Δp / Δt

For a constant mass, this reduces to F = m (v − u)/Δt = m a. The quantity F Δt is called impulse, which equals the change in momentum. This relationship is particularly useful in collision and rebound problems, where forces act over very short time intervals.

在质量不变的情况下，这简化为 F = m (v − u)/Δt = m a。乘积 F Δt 称为冲量，等于动量的变化。这一关系在碰撞和反弹问题中尤其有用，因为这类问题中力作用的时间间隔非常短。

When calculating impulse, remember that velocity and momentum are vectors. A change in direction must be accounted for by considering the sign. For example, a ball bouncing off a wall at speed v and returning at speed v has a momentum change of m(v − (−v)) = 2mv.

在计算冲量时，要记住速度和动量是矢量。方向的改变必须通过正负号来体现。例如，一个球以速度 v 撞墙并以速度 v 返回，其动量变化为 m(v − (−v)) = 2mv。

12. Common Misconceptions and Exam Tips | 常见误解与考试技巧

Several misconceptions can lead to lost marks. First, a common error is thinking that an object moving at constant velocity must have a net force acting on it; actually, constant velocity means zero resultant force. Similarly, an object can be at rest yet have forces acting on it, as long as they cancel.

一些常见的误解可能会导致失分。首先，一个普遍的错误是认为匀速运动的物体一定受到合外力的作用；实际上，匀速运动意味着合外力为零。同样，物体可以静止但受到多个力作用，只要这些力相互抵消即可。

Another pitfall is confusing mass and weight. Mass is a scalar measured in kilograms; weight is a force (W = mg) measured in newtons. Also, students often treat action-reaction pairs as if they cancel, but they act on different bodies and cannot cancel each other out in the equilibrium of a single body.

另一个易错点是混淆质量和重量。质量是标量，以千克为单位；重量是一种力（W = mg），以牛顿为单位。此外，学生常常认为作用力与反作用力可以相互抵消，但它们作用在不同物体上，不能在单一物体的平衡中互相抵消。

Tips for the exam: always draw a large, clearly labelled free-body diagram. Choose a convenient coordinate system and resolve all forces. Write down F = ma separately for each direction, and be meticulous with signs. Check that your final answer is physically sensible, e.g., tension cannot be negative, and acceleration of a pulley system must be less than g.

考试技巧：始终画一个大的、标注清晰的受力分析图。选择合适的坐标系，分解所有的力。分别对每个方向写出 F = ma 的方程，并仔细处理正负号。检查最终答案在物理上是否合理，例如张力不能为负值，滑轮系统的加速度必须小于 g。

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Newton’s Laws of Motion | 牛顿运动定律