Applications of Forces | 力的应用

📚 Applications of Forces | 力的应用

In Edexcel A-Level Mechanics, applications of forces bring together vector resolution, Newton’s laws, friction, connected systems, inclined planes and moments. A strong grasp of these ideas is essential for modelling real-world situations such as lifts, pulleys, towing vehicles and objects on slopes.

在 Edexcel A-Level 力学中，力的应用将向量分解、牛顿定律、摩擦、连接体、斜面和力矩等知识综合在一起。熟练掌握这些概念对于建立电梯、滑轮、牵引车辆和斜面上物体等现实情境的模型至关重要。

This article covers the key techniques you need for Edexcel Mechanics questions. It includes resolving forces, equilibrium, Newton’s second law, connected particles, friction, inclined planes, moments, rigid-body equilibrium and common exam applications.

本文涵盖 Edexcel 力学题目所需的关键方法，包括力的分解、平衡、牛顿第二定律、连接体、摩擦、斜面、力矩、刚体平衡以及常见考试应用。

1. Resolving Forces into Components | 力的分解与分量

A force is a vector quantity, so it has both magnitude and direction. When several forces act at a point, the cleanest method is to resolve each force into perpendicular components, usually horizontal and vertical. If a force F acts at an angle θ to the horizontal, then its horizontal component is F cos θ and its vertical component is F sin θ.

力是一个矢量，既有大小又有方向。当多个力作用于同一点时，最清晰的方法是将每个力分解为互相垂直的分量，通常取水平和竖直方向。如果一个力 F 与水平方向成 θ 角，那么它的水平分量为 F cos θ，竖直分量为 F sin θ。

Always draw a large, labelled force diagram before resolving. Choose the positive direction for each axis, and write all components consistently. Missing one component is a common mistake in Edexcel exams, especially when a force is not aligned with the slope.

在分解之前，一定要画出清晰且标注完整的受力图。为每个坐标轴选定正方向，并一致地写出所有分量。在 Edexcel 考试中，漏掉某个分力是常见错误，尤其是当力不平行于斜面时。

For example, a force of 50 N acting at 30° to the horizontal has components 50 cos 30° and 50 sin 30°. Since cos 30° = √3/2 and sin 30° = 1/2, the components are approximately 43.3 N horizontally and 25 N vertically.

例如，一个 50 N 的力与水平方向成 30° 角，其分量为 50 cos 30° 和 50 sin 30°。因为 cos 30° = √3/2，sin 30° = 1/2，所以水平分量约为 43.3 N，竖直分量为 25 N。

2. Equilibrium and Newton’s First Law | 平衡与牛顿第一定律

An object is in equilibrium when the vector sum of all forces acting on it is zero. By Newton’s first law, if the resultant force is zero, the acceleration is zero; the object either remains at rest or moves with constant velocity.

当作用在物体上的所有力的矢量和为零时，物体处于平衡状态。根据牛顿第一定律，如果合力为零，加速度就为零；物体要么保持静止，要么以恒定速度运动。

For equilibrium problems, resolve forces in two perpendicular directions and set each algebraic sum equal to zero. This gives two independent equations, allowing you to find up to two unknown forces or angles.

对于平衡问题，在两个互相垂直的方向上分解力，并使每个方向的代数和为零。这样就可以得到两个独立方程，从而求出最多两个未知力或角度。

Σ F (horizontal) = 0 and Σ F (vertical) = 0

For instance, a particle of weight 20 N held in equilibrium by two strings could be solved by resolving horizontally and vertically. The two equations link the string tensions and the given angles, so the unknowns can be found by simultaneous equations.

例如，一个重力为 20 N 的质点由两根绳悬挂并保持平衡，可以通过水平和竖直分解求解。两个方程将绳子张力与给定角度联系起来，因此可以通过联立方程求出未知量。

Equilibrium does not mean the object must be stationary. A car moving at constant speed along a straight road is also in equilibrium because its resultant force is zero, even though it is moving.

平衡并不意味着物体必须静止。一辆沿直线匀速行驶的汽车也处于平衡状态，因为它的合力为零，尽管它在运动。

3. Newton’s Second Law and Resultant Force | 牛顿第二定律与合力

When forces are not balanced, the resultant force causes acceleration. Newton’s second law states that the resultant force is equal to the mass of the object multiplied by its acceleration.

当力不平衡时，合力会产生加速度。牛顿第二定律指出，合力等于物体的质量乘以其加速度。

F = ma

The resultant force F is measured in newtons (N), mass m in kilograms (kg), and acceleration a in metres per second squared (m s⁻²). You must clearly identify the direction of acceleration and treat it as positive when resolving forces.

合力 F 的单位是牛顿 (N)，质量 m 的单位是千克 (kg)，加速度 a 的单位是米每二次方秒 (m s⁻²)。你必须清楚地确定加速度的方向，并在分解力时将其视为正方向。

For example, if a 3 kg block is pushed by a 15 N force against a 6 N resistance, the resultant force is 15 N − 6 N = 9 N, so its acceleration is a = F ÷ m = 9 ÷ 3 = 3 m s⁻².

例如，一个 3 kg 的物块受到 15 N 的推力，同时受到 6 N 的阻力，合力为 15 N − 6 N = 9 N，因此其加速度为 a = F ÷ m = 9 ÷ 3 = 3 m s⁻²。

In harder problems, the applied force may act at an angle. Only the component of the force in the direction of motion contributes to the acceleration, while the perpendicular component may affect the normal reaction and therefore friction.

在较难的题目中，作用力可能与运动方向成一定角度。只有沿运动方向的分力才会产生加速度，

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