📚 Resolving Forces | 力的分解
Resolving forces is the process of splitting a single force into two perpendicular components, making it easier to analyse equilibrium and motion in mechanics. In Edexcel A‑Level Mathematics and Physics, this technique underpins almost every problem involving inclined planes, connected particles and friction. Understanding how to resolve correctly allows you to apply Newton's laws and conditions for equilibrium systematically.
力的分解是将一个力拆分为两个互相垂直的分量的过程，便于分析力学中的平衡与运动。在 Edexcel A‑Level 数学和物理中，几乎所有涉及斜面、连接体和摩擦力的题目都离不开这项技术。正确掌握分解方法，才能系统地应用牛顿定律和平衡条件。
1. The Principle of Resolving Forces | 力的分解原理
A single force can be replaced by two perpendicular components that have exactly the same effect. The components are usually taken along horizontal and vertical axes, or parallel and perpendicular to an inclined surface. The choice of axes is arbitrary, but a smart choice simplifies calculations.
一个力可以用两个相互垂直的分量来替代，效果完全相同。分量通常取在水平和竖直方向，或沿着平行和垂直于斜面的方向。坐标轴的选取是任意的，但巧妙的选择会大大简化计算。
If a force F makes an angle θ with the horizontal, its horizontal component is F cosθ and its vertical component is F sinθ. These components are independent, so we can apply equations of motion separately in each direction.
若力 F 与水平方向夹角为 θ，则水平分量为 F cosθ，竖直分量为 F sinθ。这两个分量彼此独立，因此可以在每个方向上分别应用运动方程。
F_horizontal = F cosθ, F_vertical = F sinθ
2. Resolving into Perpendicular Axes | 分解到垂直轴上
Choose two perpendicular directions and resolve every force into components along those axes. It is essential to treat the sign of each component consistently – usually one direction is taken as positive. For example, take rightwards and upwards as positive; then a force pointing left or down will have a negative component.
先选取两个互相垂直的方向，再将每个力沿这些轴分解为分量。各分量的正负号必须保持一致——通常取向右和向上为正；那么指向左或向下的力其分量就为负。
When a force acts at an angle, draw a right‑angled triangle. The original force is the hypotenuse. The adjacent side gives the component next to the angle (cos), and the opposite side gives the component away from the angle (sin). Always label the angle clearly to avoid mixing up sin and cos.
当力成一定角度作用时，画一个直角三角形。原来的力是斜边，邻边给出与角相邻的分量（乘以 cos），对边给出远离角的分量（乘以 sin）。一定要清晰地标出角度，以免混淆 sin 和 cos。
3. Resolving Forces on an Inclined Plane | 斜面上的力的分解
For a body on a slope, the most convenient axes are parallel to the plane and perpendicular to the plane. The weight mg acts vertically downwards. It must be resolved into a component down the slope, mg sinθ, and a component into the plane, mg cosθ, where θ is the angle of inclination.
对于斜面上的物体，最方便的坐标轴是平行于斜面和垂直于斜面。重力 mg 竖直向下，必须分解为一个沿斜面向下的分量 mg sinθ 和一个垂直于斜面的分量 mg cosθ，其中 θ 是斜面的倾角。
The normal reaction R acts perpendicular to the slope, balancing mg cosθ if there is no other vertical force. Friction acts along the slope, opposing motion or tendency to move. The net force parallel to the slope determines the acceleration down the plane.
法向反力 R 垂直于斜面作用，在没有其他竖直力时与 mg cosθ 平衡。摩擦力沿斜面作用，阻碍运动或运动趋势。沿斜面方向的合力决定了沿斜面向下的加速度。
Component down slope = mg sinθ, Component into slope = mg cosθ
4. Equilibrium Conditions Using Resolution | 利用分解建立平衡条件
An object is in equilibrium when the resultant force in any direction is zero. By resolving forces in two perpendicular directions, we obtain two independent equations: ΣFx = 0 and ΣFy = 0. Solving these yields unknown forces such as tension, reaction or friction.
当物体在任何方向上的合力均为零时，物体处于平衡。将力沿两个相互垂直的方向分解，即可得到两个独立的方程：ΣFx = 0 和 ΣFy = 0。求解这些方程就能得出未知力，比如张力、反力或摩擦力。
For instance, a particle suspended by two strings at different angles can be analysed by resolving horizontally and vertically. The tensions are found from the simultaneous equations formed by the zero‑resultant conditions.
例如，一个质点由两根不同角度的绳子悬挂，可以通过水平和竖直方向的分解来分析。通过合力为零的条件建立联立方程组，即可求出两根绳子的张力。
5. Resolving Forces with Newton's Second Law | 牛顿第二定律与力的分解
When a resultant force acts, the body accelerates. Resolving is essential to apply Fnet = m a in each axis separately. The net force along the x‑axis equals m ax, and similarly for the y‑axis. This is extremely powerful for problems where forces act in different directions.
当存在合力时，物体会加速。要分别在每个轴上应用 Fnet = m a，就必须进行分解。x 轴方向的合力等于 m ax，y 轴同理。对于各个力方向不同的问题，这一方法非常有效。
Perpendicular to the slope: R − mg cosθ = 0, since there is no movement in that direction.
对于一个沿粗糙斜面下滑的物块：沿斜面分解求加速度：mg sinθ − F_friction = m a。垂直于斜面方向：R − mg cosθ = 0，因为该方向无运动。
6. Friction and Resolved Components | 摩擦力与分解分量
is the coefficient of friction and R is the normal reaction. Resolving forces perpendicular to the contact surface is needed to find R. Only when R is known can the maximum friction be calculated.
摩擦力用 F = μR 描述，其中 μ 是摩擦系数，R 是法向反力。要求出 R，就需要对垂直于接触面的力进行分解。只有知道 R，才能计算最大静摩擦力或动摩擦力。
If an object is on the point of sliding, the friction reaches its limiting value μR. The equilibrium equation parallel to the surface can then be used to find unknown forces or the critical angle.
当物体处于即将滑动的临界状态时，摩擦力达到极限值 μR。此时可利用沿表面方向的平衡方程求出未知力或临界角度。
7. Tension, Pulleys and Resolving | 张力、滑轮与分解
In connected particle problems, a light inextensible string transmits tension unchanged. When a string passes over a smooth pulley, the direction of tension changes. Resolving forces on each particle separately, with careful sign conventions, yields the accelerations and tensions.
在连接体问题中，轻质且不可伸长的绳子传递恒定的张力。绳子绕过光滑滑轮时，张力的方向改变。对每个质点分别进行力的分解，并注意正负号规定，即可求出加速度和张力。
For a particle hanging vertically and another on a horizontal table, resolve vertically for the hanging mass, and horizontally for the mass on the table. The tension T and acceleration a appear in both sets of equations, allowing for simultaneous solution.
对于一质点悬挂在空中、另一质点放在水平桌上的情况，对悬挂质量做竖直分解，对桌上质量做水平分解。张力 T 和加速度 a 会同时出现在两套方程中，从而可以联立求解。
8. Resolving Multiple Forces at a Point | 多个共点力的分解
When three or more forces act at a point, resolve each one into the chosen x‑ and y‑components. Sum the components to find ΣFx and ΣFy. The magnitude of the resultant is given by √ΣFx² + ΣFy², and its direction by tan⁻¹(ΣFy / ΣFx).
当三个或更多的力作用在同一点时，将每个力分别沿选定的 x 轴和 y 轴分解。将所有分量累加得到 ΣFx 和 ΣFy。合力的大小为 √ΣFx² + ΣFy²，方向为 tan⁻¹(ΣFy / ΣFx)。
For equilibrium, both sums must be zero. This vector approach avoids drawing a scale diagram and works for any number of forces, making it highly systematic.
要满足平衡，这两个分量之和都必须为零。这种矢量方法无需绘制比例图，对任意数量的力都适用，极具系统性。
9. Worked Example – Object on a Rough Slope | 例题讲解 – 粗糙斜坡上的物体
etermine whether the block slides and, if so, find the acceleration. (Take g = 9.8 m s⁻².)
一质量为 5 kg 的物块静止在与水平面成 30° 的斜面上。物块与斜面间的静摩擦系数为 0.4。判断物块是否会滑动，如果会，求出加速度。（取 g = 9.8 m s⁻²）
lel: 49 sin30° = 24.5 N down the plane. Perpendicular: 49 cos30° ≈ 42.44 N into the plane.
步骤 1：分解重力。mg = 49 N。平行于斜面：49 sin30° = 24.5 N 向下；垂直于斜面：49 cos30° ≈ 42.44 N 压向斜面。
rium perpendicular to slope, R = 42.44 N. Maximum static friction = 0.4 × 42.44 ≈ 16.98 N.
步骤 2：求 R。因垂直于斜面方向平衡，R = 42.44 N。最大静摩擦力 = 0.4 × 42.44 ≈ 16.98 N。
Step 3: Compare. The component down the plane (24.5 N) exceeds maximum friction. Hence the block accelerates. Use Fnet = m a: 24.5 − 16.98 = 5 a → a = 1.504 m s⁻².
步骤 3：比较。沿斜面向下的分量（24.5 N）大于最大静摩擦力，所以物块会加速。用 Fnet = m a：24.5 − 16.98 = 5 a → a = 1.504 m s⁻²。
10. Common Pitfalls and How to Avoid Them | 常见错误及避免方法
Many students confuse sin and cos when resolving on an incline. Remember: the component next to the angle is cos; the component opposite the angle is sin. Drawing the right‑angled triangle with the angle clearly marked helps enormously.
许多学生在斜面分解时会混淆 sin 和 cos。记住：与角相邻的分量用 cos；与角相对的分量用 sin。画直角三角形并清楚标记角度，对解题大有帮助。
Another error is forgetting to include all forces, such as tension or reaction. Always draw a free‑body diagram with all forces acting on that single body, then resolve. Never include forces acting on other objects in the same diagram.
另一个错误是漏掉力，比如张力或反力。一定要绘制该单个物体的受力图，标出所有作用在该物体上的力，再进行分解。切勿在同一个受力图中包含作用在其他物体上的力。
o use the correct normal reaction, which comes from resolving perpendicular to the contact surface, not simply mg. The presence of vertical components of other forces will change R.
最后，使用 F = μR 时，一定要用正确的法向反力，它来自垂直于接触面的分解，而不是简单的 mg。其他力的竖直分量会改变 R 的大小，务必注意。
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