📚 Resolving Forces | 力的分解
In A-Level Mechanics, resolving forces is the process of replacing a single force with two perpendicular components. This technique allows us to apply Newton’s laws separately in the horizontal and vertical directions, making it much easier to solve equilibrium and motion problems. Whether you are working with a block on a rough slope or a particle held by two strings, resolving forces is often the first step.
在 A-Level 力学中，力的分解是将一个力替换为两个互相垂直分力的过程。这种方法使我们能够分别在水平方向和竖直方向应用牛顿定律，从而更容易解决平衡和运动问题。无论是粗糙斜面上的物块，还是由两根绳子悬挂的质点，力的分解通常都是解题的第一步。
1. What is Resolving Forces? | 什么是力的分解？
Resolving a force means splitting it into two components that act at right angles to each other. The original force is the vector sum of these components, so the effect on the object is unchanged. This is useful because horizontal and vertical motions can be treated independently under Newton’s second law.
分解力是指将一个力拆分为两个相互垂直的分力。原来的力是这两个分力的矢量和，因此对物体的作用效果不变。这样做的用处在于，根据牛顿第二定律，水平和竖直方向的运动可以独立处理。
F = Fx + Fy, where Fx = F cos θ and Fy = F sin θ
公式：F = Fₓ + F_y，其中 Fₓ = F cos θ，F_y = F sin θ。
The angle θ is measured from the direction of the component you are finding to the original force. If you resolve in the direction of the force, the component is F cos 0° = F, and perpendicular to it the component is F cos 90° = 0.
角度 θ 是从你所求分力的方向到原力方向之间的夹角。如果你沿力的方向分解，则分力为 F cos 0° = F；垂直于力的方向分力为 F cos 90° = 0。
2. Resolving a Single Force into Components | 将单个力分解为分量
To resolve a force F at an angle θ to the horizontal, draw a right-angled triangle with F as the hypotenuse. The horizontal component is adjacent to θ, so it is F cos θ. The vertical component is opposite θ, so it is F sin θ.
要分解一个与水平方向成 θ 角的力 F，可以以 F 为斜边画一个直角三角形。水平分力与 θ 相邻，因此为 F cos θ；竖直分力与 θ 相对，因此为 F sin θ。
Horizontal component: Fx = F cos θ
水平分力：Fₓ = F cos θ。
Vertical component: Fy = F sin θ
竖直分力：F_y = F sin θ。
Always check whether the component is adjacent or opposite to the angle. If θ is small, the horizontal component should be close to the full force F, while the vertical component should be small. This provides a quick numerical check.
始终检查分力是与角度相邻还是相对。如果 θ 很小，水平分力应接近原力 F，而竖直分力应很小。这可以作为一个快速的数值检验方法。
3. Choosing Perpendicular Directions | 选择垂直方向
You are free to choose any pair of perpendicular directions. The best choice is usually one that makes the most forces lie along the axes, reducing the number of trigonometric terms. For a slope problem, use directions parallel and perpendicular to the plane.
你可以自由选择任意一对互相垂直的方向。最佳选择通常是让尽可能多的力沿坐标轴方向，从而减少三角函数的数量。对于斜面问题，应使用平行于斜面和垂直于斜面的方向。
For a block on a smooth horizontal table pulled by a rope at an angle, the natural axes are horizontal and vertical. The tension then needs to be resolved, while the weight and normal reaction stay on the vertical axis.
对于放在光滑水平桌面上、被绳子以一定角度拉动的物块，自然坐标轴是水平和竖直方向。此时绳的拉力需要分解，而重力和法向反作用力保持在竖直轴上。
4. Equilibrium Conditions | 平衡条件
A particle is in equilibrium if the resultant force is zero. Resolving in two perpendicular directions gives two independent equations: the sum of components in each direction must be zero.
若质点处于平衡状态，则合力为零。在两个互相垂直的方向上分解，可得到两个独立的方程：每个方向上的分力之和必须为零。
Σ Fx = 0 and Σ Fy = 0
公式：Σ Fₓ = 0 且 Σ F_y = 0。
If the object is accelerating, each direction follows Newton’s second law instead: the total component in a direction equals mass times the acceleration in that direction.
如果物体在加速，则每个方向遵循牛顿第二定律：该方向上的合力分量等于质量乘以该方向上的加速度。
Σ Fx = max and Σ Fy = may
公式：Σ Fₓ = maₓ，Σ F_y = ma_y。
5. Forces on an Inclined Plane | 斜面上的力
For a block of mass m on a plane inclined at angle θ to the horizontal, the weight mg acts vertically downwards. Resolve the weight parallel and perpendicular to the plane. The component down the slope is mg sin θ, and the component into the plane is mg cos θ.
对于质量为 m、放在与水平面成 θ 角的斜面上的物块，重力 mg 竖直向下。将重力沿斜面方向和垂直于斜面方向分解。沿斜面向下的分力为 mg sin θ，垂直于斜面压入斜面的分力为 mg cos θ。
Weight component down the slope = mg sin θ
沿斜面向下的重力分量 = mg sin θ。
Weight component perpendicular to the slope = mg cos θ
垂直于斜面的重力分量 = mg cos θ。
If the block is stationary and no other force acts perpendicular to the plane, the normal reaction R balances the perpendicular weight component, so R = mg cos θ. The friction force then balances the down-slope component if the block is in equilibrium.
如果物块静止，且没有其他力垂直于斜面作用，则法向反作用力 R 与垂直于斜面的重力分量平衡，因此 R = mg cos θ。若物块处于平衡状态，摩擦力则与沿斜面向下的分量平衡。
6. Friction and the Normal Reaction | 摩擦力与法向反作用力
The normal reaction R acts perpendicular to the contact surface. For an object on a slope with no other perpendicular forces, R balances the perpendicular weight component, so R = mg cos θ. Friction acts along the surface to oppose motion or potential motion.
法向反作用力 R 垂直于接触面。对于斜面上的物体，若没有其他垂直力，R 与垂直于斜面的重力分量平衡，因此 R = mg cos θ。摩擦力沿接触面方向，阻碍运动或运动趋势。
The friction force F satisfies F ≤ μR. At the point of slipping, it reaches the limiting value F = μR, where μ is the coefficient of friction. If the block is in equilibrium, F is whatever value is needed up to μR, not automatically μR.
摩擦力 F 满足 F ≤ μR。在即将滑动时，摩擦力达到极限值 F = μR，其中 μ 为摩擦系数。如果物块处于平衡状态，F 可以是任何所需的值，最大不超过 μR，而不是一定等于 μR。
When an object is on a rough slope and is about to slip, use F = μR together with the resolved components to find the angle of friction or the coefficient of friction.
当物体在粗糙斜面上即将滑动时，将 F = μR 与分解后的分量联立，可以求出摩擦角或摩擦系数。
7. Resolving Multiple Forces | 分解多个力
When several forces act at different angles, resolve each force into the chosen x- and y-components. Then add all the x-components to get the total x-component, and all the y-components to get the total y-component. Use Pythagoras to find the magnitude of the resultant if needed.
当多个力以不同角度作用时，将每个力分解到选定的 x 轴和 y 轴方向。然后将所有 x 分量相加得到总 x 分量，将所有 y 分量相加得到总 y 分量。如果需要求合力大小，可使用勾股定理。
Rx = Σ F cos θ, Ry = Σ F
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