Static Rigid Bodies | 静态刚体

📚 Static Rigid Bodies | 静态刚体

Static rigid bodies appear throughout Edexcel A-Level Mechanics. Unlike particles, a rigid body has size and shape, so forces acting at different points can cause rotation. This topic brings together resolving forces, moments, friction and limiting equilibrium. A clear diagram and a consistent sign convention are usually the fastest route to full marks.

静态刚体是爱德思 A-Level 力学中的重要内容。与质点不同，刚体有大小和形状，因此作用在不同点上的力可以引起转动。本主题综合考查力的分解、力矩、摩擦和极限平衡。清晰的受力图和一致的符号约定通常是拿到满分的最快途径。

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1. Rigid Body Assumptions | 刚体假设

A rigid body is modelled as an object that does not deform under the applied forces. The distances between points in the body remain fixed, so the line of action of every force matters. Common models include uniform rods, non-uniform rods, laminas and blocks.

刚体被建模为在外力作用下不发生形变的物体。刚体内各点之间的距离保持不变，因此每个力的作用线都很重要。常见模型包括均匀杆、非均匀杆、薄板和方块。

Because the body does not deform, a force can be moved along its line of action without changing its external effect. However, it cannot be moved sideways, because that would alter the perpendicular distance from a pivot and therefore change the moment.

由于刚体不发生形变，一个力可以沿其作用线滑动而不改变外部效果。但力不能横向平移，因为这会改变力到支点的垂直距离，从而改变力矩。

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2. Moment of a Force | 力的力矩

The moment of a force about a point measures its turning effect. It is calculated by multiplying the magnitude of the force by the perpendicular distance from the point to the line of action of the force.

力对某点的力矩衡量该力的转动效果。它等于力的大小乘以该点到力作用线的垂直距离。

M = F × d

If the force is not perpendicular to the distance, use the perpendicular component of the force. In many Edexcel problems the distance is given along a rod, so for an angle θ between the force and the rod, the moment is F × r × sin θ where r is the distance along the rod.

如果力与距离不垂直，应使用力的垂直分量。在许多爱德思题目中，距离沿杆给出，因此当力与杆的夹角为 θ 时，力矩为 F × r × sin θ，其中 r 是沿杆的距离。

Moments are usually taken as positive anticlockwise or positive clockwise. Choose one convention and use it for every moment in the same equation. Units are newton metres, written N m.

力矩通常以逆时针为正或以顺时针为正。选择一种约定并在同一个方程中对所有力矩一致使用。单位为牛顿米，写作 N m。

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3. Couples and Torque | 力偶与转矩

A couple consists of two equal and opposite parallel forces whose lines of action do not coincide. The resultant force of a couple is zero, but the resultant moment is not zero, so a couple produces pure rotation without translation.

力偶由两个大小相等、方向相反且作用线不重合的平行力组成。力偶的合力为零，但合力矩不为零，因此力偶只产生转动而不产生平动。

The moment of a couple is F × d, where d is the perpendicular distance between the two force lines. This moment is the same about any point, which makes couples useful when calculating net turning effects on a rigid body.

力偶矩等于 F × d，其中 d 是两条力作用线之间的垂直距离。该力矩对任意点都相同，因此在计算刚体的净转动效果时，力偶非常有用。

In equilibrium, any applied couple must be balanced by another couple or by a combination of forces that produces an equal and opposite moment.

在平衡状态下，任何施加的力偶都必须被另一个力偶或一组产生等大反向力矩的力所平衡。

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4. Conditions for Equilibrium | 平衡条件

For a rigid body in static equilibrium under coplanar forces, three independent conditions must be satisfied. The vector sum of all horizontal forces must be zero, the vector sum of all vertical forces must be zero, and the sum of moments about any point must be zero.

对于受共面力作用的静态平衡刚体，必须满足三个独立条件：所有水平力的矢量和为零，所有竖直力的矢量和为零，以及对任意点的力矩之和为零。

ΣFₓ = 0, ΣFᵧ = 0, ΣM = 0

You can take moments about any point, but the best choice is usually a point where two or more unknown forces intersect. This eliminates those unknowns from the moment equation and often gives a one-step solution.

可以对任意点取矩，但最佳选择通常是两个或多个未知力作用线的交点。这样可以将这些未知力从力矩方程中消去，常常一步即可求解。

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5. Support Reactions and Contact Forces | 支撑反力与接触力

Different supports produce different reaction patterns. A smooth surface produces only a normal reaction perpendicular to the surface. A rough surface produces a normal reaction plus a friction force parallel to the surface. A hinge usually produces two perpendicular components of reaction.

不同的支撑会产生不同的反力模式。光滑表面只产生垂直于表面的法向反力。粗糙表面产生法向反力和平行于表面的摩擦力。铰链通常产生两个互相垂直的反力分量。

Support type | 支撑类型	Reaction | 反力	Notes | 说明
Smooth surface | 光滑表面	Normal only | 仅法向	No friction | 无摩擦
Rough surface | 粗糙表面	Normal + friction | 法向 + 摩擦	Friction opposes motion | 摩擦阻碍运动
Hinge | 铰链	Two components | 两个分量	Fixed position | 固定位置
Roller | 滚轴	Normal to plane | 垂直于平面	One reaction only | 仅一个反力

Always draw the reaction arrow in the correct direction before resolving. If the direction is unknown, assign a positive direction and keep the sign convention consistent throughout the working.

在进行分解前，务必先沿正确方向画出反力箭头。如果方向未知，可先假定一个正方向，并在整个计算过程中保持符号约定一致。

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6. Uniform and Non-uniform Rods | 均匀与非均匀杆

For a uniform rod, the weight acts at the midpoint of the rod. This simplification is used in nearly every rigid body problem involving a rod, ladder or beam. Label the weight as W and place it at the centre of the rod.

对于均匀杆，重力作用在杆的中点。几乎所有涉及杆、梯子或梁的刚体问题都会用到这一简化。将重力标为 W，并将其画在杆的中点。

For a non-uniform rod, the weight acts at the centre of mass, which may be given or may be the quantity you are asked to find. If a rod is in equilibrium under known forces, taking moments about a support can locate the centre of mass.

对于非均匀杆，重力作用在质心上，质心位置可能给出，也可能是题目要求求解的量。如果杆在已知力作用下平衡，可以对某一支撑点取矩来求出质心位置。

For example, a uniform rod AB of length 4 m and weight 30 N has its weight 2 m from each end. If the rod rests on a support at A, the moment of the weight about A is 30 N × 2 m = 60 N m.

例如，一根长 4 m、重 30 N 的均匀杆 AB，其重力作用在距两端各 2 m 处。如果杆在 A 端有支撑，则重力对 A 点的力矩为 30 N × 2 m = 60 N m。

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7. Friction and Limiting Equilibrium | 摩擦与极限平衡

When two rough surfaces are in contact, the friction force F opposes sliding and satisfies F ≤ μR, where R is the normal reaction and μ is the coefficient of friction. In a static problem the friction may be less than its maximum value.

当两个粗糙表面接触时，摩擦力 F 阻碍滑动，并满足 F ≤ μR，其中 R 是法向反力，μ 是摩擦系数。在静力问题中，摩擦力可能小于其最大值。

Limiting equilibrium means the body is just about to slide. In this special case the friction reaches its maximum value and F = μR. If the question says ‘just about to move’ or ‘on the point of sliding’, use the equality.

极限平衡指刚体恰好即将滑动。在这种特殊情况下，摩擦力达到最大值，即 F = μR。如果题目中说“即将运动”或“即将滑动”，应使用等号。

The angle of friction is also useful on an inclined plane. At limiting equilibrium on a plane inclined at angle θ to the horizontal, tan θ = μ. This links the geometry of the slope to the roughness of the surfaces.

摩擦角在斜面上也很有用。在倾角为 θ 的斜面上处于极限平衡时，tan θ = μ。这一关系将斜面的几何角度与表面粗糙程度联系起来。

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8. Ladder Problems | 梯子问题

A ladder resting against a rough floor and a smooth wall is a standard Edexcel model. The floor produces a normal reaction and friction, while the smooth wall produces only a horizontal normal reaction. The weight of the ladder acts at its midpoint if the ladder is uniform.

靠在粗糙地面和光滑墙上的梯子是爱德思的标准模型。地面产生法向反力和摩擦力，而光滑墙只产生水平法向反力。如果梯子均匀，其重力作用在中点。

Let the ladder length be L and the wall contact point be at height h. Resolving vertically gives R_floor = W. Resolving horizontally gives F_floor = R_wall. Taking moments about the base of the ladder often eliminates the floor contact forces and solves R_wall.

设梯子长度为 L，墙接触点高度为 h。竖直方向分解得 R_floor = W。水平方向分解得 F_floor = R_wall。对梯子底部取矩通常能消去地面接触力并求出 R_wall。

For a uniform ladder of length L making an angle θ with the horizontal, taking moments about the base gives the wall reaction from R_wall × L sin θ = W × (L/2) × cos θ. The length L cancels, and the ratio of R_wall to W can be found from θ.

对于与水平面成 θ 角、长度为 L 的均匀梯子，对底部取矩可得 R_wall × L sin θ = W × (L/2) × cos θ。长度 L 可消去，R_wall 与 W 的比值可由 θ 求得。

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9. Toppling and Sliding | 翻倒与滑动

An object on a rough inclined plane can fail in two different ways: it can slide down the plane or topple over. These conditions must be checked separately, and the one that occurs first is the limiting condition.

粗糙斜面上的物体可能以两种不同方式失稳：沿斜面下滑或绕某边翻倒。这两种情况必须分别检验，先发生的一种即为极限条件。

Sliding occurs when the component of weight down the plane exceeds the maximum friction. For a block of mass m on a plane inclined at θ, sliding begins when mg sin θ > μ mg cos θ, which simplifies to tan θ > μ.

当下滑方向的重力分量超过最大摩擦力时，物体开始滑动。对于质量为 m 的方块在倾角为 θ 的斜面上，滑动条件为 mg sin θ > μ mg cos θ，简化为 tan θ > μ。

Toppling occurs when the line of action of the weight falls outside the base of the block. If the block has height h and base width b, toppling is possible when tan θ > b/h. Compare this with tan θ > μ to determine whether the block slides or topples first.

当重力作用线落在方块底面外侧时，物体会翻倒。如果方块高度为 h、底面宽度为 b，则翻倒条件为 tan θ > b/h。将此条件与 tan θ > μ 进行比较，即可判断方块先滑动还是先翻倒。

Condition | 条件	Threshold | 临界值	Meaning | 含义
Sliding | 滑动	tan θ = μ	Friction limit | 摩擦极限
Toppling | 翻倒	tan θ = b/h	Weight line leaves base | 重力线离开底面

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10. Exam Strategy and Common Errors | 考试策略与常见错误

Start by drawing a large, labelled diagram with every force shown at its correct point of application. Mark angles clearly and decide immediately whether the body is uniform or non-uniform. This simple step prevents most mistakes.

首先要画一个较大的标注受力图，将每个力画在正确的作用点上。清楚标出角度，并立即判断刚体是均匀还是非均匀。这一简单步骤能避免大多数错误。

Resolve forces in two perpendicular directions and take moments about a point that removes the largest number of unknowns. In many questions one moment equation alone is enough to find the required quantity, so do not solve three equations unnecessarily.

沿两个互相垂直的方向分解力，并对能够消去最多未知力的点取矩。在许多题目中，仅一个力矩方程就足以求出所需量，因此不要不必要地联立三个方程求解。

Common errors include using the wrong perpendicular distance, taking the weight at one end of a rod, assuming friction is limiting when it is not, and mixing clockwise and anticlockwise moment signs. Always write the equilibrium conditions explicitly before substituting numbers.

常见错误包括使用错误的垂直距离、把重力画在杆的一端、在摩擦未达到极限时误用 F = μR，以及混淆顺时针与逆时针力矩符号。代入数值前，务必先明确写出平衡条件。

Finally, check that your answer is physically sensible. A reaction force should not be negative unless the body has lost contact, and a coefficient of friction should normally be between 0 and 1, although values above 1 are possible for very rough surfaces.

最后，检查答案是否符合物理实际。除非物体已经脱离接触，否则反力不应为负；摩擦系数通常应在 0 到 1 之间，但非常粗糙的表面也可能出现大于 1 的值。

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