Moment of a Force in 3D: Vector Calculation | 三维空间中力矩的向量计算

📚 Moment of a Force in 3D: Vector Calculation | 三维空间中力矩的向量计算

In three-dimensional mechanics, the moment of a force about a point or an axis is a fundamental concept. Unlike 2D problems where the moment is simply a signed scalar, 3D problems require a full vector treatment using the cross product. This article explains the vector formulation, physical interpretation, practical calculation methods, and common applications of moments in 3D.

在三维力学中，力关于一点或一轴的力矩是核心概念。与二维问题中力矩只是一个带符号的标量不同，三维问题需要用叉积进行完整的向量处理。本文将讲解三维力矩的向量公式、物理意义、实际计算方法以及常见应用。

1. Definition of Moment | 力矩的定义

The moment of a force F about a point O is defined as the cross product of the position vector r and the force vector F:

力 F 关于点 O 的力矩定义为位矢 r 与力向量 F 的叉积：

M = r × F

Here r is the vector from O to any point on the line of action of the force. The SI unit of moment is the newton-metre (N·m). The vector M is perpendicular to the plane containing r and F, and its direction is determined by the right-hand rule.

其中 r 是从 O 到力的作用线上任意一点的向量。力矩的 SI 单位是牛·米（N·m）。向量 M 垂直于由 r 和 F 构成的平面，其方向由右手定则确定。

Physically, the moment vector measures the tendency of the force to rotate a body about O. The larger the force or the farther its line of action from O, the greater the rotation effect. In 3D, the moment is not a single number but has three components, each describing rotation about one coordinate axis.

从物理上讲，力矩向量衡量力使物体绕 O 点转动的趋势。力越大，或力的作用线离 O 越远，转动效应就越强。在三维中，力矩不是一个单独的数，而是有三个分量，每个分量描述绕一个坐标轴的转动。

2. Cross Product in Cartesian Coordinates | 笛卡尔坐标中的叉积

Let the position vector be expressed in Cartesian components as r = x i + y j + z k, and the force as F = Fx i + Fy j + Fz k. The cross product is:

设位矢在笛卡尔坐标中的分量为 r = x i + y j + z k，力为 F = Fx i + Fy j + Fz k，则叉积为：

M = r × F = (y Fz − z Fy) i + (z Fx − x Fz) j + (x Fy − y Fx) k

Each component can be remembered as follows:

各分量可这样记忆：

Mx = y Fz − z Fy: this is the moment about the x-axis, produced by the y- and z-components of r and F.
Mx = y Fz − z Fy：这是关于 x 轴的力矩，由 r 和 F 的 y、z 分量产生。
My = z Fx − x Fz: notice the order is reversed in the second term, which is a common source of sign errors.
My = z Fx − x Fz：注意第二项顺序相反，这是常见的符号错误来源。
Mz = x Fy − y Fx: this is the familiar 2D moment about the z-axis when only x and y components exist.
Mz = x Fy − y Fx：当只有 x、y 分量时，这就是熟悉的绕 z 轴的二维力矩。

An alternative compact representation uses the determinant:

另一种紧凑表示使用行列式：

M = r × F = | i j k; x y z; Fx Fy Fz |

Expanding along the first row gives exactly the three component formulas above. This determinant method is usually the fastest and least error-prone for numerical work.

按第一行展开恰好得到上面的三个分量公式。在数值计算中，行列式法通常最快且不易出错。

3. Choosing the Position Vector | 位矢的选取

Any point on the line of action of the force may be used as the tip of r. Suppose the force acts at point A, so r = OA. If instead you choose another point B on the same line of action, then the new position vector is r’ = OB. Since AB is parallel to F, the change in moment is:

力的作用线上的任意一点都可以作为 r 的终点。假设力作用于点 A，则 r = OA。若选择同一直线上另一点 B，新的位矢是 r’ = OB。由于 AB 平行于 F，力矩的变化为：

(r’ − r) × F = (AB) × F = 0

Therefore the moment about O is independent of which point on the line of action is chosen. This property allows us to slide a force along its line of action without changing its moment about a given point.

因此，力关于 O 的力矩与作用线上选取哪一点无关。这一性质允许我们沿作用线滑动力，而不改变它关于给定点的力矩。

In practical calculations, choose the point whose coordinates are simplest, or the point where the force is actually applied, to minimize arithmetic and reduce mistakes.

在实际计算中，选择坐标最简单的点，或者力的实际作用点，可以最小化运算量并减少错误。

4. Moment about an Axis | 关于轴的力矩

Often we need the moment of a force about an axis rather than a point. Let the axis pass through a point O and have a unit direction vector u. The moment of force F about this axis is the scalar projection of the moment vector M = r × F onto u:

通常我们需要的是力关于某条轴而不是某个点的力矩。设轴过点 O，单位方向向量为 u。力 F 关于该轴的力矩等于力矩向量 M = r × F 在 u 上的标量投影：

M_axis = M · u = (r × F) · u

This scalar is the signed magnitude of the component of M along the axis. It measures the tendency of the force to rotate the body about that particular axis.

这个标量是 M 沿轴方向分量的带符号大小，它衡量力使物体绕该特定轴转动的趋势。

The scalar triple product (r × F) · u has a geometric interpretation: its absolute value is the volume of the parallelepiped formed by r, F and u. If this volume is zero, the moment about the axis is zero. This happens when the three vectors are coplanar, which means the line of action of F intersects the axis or is parallel to it.

混合积 (r × F) · u 有几何意义：其绝对值等于由 r、F、u 构成的平行六面体的体积。若体积为零，则轴矩为零。这发生在三个向量共面时，即力的作用线与轴相交或平行。

When computing M_axis, the order of the scalar triple product can be cyclically permuted without changing the result: (r × F) · u = (F × u) · r = (u × r) · F. This flexibility can simplify calculations.

计算 M_axis 时，混合积可以循环置换而不改变结果：(r × F) · u = (F × u) · r = (u × r) · F。这种灵活性可以简化计算。

5. Step-by-Step Determinant Calculation | 行列式法的逐步计算

To compute M = r × F numerically, follow these steps:

要数值计算 M = r × F，可按以下步骤：

Write r and F in component form: r = (x, y, z), F = (Fx, Fy, Fz).
写出 r 和 F 的分量形式：r = (x, y, z)，F = (Fx, Fy, Fz)。
Set up the determinant: | i j k; x y z; Fx Fy Fz |.
建立行列式：| i j k; x y z; Fx Fy Fz |。
Expand: Mx = y Fz − z Fy, My = z Fx − x Fz, Mz = x Fy − y Fx.
展开：Mx = y Fz − z Fy，My = z Fx − x Fz，Mz = x Fy − y Fx。
Write the result as M = Mx i + My j + Mz k, with units N·m.
将结果写成 M = Mx i + My j + Mz k，单位为 N·m。

Example: Let r = 2 i + 3 j − k and F = 4 i − 5 j + 2 k. Then:

例：设 r = 2 i + 3 j − k，F = 4 i − 5 j + 2 k，则：

Mx = 3×2 − (−1)×(−5) = 6 − 5 = 1

My = (−1)×4 − 2×2 = −4 − 4 = −8

Mz = 2×(−5) − 3×4 = −10 − 12 = −22

So M = i − 8 j − 22 k N·m. A quick check: r · M = 2×1 + 3×(−8) + (−1)×(−22) = 2 − 24 + 22 = 0, and F · M = 4×1 + (−5)×(−8) + 2×(−22) = 4 + 40 − 44 = 0. Both dot products are zero, confirming that M is perpendicular to both r and F.

因此 M = i − 8 j − 22 k N·m。快速检验：r · M = 2×1 + 3×(−8) + (−1)×(−22) = 2 − 24 + 22 = 0，F · M = 4×1 + (−5)×(−8) + 2×(−22) = 4 + 40 − 44 = 0。两个点积均为零，证明 M 同时垂直于 r 和 F。

6. Right-Hand Rule and Direction | 右手定则与方向

The direction of the moment vector is determined by the right-hand rule. If you point your right hand along r and curl your fingers toward F, your thumb points in the direction of M. This direction is perpendicular to both r and F.

力矩向量的方向由右手定则确定。若右手沿 r 方向指向，四指朝 F 方向弯曲，则拇指指向 M 的方向。该方向同时垂直于 r 和 F。

In Cartesian unit vectors, the cross product obeys:

在笛卡尔单位向量中，叉积满足：

i × j = k, j × k = i, k × i = j
i × j = k，j × k = i，k × i = j
j × i = −k, k × j = −i, i × k = −j
j × i = −k，k × j = −i，i × k = −j
i × i = j × j = k × k = 0
i × i = j × j = k × k = 0

These rules are useful when checking the sign of each component. Notice that the cross product is anti-commutative: r × F = − (F × r). Therefore, the order of multiplication is crucial. Always write the position vector first when defining moment about a point.

这些规则可用于检验各分量的符号。注意叉积是反交换的：r × F = − (F × r)。因此乘法顺序至关重要。在定义关于某点的力矩时，始终先写位矢。

If the force vector is parallel to the position vector, then θ = 0 or 180° and the moment is zero; the force passes through the reference point and produces no rotation about it.

若力向量与位矢平行，则 θ = 0° 或 180°，力矩为零；即力通过参考点，不产生绕该点的转动。

7. Magnitude and Perpendicular Distance | 力矩的大小与垂直距离

The magnitude of the moment vector is:

力矩向量的大小为：

|M| = |r||F| sin θ

where θ is the angle between r and F. This can also be seen as:

其中 θ 是 r 与 F 的夹角。这也可以看作：

|M| = |F| × d

where d = |r| sin θ is the perpendicular distance from the point O to the line of action of the force. This distance is often called the lever arm or moment arm.

其中 d = |r| sin θ 是从点 O 到力的作用线的垂直距离，通常称为力臂或力矩臂。

In 3D, the lever arm is not simply the distance between two points; it is the shortest distance from O to the infinite line containing the force vector. This distance can be found by projecting r onto a direction perpendicular to F:

在三维中，力臂不是两点之间的简单距离，而是从 O 到包含力向量的无限直线的最近距离。可通过将 r 投影到垂直于 F 的方向上得到：

d = |r| sin θ = |r × F| / |F|

This formula is especially useful when the geometry is complicated and the perpendicular distance is not obvious.

当几何关系复杂且垂直距离不明显时，此公式尤其有用。

8. Worked Example: Single Force | 例题：单个力

Consider a force F = (3, −2, 4) N acting at point A(1, 2, −1) m. Find the moment about the origin O(0, 0, 0).

考虑力 F = (3, −2, 4) N 作用于点 A(1, 2, −1) m，求关于原点 O(0,0,0) 的力矩。

Here r = OA = (1, 2, −1). Using the determinant:

这里 r = OA = (1, 2, −1)。使用行列式：

M = | i j k; 1 2 −1; 3 −2 4 |

Compute each component:

计算每个分量：

Mx = 2×4 − (−1)×(−2) = 8 − 2 = 6
Mx = 2×4 − (−1)×(−2) = 8 − 2 = 6
My = (−1)×3 − 1×4 = −3 − 4 = −7
My = (−1)×3 − 1×4 = −3 − 4 = −7
Mz = 1×(−2) − 2×3 = −2 − 6 = −8
Mz = 1×(−2) − 2×3 = −2 − 6 = −8

So M

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Moment of a Force in 3D: Vector Calculation | 三维空间中力矩的向量计算