Let's assume that forces \(P_{1}, P_{2}, \ldots, P_{n}\) are attached at certain points - this set of forces is called a system of forces [it may be planar or spatial) (Fig.1). If we choose an arbitrary point called the pole, for example point A, we determine the so-called radius vectors. They start at point A and end at the point of force attachment. There are as many vectors as there are forces. \(r_{1}, r_{2}, \ldots, r_{n}\), where \(r_{1}=\overline{A A_{1}}\) etc.

We determine the geometric sum of moments of all the forces in the system with respect to the pole A, that is:
$$
\overline{M_{A}}=\overline{M_{A}}\left(\overline{P_{1}}\right)+\overline{M_{A}}\left(\overline{P_{2}}\right)+\ldots+\overline{M_{A}}\left(\overline{P_{n}}\right)=\sum_{i=1}^{n} \overline{M_{A}}\left(\overline{P_{i}}\right)
$$

(Fig.1 any system of forces)

This equation can also be written as $$ \overline{M_{A}}=\overline{r_{1}} \times \overline{P_{1}}+\overline{r_{2}} \times \overline{P_{2}}+\ldots+\overline{r_{n}} \times \overline{P_{n}}=\sum_{i=1}^{n}\left(\overline{r_{i}} \times \overline{P_{i}}\right) $$

These equations define the so-called general moment vector of the system of forces with respect to the pole A. It is the geometric sum of moments of all the component forces with respect to the same pole. The general moment of the system of forces defined with respect to different poles gives different results.

Fig.2 any system of forces

The vector sum \(\bar{S}\) of the system of forces is defined by the equation: $$ \bar{S}=\overline{P_{1}}+\overline{P_{2}}+\ldots+\overline{P_{n}}=\sum_{i=1}^{n} \overline{P_{i}} $$ If we choose a new pole, for example B, such that the position of point A with respect to point B determines the radius vector \(\overline{r_{A}}\) (Fig.2), then the general moment with respect to this pole is determined as $$ \overline{M_{B}}=\overline{r_{1} \prime} \times \overline{P_{1}}+\overline{r_{2} \prime} \times \overline{P_{2}}+\ldots+\overline{r_{n}{ }^{\prime}} \times \overline{P_{n}} $$ Since it follows from the respective triangles that $$ \left\{\begin{array}{l} \overline{r_{1} \prime}=\overline{r_{A}}+\overline{r_{1}} \\ \overline{r_{2} \prime}=\overline{r_{A}}+\overline{r_{2}} \\ \vdots \\ \overline{r_{n}}=\overline{r_{A}}+\overline{r_{n}} \end{array}\right. $$ Introducing dependency (5) into equation (4) we obtain: $$ \overline{M_{B}}=\overline{r_{1}} \times \overline{P_{1}}+\overline{r_{2}} \times \overline{P_{2}}+\ldots+\overline{r_{n}} \times \overline{P_{n}}+\overline{r_{A}} \times\left(\overline{P_{1}}+\overline{P_{2}}+\ldots+\overline{P_{n}}\right) $$Taking into account relationship (3) and (2), equation (6) can be written as $$ \overline{M_{B}}=\overline{M_{A}}+\overline{r_{A}} \times \bar{S} $$

This equation shows that the general moment vector of the system of forces, defined with respect to the new pole B, is equal to the geometric sum of the general moment vector defined with respect to the old pole A and the moment vector that the sum vector attached to the old pole gives with respect to the new pole.

This relationship (7) is called the theorem of change of moment pole.

If \(\bar{S}=0\), then from equation [7] it follows that \(\overline{M_{B}}=\overline{M_{A}}=\) constant. In this case, the general moment vector of the system of forces is a so-called free vector, which means that it can be attached at any point while preserving its direction, sense and value; in this case, we say that the general moment vector is an invariant of the system.

Let's consider the set of forces \(P_{1}, \ldots, P_{n}\) in the xy plane attached at points \(A_{1}, \ldots, A_{n}\) (Fig.3). By means of the so-called reduction of the system of forces, the action of such a system can be replaced by a simpler system.

In carrying out the reduction of the system of forces, we assume a so-called reduction pole, i.e. an arbitrary point lying in the xy plane, for example point A, and replace the action of the system of forces with the following vectors: - the sum vector \(\bar{S}\), - the moment vector \(\overline{M_{A}}\).

Fig.3 planar arbitrary system of forces

The sum vector \(\bar{S}\) of the system of forces \(\bar{S}=\sum_{i=1}^{n} \overline{P_{i}}\) - during the reduction, this vector is called the main vector, it is an invariant of the system (Fig.4)

Fig.4 main vector of the system of forces

Its value is \(S=\sqrt{\left(S_{x}\right)^{2}+\left(S_{y}\right)^{2}}\), where $$ \begin{aligned} &S_{x}=\sum_{i=1}^{n} P_{ix} \\ &S_{y}=\sum_{i=1}^{n} P_{iy} \end{aligned} $$

The general moment \(\overline{M_{A}}\) of the system of forces $$ \overline{M_{A}}=\sum_{i=1}^{n}\left(\overline{r_{i}} \times \overline{P_{i}}\right) $$ This vector is called the principal moment during the reduction. It is not an invariant of the system, it is attached at the reduction pole, its direction is perpendicular to the plane in which the forces lie, i.e. to the xy plane

Fig.5 main moment vector of the system of forces

The result of the reduction

As a result of the reduction, the action of the planar arbitrary system of forces was replaced by the main vector and the main moment. This system of two vectors is shown in the figure below.

Fig.6 main vector and main moment

This system of two vectors produces the same effects as the reduced system of forces.

Reduction cases

Since the reduction pole can be chosen arbitrarily, the following cases can occur:

1) \(\bar{S} \neq 0, \overline{M_{A}}=0\) - in this case, the action of the system of forces is replaced by a single resulting force \(\bar{S}=\bar{W}\) - resultant force vector (Fig.7)

Fig.7 resultant force vector as a result of reduction; 1-line of action of the resultant force

\(\text { 2) } \bar{S}=0, \overline{M_{A}} \neq 0 \text { - the action of the system of forces is replaced by the so-called couple, } \overline{M_{A}} \text { - moment vector of the couple (Fig.8) }\)

Fig.8 moment vector of the couple as a result of reduction

3] $\bar{S} \neq 0, \overline{M_{A}} \neq 0$ - in this case, we can replace the main vector and the main moment with an even simpler system; we replace the moment vector with a couple lying in the xy plane, and we move, rotate, and adjust the value of the force of the couple so that one of the couple forces together with the vector \(\bar{S}\) forms a zero pair \(\bar{S}+\overline{W^{\prime}}=0\) [a pair of forces lying on one line with opposite directions and equal magnitudes], as shown in the following figure

Fig.9 resultant force vector

\(W=W^{\prime}=S\). Then the arm of the couple \(h=\frac{M_{A}}{S}\)

In this case, the action of the system of forces is replaced by a single force, which is the resultant force of the system of forces shifted from the reduction pole by a distance h.

\(\text { 4) } \bar{S}=0, \overline{M_{A}}=0 \text { - the system of forces remains in static equilibrium. }\)