Reduction of a flat arbitrary system of forces

General moment of the force system

Assume that forces \(P_{1}, P_{2}, \ldots, P_{n}\) are attached at certain points - this is a set of forces called a force system (it can be planar or spatial) (fig.1). If we choose any point called the pole, e.g., point A, we will define the so-called radius vectors. They start at point A and end at the point of force attachment. There are as many such vectors as there are forces. \(r_{1}, r_{2}, \ldots, r_{n}\), where \(r_{1}=\overline{A A_{1}}\) etc.

We define the geometric sum of the moments of all forces in the system with respect to pole A, that is:

$$ \overline{M_{A}}=\overline{M_{A}}\left(\overline{P_{1}}\right)+\overline{M_{A}}\left(\overline{P_{2}}\right)+\ldots+\overline{M_{A}}\left(\overline{P_{n}}\right)=\sum_{i=1}^{n} \overline{M_{A}}\left(\overline{P_{i}}\right) $$

This equation can also be written in the form

$$ \overline{M_{A}}=\overline{r_{1}} \times \overline{P_{1}}+\overline{r_{2}} \times \overline{P_{2}}+\ldots+\overline{r_{n}} \times \overline{P_{n}}=\sum_{i=1}^{n}\left(\overline{r_{i}} \times \overline{P_{i}}\right) $$

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

Change of the moment pole

The vector sum of the force system \(\bar{S}\) is defined by the relationship:

$$ \bar{S}=\overline{P_{1}}+\overline{P_{2}}+\ldots+\overline{P_{n}}=\sum_{i=1}^{n} \overline{P_{i}} $$

If we choose a new pole, e.g., B, such that the position of point A with respect to point B is defined by the radius vector \(\overline{r_{A}}\) (fig2), then the general moment with respect to this pole will be defined 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 appropriate triangles that

$$ \left\{\begin{array}{l} \overline{r_{1} \prime}=\overline{r_{A}}+\overline{r_{1}} \\ \overline{r_{2} \prime}=\overline{r_{A}}+\overline{r_{2}} \\ \vdots \\ \overline{r_{n}}=\overline{r_{A}}+\overline{r_{n}} \end{array}\right. $$

Substituting relationship (5) into equation (4) we obtain:

$$ \overline{M_{B}}=\overline{r_{1}} \times \overline{P_{1}}+\overline{r_{2}} \times \overline{P_{2}}+\ldots+\overline{r_{n}} \times \overline{P_{n}}+\overline{r_{A}} \times\left(\overline{P_{1}}+\overline{P_{2}}+\ldots+\overline{P_{n}}\right) $$

Considering relationships (3) and (2), equation (6) can be written in the form

$$ \overline{M_{B}}=\overline{M_{A}}+\overline{r_{A}} \times \bar{S} $$

From this equation, it follows that the vector of the general moment of the force system, defined with respect to the new pole B, is equal to the geometric sum of the vector of the general moment defined with respect to the old pole A and the vector of the moment that the vector sum gives with respect to the new pole anchored at the old pole.

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

If \(\bar{S}=0\), then from equation (7) it follows that \(\overline{M_{B}}=\overline{M_{A}}=\) const. Then the vector of the general moment of the force system is called a free vector, meaning it can be anchored at any point while maintaining its direction, sense, and value; we then say that the vector of the general moment is an invariant of the system.

Reduction of a planar arbitrary force system

Consider a set of forces \(P_{1}, \ldots, P_{n}\) in the xy plane attached at points \(A_{1}, \ldots, A_{n}\) (fig3). Through the so-called reduction of the force system, the action of such a system can be replaced by a simpler system

When performing the reduction of the force system, we adopt the so-called reduction pole, i.e., any point lying in the xy plane, e.g., point A, and replace the action of the force system with the following vectors:

• the sum vector \(\bar{S}\),
• the moment vector \(\overline{M_{A}}\).



a) The sum vector \(\bar{S}\) of the force system

\(\bar{S}=\sum_{i=1}^{n} \overline{P_{i}}\) - during reduction, this vector is called the main vector, it is an invariant of the system (fig4)



Its value is \(S=\sqrt{\left(S_{x}\right)^{2}+\left(S_{y}\right)^{2}}\), where

$$ \begin{aligned} &S_{x}=\sum_{i=1}^{n} P_{ix} \\ &S_{y}=\sum_{i=1}^{n} P_{iy} \end{aligned} $$

b) The general moment of the force system \(\overline{M_{A}}\)

$$ \overline{M_{A}}=\sum_{i=1}^{n}\left(\overline{r_{i}} \times \overline{P_{i}}\right) $$

This vector during reduction is called the main moment. It is not an invariant of the system, we anchor it at the reduction pole, its direction is perpendicular to the plane in which the forces lie, that is, to the xy plane



c) Result of the reduction

As a result of the performed reduction, we replaced the action of the planar arbitrary force system with the main vector and moment. This system of two vectors is shown in the figure below.



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

Cases of reduction

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

1) \(\bar{S} \neq 0, \overline{M_{A}}=0\)

In this case, the action of the force system is replaced by a single force, which is the resultant force \(\bar{S}=\bar{W}\) - the vector of the resultant force (fig7)



2) \(\bar{S}=0, \overline{M_{A}} \neq 0\)

The action of the force system is replaced by the so-called resultant couple, \(\overline{M_{A}}\) - the moment vector of the resultant couple (fig8)



3) \(\bar{S} \neq 0, \overline{M_{A}} \neq 0\)

In this case, we can replace the main vector and main moment with an even simpler system; we replace the moment vector with a couple of forces lying in the xy plane and we move, rotate, and select the value of the force of the couple so that one force of the couple with the vector \(\bar{S}\) creates a zero couple \(\bar{S}+\overline{W^{\prime}}=0\) (a couple of forces lying on the same line with opposite directions and the same values), as shown in the figure below



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

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

4) \(\bar{S}=0, \overline{M_{A}}=0\)

The force system remains in static equilibrium.