There are two simple ways of movement in a rigid system: translation and rotation. Any other possible form of movement, as complex it may be, can always be considered as the combination of a rotation and a translation.
It is not always possible to consider a body as a point particle, in general, when we are interested not only in the displacement of an object, but also in its rotation. The following definition is important:
The two possible types of motion of a rigid body can be defined as:
For a solid to be in equilibrium in an inertial frame, it is necessary to satisfy two conditions: one relating to the balance of translation and other related to the rotation balance, defined as follows:
The figure above illustrates an extensive body, the bridge, suffering the action of various forces. Whereas the system of interest is the bridge, since we desire it to stay in a static equilibrium, the forces acting in it must meet the following conditions: to have no translation, the resultant should be zero, \begin{align} +& \vec{N}^{(1)} + \vec{N}^{(2)} + \vec{F}_{T,B}^{(1)} +\\ +& \vec{F}_{T,B}^{(2)} + \vec{P} = 0, \end{align} and so, to not have a rotation, the torque on the bridge must also be zero \begin{align} -& N^{(1)} d_{1} + N^{(2)} d_{2} + F_{T,B}^{(1)} d_{3} +\\ +& F_{T,B}^{(2)} d_{4}+ P d_{5}= 0, \end{align} where each \(d_i\) is the distance of each force in respect to the bridge's center of mass (position of vector \(P\)), then \(d_5 = 0\).
If a rigid system is in equilibrium under the action of only three external forces, \(F_1\), \(F_2\) and \(F_3\), not parallel, the module of each is proportional to the sine of the angle between the other two, namely: $$ \frac{F_1}{sen(a)} = \frac{F_2}{sen(b)} = \frac{F_3}{sen(c)},$$ where \(a\), \(b\) and \(c\) are the angles between the forces, as shown in the figure below.
Any system of forces, as complex as it seems, can always be reduced to a single force, known as the net force, and a torque. The torque and net force are orthogonal, always.