Coplanar System of Forces

Basic Terms

The static moment of force. If line on which force lies goes through the center of gravity than force tends to cause translation (parallel displacement) of the body. If line of which force lies does not go through the center of gravity, besides the translation, the force tends to cause the rotation of the body around the center of gravity. The tendency of a force to cause the rotation is called the static moment of force of shortly the moment of force.
The definition of the force moment is: The static moment of force around some point is equal to the product of perpendicular distance from the point to the line of action of the force and magnitude of the force.
The static force is calculated as: $$ M_0^F = h\cdot F$$ where \(h\) is the perpendicular distance from point to he line of action of the force \(F\). The static moment of force around the point O is shown in Figure 5.1. T
Figure 5.1 - A force is a sliding vector because the moment of force around a point does not change if the force moves in the direction of its action.
Force couple consist of two forces that are equal and oppositely directed and do not lie on the same line. If force couple are acting on the rigid plate as shown in Figure 5.2 then the force couple is causing the plate rotation without translation. Figure 5.2 - The fource coupling - two equal, non-colinear and oppositely directed forces. The moment of force couple \(M\) is equal to the product of force from force couple and the mutual distance between forces. This can be written in mathematical from as: $$ M = aF.$$ The sine of the force couple is positive if it is acting in counterclockwise direction. The moment of the force and moment of force couple can be defined as vectors: \begin{eqnarray} \vec{M}_O^F &=& \vec{r} \times \vec{F}\\ \vec{M} &=& \vec{a} \times \vec{F}. \end{eqnarray} The force moment vector is perpendicular to the plane passing through the point O and the direction of the force as shown in Figure 5.3. Figure 5.3 The force moment and moment of the force coupling. The force moment is vector connected to the point. The force coupling moment is free vector. The force coupling vector is perpendicular to the coupling force plane. The absolute value of the moment vector is equal to the magnitude of the moment, and the meaning is determined by the rule of the right hand. The coupling moment does not depend on the choice of the point according to which it is calculated. Figure 5.4 shows a system with force coupling, and it is necessary to determine the torque at points A, B and C. Figure 5.4 - The system with force couple
The moments in point A, B and C can be calculated as follows: \begin{eqnarray} M_A &=& Fa \\ M_B &=& F(a+b)-Fb = Fa + Fb - Fb = Fa\\ M_C &=& F(a+b+c) - F(b+c) = Fa + Fb + Fc - Fb - Fc = Fa \end{eqnarray} From previous example it can be concluded that force couple is a free vector.

Varignon's theorem

The Varignon's theorem was originally developed by the French mathematician Varignon (1654 - 1722). This theorem states that the moment of a force about a point is equal to the sum of the moments of the components of the force about the point. The proof of this theorem can be achieved using the vecotr cross porduct since the cross product obeys the distrubutive law. Consider the moments of the force \(\vec{F}\) and two of its components about the point O as shown in Figure 5.5.
Figure 5.5 - Example for proof of Varignon's theorem Since \(\vec{F} = \vec{F}_1 +\vec{F}_2\) we have: \begin{eqnarray} \vec{M}_O = \vec{r} \times \vec{F} = \vec{r} \times (\vec{F}_1 + \vec{F}_2) = \vec{r} \times \vec{F}_1 + \vec{r} \times \vec{F}_2. \end{eqnarray} For two dimensional problem as shown in Figure 5.6 the principle of moments can be used by resolving the force into its rectangular components and then determine the moment using a sclar analysis. \begin{eqnarray} M_O &=& F_xy - F_yx \end{eqnarray} Figure 5.6 - The moment around the point O in two dimensional probelm. This method is gnerally easier than finding the same moment using \(M_O = Fd\).

Force system reduction

The reduction of the plane system of forces on one point is shown in figure Figure 5.7. Figure 5.7 - The reduction of coplanar system of forces. On the plane \(n\) forces \(\vec{F}_1, \vec{F}_2,...,\vec{F}_n\), are acting. According to the rule of parallel force displacement, all forces may be moved in parallel to one point called the reduction pole P. the corresponding moment is added for each force: \(M_1, M_2,...,M_n\). The resultant \(\vec{F}_R\) is determined in the usual vector way, i.e.: \begin{eqnarray} \vec{F}_R &=& \vec{F}_1 +\vec{F}_2 + \cdots + \vec{F}_n = \sum_{i=1}^{n} \vec{F}_i \end{eqnarray} or analyticly: \begin{eqnarray} F_{Rx} &=& \sum F_{ix}, \quad F_{Ry} = \sum F_{iy}\\ F_R &=& \sqrt{F_{Rx}^2 + F_{Ry}^2}, \quad \tan \alpha = \frac{F_{Ry}}{F_{Rx}} \end{eqnarray} The resulting moment \(M_R\) is equal to the algebraic sum of the moments of individual forces, ie: \begin{eqnarray} M_R = \sum M_i. \end{eqnarray}

Invariant of coplanar system of forces

By changing the reduction pole P changes the value of the resulting moment \(M_R\) while the value of the resultant \(F_R\) remains unchanged. Therefore, the resultant \(F_R\) is the invariant of the coplanar system of forces. By a suitable choice of the pole P_1, the coplanar system of forces can be reduced to only one force, i.e. the resultant \(F_R\).
Figure 5.8 Reduction of a plane system of forces to one force
The position of the new pole \(P_1\) relative to the old pole of reduction P is determined by the distance \(d = M_R/F_R\). The pole \(P_1\) must be selected so that the resultant \(F_R\) in the new position cancels the moment \(M_R\).

Equilibrium of a rigid plate in a plane

Rigid plate connections

The rigid plate in a plane can make three independent movements and these are two translations and one rotation i.e. the plate has three degrees of freedom. To make plate immovable, three connections should be added to the plate as shown in Figure xxxx.
However when adding connections to the plate you have to avoid:

• The axes of the rods must not intersect at one point S.
• The rods must not be parallel, as in this case the plate could move perpendicular to the direction of the rods.

Canonical conditions of equilibrium

For a rigid plate to be in equilibrium, the resultant F_R and the resultant coupling M_R of all forces acting on the plate, including the bond reactions, must be equal to zero, ie it must be: \begin{eqnarray} \vec{F}_R &=& 0 \quad \vec{M}_R =0\\ \end{eqnarray} Expressed analytically, the term reads: \begin{eqnarray} F_{Rx} &=& \sum F_{ix} = 0 \\ F_{Ry} &=& \sum F_{iy} = 0 \\ M_R &=& \sum M_i = 0. \end{eqnarray} The above three equations are how the equilibrium conditions are. The equations thus set up are always linearly independent. The first two equations of expression are called the equations of forces, and the third the equation of moments. Equilibrium conditions can be set differently. Namely, each force equation can be replaced by a complementary moment equation. The reversal is not valid: the moment equation must not be replaced by the complementary force equation.