If the directions of action of all the forces of a system of forces intersect at one point, the system is a system of concurrent or opposing forces. When all forces, lie in one plane, it is a planar-concurrent system of forces.
The force \(\vec{F}\) can be easily broken down to two components \(\vec{F}_1\) and \(\vec{F}_2\). Graphically, this is done by drawing a parallel with the direction of the component \(\vec{F}_1\) through one end of the vector \(\vec{F}\) and a parallel with the direction of the component \(\vec{F}_2 \) through the other end. This leads to a triangle of forces which is shown in Figure 4.1.
Figure 4.1 The graphical solution using triangle of forces.
The resultant force. The resultant of a concurrent coplanar system of forces can be determined graphically and analytically. On Figure 4.2 a plan of the position of several forces acting on a single point is shown. The resultant force is determined in the plan of forces using polygon of forces. The forces are applied one after the other by direction and size, starting from a pole P. The order of appliaction of forces can be changed, it does not affect the size of the resultant force.
The resultant force can be determined using analytical approach so every force is projected on coordiantes \(x\) and \(y\). The example of projected forces is shown in Figure 4.2.
Figure 4.2 Force projections on coordinate axis.
From Figure 4.2 it can be noticed that projection of forces on \(x\) and \(y\) axes are equal to algebraic addition of component projections which can be written as: \begin{eqnarray} F_{Rx} &=& F_{1x} + F_{2x} + \cdots + F_{ix} + \cdots + F_{nx} = \sum_{i=1}^n F_{ix},\\ F_{Ry} &=& F_{1y} + F_{2y} + \cdots + F_{iy} + \cdots + F_{nx} = \sum_{i=1}^n F_{iy}, \end{eqnarray} where \(F_{Rx}\) and \(F_{Ry}\) are resultant components, \(F_{1x}, F_{1y}, F_{2x}, F_{2y}\) and etc are projections of defined forces. The resultant component can be determined from equation: \begin{equation} F_R = \sqrt{F_{Rx}^2 + F_{Ry}^2}, \end{equation} and the angle \(\alpha\) between components \(F_{Rx}\) and \(F_{Ry}\) can be determined from equation: \begin{equation} \tan \alpha = \frac{F_{Ry}}{F_{Rx}}. \end{equation} Equilibrium Conditions The particle is moving in a plane and it has two independent displacements in the direction of two mutually perpendicular directions, i.e. the particle has two degrees of freedom. To make particle fixed, the particle should be added two connections. The reactions of these connections can be determined from equilibrium condition. The equilibrium condition for particle can be written as: \begin{equation} \vec{F}_R = F_{Rx}\vec{i} + F_{Ry}\vec{j} = 0, \end{equation} where \(\vec{F}_R\) is resultant of all forces that are acting on a particle. Two algebraic equations correspond to the previous equation which can be written in the following form: \begin{eqnarray} F_{Rx} &=& \sum_{i=1}^nF_{ix} = 0,\\ F_{Ry} &=& \sum_{i=1}^nF_{iy} = 0. \end{eqnarray} The process of solving equilibrium problems To solve the euqilibrium problems the procedure is:
Breaking the forces on components
The force \(\vec{F}\) can be easily broken down to two components \(\vec{F}_1\) and \(\vec{F}_2\). Graphically, this is done by drawing a parallel with the direction of the component \(\vec{F}_1\) through one end of the vector \(\vec{F}\) and a parallel with the direction of the component \(\vec{F}_2 \) through the other end. This leads to a triangle of forces which is shown in Figure 4.1.on a system with two rods |
The resultant force. The resultant of a concurrent coplanar system of forces can be determined graphically and analytically. On Figure 4.2 a plan of the position of several forces acting on a single point is shown. The resultant force is determined in the plan of forces using polygon of forces. The forces are applied one after the other by direction and size, starting from a pole P. The order of appliaction of forces can be changed, it does not affect the size of the resultant force.
The resultant force can be determined using analytical approach so every force is projected on coordiantes \(x\) and \(y\). The example of projected forces is shown in Figure 4.2.
From Figure 4.2 it can be noticed that projection of forces on \(x\) and \(y\) axes are equal to algebraic addition of component projections which can be written as: \begin{eqnarray} F_{Rx} &=& F_{1x} + F_{2x} + \cdots + F_{ix} + \cdots + F_{nx} = \sum_{i=1}^n F_{ix},\\ F_{Ry} &=& F_{1y} + F_{2y} + \cdots + F_{iy} + \cdots + F_{nx} = \sum_{i=1}^n F_{iy}, \end{eqnarray} where \(F_{Rx}\) and \(F_{Ry}\) are resultant components, \(F_{1x}, F_{1y}, F_{2x}, F_{2y}\) and etc are projections of defined forces. The resultant component can be determined from equation: \begin{equation} F_R = \sqrt{F_{Rx}^2 + F_{Ry}^2}, \end{equation} and the angle \(\alpha\) between components \(F_{Rx}\) and \(F_{Ry}\) can be determined from equation: \begin{equation} \tan \alpha = \frac{F_{Ry}}{F_{Rx}}. \end{equation} Equilibrium Conditions The particle is moving in a plane and it has two independent displacements in the direction of two mutually perpendicular directions, i.e. the particle has two degrees of freedom. To make particle fixed, the particle should be added two connections. The reactions of these connections can be determined from equilibrium condition. The equilibrium condition for particle can be written as: \begin{equation} \vec{F}_R = F_{Rx}\vec{i} + F_{Ry}\vec{j} = 0, \end{equation} where \(\vec{F}_R\) is resultant of all forces that are acting on a particle. Two algebraic equations correspond to the previous equation which can be written in the following form: \begin{eqnarray} F_{Rx} &=& \sum_{i=1}^nF_{ix} = 0,\\ F_{Ry} &=& \sum_{i=1}^nF_{iy} = 0. \end{eqnarray} The process of solving equilibrium problems To solve the euqilibrium problems the procedure is:
- The particle must be liberated of all connections by adding the reaction connections.
- The coorinate system must be selected by taking into account that both or at least one unkown reaction coincides with coordinate axis.
- The equilibrium conditions are set
- The equations of equilibrium conditions are solved to determine unkown reactions.
The spatially concurent system of forces.
The force decompositions to components - The orientation of the force in space can be defined on multiple ways: using angles that make the line on which force vector lies, using two angles \(\varphi\) and \(\psi\) or using sides of parallelepiped that main diagonal determines the line of force.
Figure 4.3 - Three different types of force decomposition in space on components.
In first case the following expressions are valid: \begin{eqnarray} F_x &=& F\cos \alpha, \quad F_y = F\cos\beta, \quad F_z = F\cos \gamma. \end{eqnarray} The angles \(\alpha, \beta,\) and \(\gamma\) are dependable since: \begin{equation} \cos^2\alpha + \cos^2\beta + \cos^2 \gamma =1, \end{equation} so it is enough to know only two angles, and the third one can be determined from previous expression. It is necessary to know in which octant of the coordinate system the force acts. In second case the following expressions are valid: \begin{eqnarray} F_x &=& F\cos \psi \cos \varphi,\\ F_y &=& F\cos \psi \cos \varphi,\\ F_z &=& F \sin \psi. \end{eqnarray} In third case the following relations are valid: \begin{eqnarray} F_x &=& \frac{aF}{\sqrt{a^2 + b^2 + c^2}},\\ F_y &=& \frac{bF}{\sqrt{a^2 + b^2 + c^2}},\\ F_z &=& \frac{cF}{\sqrt{a^2 + b^2 + c^2}} \end{eqnarray} Determination of the resultant force - the resultant of spatially concurent system of forces can be determined using same principle as in case of planar-concurrent system of forces. Namely, in this case, for the force as for any vector, the rule applies that the projection of the resultant on any axis is equal to the algebraic sum of the projections of its components on that same axis, i.e. \begin{eqnarray} F_{Rx} &=& \sum_{i=1}^n F_{ix} = \sum_{i=1}^n F_i \cos \alpha_i,\\ F_{Ry} &=& \sum_{i=1}^n F_{iy} = \sum_{i=1}^n F_i \cos \beta_i,\\ F_{Rz} &=& \sum_{i=1}^n F_{iz} = \sum_{i=1}^n F_i \cos \gamma_i\\ F_R &=& \sqrt{F_{Rx}^2 + F_{Ry}^2 + F_{Rz}^2},\\ \cos \alpha &=& \frac{F_{Rx}}{F_R},\\ \cos \beta &=& \frac{F_{Ry}}{F_R}, \end{eqnarray} where \(\alpha, \beta, \) and \(\gamma\) are angles that resultat \(F_R\) makes with coordinate axis \(x,y,\) and \(z\).
Particle equilibrium conditions - The particle in space has 3 independent movements i.e three degrees of freedom. Three joints, eg three rods, are required to fix the particle to the substrate. Two rods are not enough to fix the particle, and fourth rod makes the rendundatnt system i.e. makes the system statically indefinite. The equilibrium conditions can be written as: \begin{equation} \vec{F}_R = F_{Rx} \vec{i} + F_{Ry}\vec{j} + F_{Rz}\vec{k} = 0, \end{equation} or in scalar form: \begin{eqnarray} \sum F_{ix} &=& 0,\quad \sum F_{iy} = 0,\quad \sum F_{iz} = 0.\\ \end{eqnarray}
In first case the following expressions are valid: \begin{eqnarray} F_x &=& F\cos \alpha, \quad F_y = F\cos\beta, \quad F_z = F\cos \gamma. \end{eqnarray} The angles \(\alpha, \beta,\) and \(\gamma\) are dependable since: \begin{equation} \cos^2\alpha + \cos^2\beta + \cos^2 \gamma =1, \end{equation} so it is enough to know only two angles, and the third one can be determined from previous expression. It is necessary to know in which octant of the coordinate system the force acts. In second case the following expressions are valid: \begin{eqnarray} F_x &=& F\cos \psi \cos \varphi,\\ F_y &=& F\cos \psi \cos \varphi,\\ F_z &=& F \sin \psi. \end{eqnarray} In third case the following relations are valid: \begin{eqnarray} F_x &=& \frac{aF}{\sqrt{a^2 + b^2 + c^2}},\\ F_y &=& \frac{bF}{\sqrt{a^2 + b^2 + c^2}},\\ F_z &=& \frac{cF}{\sqrt{a^2 + b^2 + c^2}} \end{eqnarray} Determination of the resultant force - the resultant of spatially concurent system of forces can be determined using same principle as in case of planar-concurrent system of forces. Namely, in this case, for the force as for any vector, the rule applies that the projection of the resultant on any axis is equal to the algebraic sum of the projections of its components on that same axis, i.e. \begin{eqnarray} F_{Rx} &=& \sum_{i=1}^n F_{ix} = \sum_{i=1}^n F_i \cos \alpha_i,\\ F_{Ry} &=& \sum_{i=1}^n F_{iy} = \sum_{i=1}^n F_i \cos \beta_i,\\ F_{Rz} &=& \sum_{i=1}^n F_{iz} = \sum_{i=1}^n F_i \cos \gamma_i\\ F_R &=& \sqrt{F_{Rx}^2 + F_{Ry}^2 + F_{Rz}^2},\\ \cos \alpha &=& \frac{F_{Rx}}{F_R},\\ \cos \beta &=& \frac{F_{Ry}}{F_R}, \end{eqnarray} where \(\alpha, \beta, \) and \(\gamma\) are angles that resultat \(F_R\) makes with coordinate axis \(x,y,\) and \(z\).
Particle equilibrium conditions - The particle in space has 3 independent movements i.e three degrees of freedom. Three joints, eg three rods, are required to fix the particle to the substrate. Two rods are not enough to fix the particle, and fourth rod makes the rendundatnt system i.e. makes the system statically indefinite. The equilibrium conditions can be written as: \begin{equation} \vec{F}_R = F_{Rx} \vec{i} + F_{Ry}\vec{j} + F_{Rz}\vec{k} = 0, \end{equation} or in scalar form: \begin{eqnarray} \sum F_{ix} &=& 0,\quad \sum F_{iy} = 0,\quad \sum F_{iz} = 0.\\ \end{eqnarray}
Solved Problems
Problem 1 For a system shown in Figure 4.4determine analticlly and graphically the resultant force if \(F_1 = 50 \left[\mathrm{N}\right]\), \(F_2 = 100 \left[\mathrm{N}\right]\), \(F_3 =200 \left[\mathrm{N}\right]\),\(F_4 = 300 \left[\mathrm{N}\right]\), \(\alpha_1 = 30°\), \(\alpha_2 = 100°\), \(\alpha_3 = 80°\), and \(\alpha_4 = 160°\).
Figure 4.4 - The system of forces acting on a point.
Analythical Solution \begin{eqnarray} \sum F_x &=& 0 \\ \sum F_x &=& F_{1x} + F_{2x} + F_{3x} + F_{4x} = 0 \\ \cos \alpha_1 &=& \frac{F_{1x}}{F_1} \Rightarrow F_{1x} = F_1 \cos \alpha_1 = 50 \cdot \cos 30° = 43.3012 \left[\mathrm{N}\right]\\ \cos \alpha_2 &=& \frac{F_{2x}}{F_2} \Rightarrow F_{2x} = F_2 \cos \alpha_2 = 100 \cdot \cos 100° = -17.3648 \left[\mathrm{N}\right]\\ \cos \alpha_3 &=& \frac{F_{3x}}{F_3} \Rightarrow F_{3x} = F_3 \cos \alpha_3 = 200 \cdot \cos 80° = 34.729 \left[\mathrm{N}\right]\\ \cos \alpha_4 &=& \frac{F_{4x}}{F_4} \Rightarrow F_{4x} = F_4 \cos \alpha_4 = 300 \cdot \cos 160° = -281.9077 \left[\mathrm{N}\right]\\ \sum F_x &=& 43.3012-17.3648+34.729-281.9077 = -221.2423 \left[\mathrm{N}\right] \end{eqnarray} \begin{eqnarray} \sum F_y &=& 0\\ \sum F_y &=& F_{1y} + F_{2y} + F_{3y} + F_{4y} = 0\\ \sin \alpha_1 &=& \frac{F_{1y}}{F_1} \Rightarrow F_{1y} = F_1 \sin \alpha_1 = 50 \cdot \sin 30° = 25 \left[\mathrm{N}\right]\\ \sin \alpha_2 &=& \frac{F_{2y}}{F_2} \Rightarrow F_{2y} = F_2 \sin \alpha_2 = 100 \cdot \sin 100° = 98.4807 \left[\mathrm{N}\right]\\ \sin \alpha_3 &=& \frac{F_{3y}}{F_3} \Rightarrow F_{3y} = F_3 \sin \alpha_3 = 200 \cdot \sin 80° = 196.9615 \left[\mathrm{N}\right]\\ \sin \alpha_4 &=& \frac{F_{4y}}{F_4} \Rightarrow F_{4y} = F_4 \sin \alpha_4 = 300 \cdot \sin 160° = 102.606043 \left[\mathrm{N}\right]\\ \sum F_y &=& 25 + 98.4807 + 196.9615 + 102.606043 = 423.048243 \left[\mathrm{N}\right] \end{eqnarray} \begin{eqnarray} F_R &=& \sqrt{F_x^2 + F_y^2}\\ F_R &=& \sqrt{\left(-221.2423\right)^2 + \left(423.048243\right)^2}\\ F_R &=& \sqrt{48948.155 + 178969.8159} = \sqrt{227917.97}\\ F_R &=& 477.4075 \left[\mathrm{N}\right] \end{eqnarray} Graphical Solution To solve this problem graphically the plan of forces must be created and after drawing all forces the resultant force can be determined. However, the scale must be defined. In this problem 40 N will be equal to 1 cm, so 100 N is equal to 2.5 cm which is indicated in the Figure 4.1.
Figure 4.5 - The plan of forces.
The length of the resultnat force\(\vec{F}_R\) is 11.935 cm. By multiplying the length with the defined scale in Figure 4 the magnitude of the force \(\vec{F}_R\) can be calculated. $$F_R = 11.935 \left[\mathrm{cm}\right] \cdot \frac{100 \left[\mathrm{N}\right]}{2.5\left[\mathrm{cm}\right]} = 477.5 \left[\mathrm{N}\right]$$ Problem 2 For a given system of colinear forces shown in Figure (xx) calculate resultant force using analythical and graphical method. \(F_1 = 150 \left[\mathrm{N}\right]\), \(F_2 = 200 \left[\mathrm{N}\right]\), \(F_3 = 300 \left[\mathrm{N}\right]\).
Figure 4.6 - Action of the colinear system
Analythical solution For a given system of colinear forces the resultant force can be easily calculated.
\begin{eqnarray} F_R &=& \sum F_x + \sum F_y = ? \\ \sum F_x &=& 0\\ \sum F_y &=& 0 \\ \sum F_x &=& -F_1 + F_2 + F_3 \\ \sum F_x &=& -150 + 200 + 300 = 350 \left[\mathrm{N}\right]\\ F_R &=& 350 \left[\mathrm{N}\right] \end{eqnarray} Graphical solution To determine the length of resultant force the plan of forces must be developed first. The plan of forces is given in Figure 4.7.
Figure 4.7 - Plan of forces
From previous Figure 4.7 the length of the resulntant force is equal to 1.75 cm. To obtain the force magnitude the obtained length has to be multipled with hte measure defied in the Figure 4.7.
$$ F_R = 1.75 \left[\mathrm{cm}\right]\cdot \frac{100 \left[\mathrm{N}\right]}{0.5 \left[\mathrm{cm}\right]} = 350 \left[\mathrm{N}\right]$$
Problem 3 Homogenous cylinder of weight 120 [N] is in contact in point A with the smooth surface which is at angle of \(60°\) to the horizontal plane and is in contact in point B wit the corner of the horizontal surface. The point A and B are on the same level as shown in Figure 4.8. Solve the problem analytically and graphically.
Figure 4.8 - Schematic representation of a homogeneous cylinder resting on smooth walls
Analythical solution The first step in solving problem analyticly is to free homogenous cylinder of all connections by adding the reaction connections. Than the coordinate system must be selected by taking into account that both or at least one unknown reaction coincides with coordinate axis. The homogenous cylinder freed of all connections and with reaction connections is shown in Figure 4.9.
Figure 4.9 - Homogenous cylinder freed of all forces.
\begin{eqnarray}
\sum F_x &=& 0\\
F_{Bx} - F_{Ax} &=& 0,\\
F_B\sin \alpha - F_A \sin \alpha &=& 0 \Rightarrow F_A = F_B \\
\sum F_y &=& 0 \\
F_B \cos \alpha + F_A \cos \alpha - G &=& 0 \\
2F_B \cos \alpha - G &=& 0\\
F_B &=& \frac{G}{2\cos \alpha} = \frac{40}{2\cdot 0.5} = 40\left[\mathrm{N}\right]
\end{eqnarray}
Graphical solution The plan of forces is shown in Figure 4.10. The scale used in this figure is 10 N is 1 cm. From this data the \(F_A\) and \(F_B\) can be calculated.
Figure 4.10 - Plan of forces
$$F_A= 4 \left[\mathrm{cm}\right]\cdot \frac{10 \left[\mathrm{N}\right]}{\left[\mathrm{cm}\right]} = 40 \left[\mathrm{N}\right]$$
$$F_B = 4 \left[\mathrm{cm}\right]\cdot \frac{10\left[\mathrm{N}\right]}{\left[\mathrm{cm}\right]} = 40 \left[\mathrm{N}\right]$$
Problem 4 The ABC crane is maintained in balance by the BD rope. Determine analythically the force in the rope as well as the size, the direction and direction of the reaction in the joint C, if the weight of the suspended load is M in point A is equal to G. The part CB of the crane is vertical. \(\overline{AB} = a = 25.98 \left[\mathrm{m}\right]\), \(\overline{OB} = b = 20\left[\mathrm{m}\right]\), \(\overline{AO} = c = 39.93 \left[\mathrm{m}\right]\), \(G = 40 \left[\mathrm{kN}\right]\), and \(\varphi = 60°\). Ignore the crane weight.
Figure 4.11 - Schematic view of the crane
Analythical solution
Figure 4.12 - The crane ABO isolated
\begin{eqnarray}
\sum F_X &=& 0 \\
-F_{XO} + F_B\sin \varphi &=& 0 \\
\sum F_Y &=& 0\\
F_{YO} -G-F_B\cos\varphi &=& 0 \\
\sum M_O &=& 0 \\
Gm - F_B \cdot n &=& 0 \\
F_{XO} &=& F_B \sin \varphi = 51.96 \cdot 0.86603 = 42 \left[\mathrm{kN}\right]\\
F_{YO} &=& G + F_B \cos \varphi = 40 + 51.96\cdot 0.5 = 65.98 \left[\mathrm{kN}\right]\\
F_B &=& \frac{Gm}{n} = \frac{Ga\sin \varphi}{b\sin \varphi} = \frac{Ga}{b}\\
F_B &=& \frac{40\cdot 25.98}{20} = 51.96 \left[\mathrm{kN}\right]\\
F_O &=& \sqrt{F_{XO}^2 + F_{YO}^2} = \sqrt{45^2 + 65.98^2} = 79.68 \left[\mathrm{kN}\right]
\end{eqnarray}
Analythical Solution \begin{eqnarray} \sum F_x &=& 0 \\ \sum F_x &=& F_{1x} + F_{2x} + F_{3x} + F_{4x} = 0 \\ \cos \alpha_1 &=& \frac{F_{1x}}{F_1} \Rightarrow F_{1x} = F_1 \cos \alpha_1 = 50 \cdot \cos 30° = 43.3012 \left[\mathrm{N}\right]\\ \cos \alpha_2 &=& \frac{F_{2x}}{F_2} \Rightarrow F_{2x} = F_2 \cos \alpha_2 = 100 \cdot \cos 100° = -17.3648 \left[\mathrm{N}\right]\\ \cos \alpha_3 &=& \frac{F_{3x}}{F_3} \Rightarrow F_{3x} = F_3 \cos \alpha_3 = 200 \cdot \cos 80° = 34.729 \left[\mathrm{N}\right]\\ \cos \alpha_4 &=& \frac{F_{4x}}{F_4} \Rightarrow F_{4x} = F_4 \cos \alpha_4 = 300 \cdot \cos 160° = -281.9077 \left[\mathrm{N}\right]\\ \sum F_x &=& 43.3012-17.3648+34.729-281.9077 = -221.2423 \left[\mathrm{N}\right] \end{eqnarray} \begin{eqnarray} \sum F_y &=& 0\\ \sum F_y &=& F_{1y} + F_{2y} + F_{3y} + F_{4y} = 0\\ \sin \alpha_1 &=& \frac{F_{1y}}{F_1} \Rightarrow F_{1y} = F_1 \sin \alpha_1 = 50 \cdot \sin 30° = 25 \left[\mathrm{N}\right]\\ \sin \alpha_2 &=& \frac{F_{2y}}{F_2} \Rightarrow F_{2y} = F_2 \sin \alpha_2 = 100 \cdot \sin 100° = 98.4807 \left[\mathrm{N}\right]\\ \sin \alpha_3 &=& \frac{F_{3y}}{F_3} \Rightarrow F_{3y} = F_3 \sin \alpha_3 = 200 \cdot \sin 80° = 196.9615 \left[\mathrm{N}\right]\\ \sin \alpha_4 &=& \frac{F_{4y}}{F_4} \Rightarrow F_{4y} = F_4 \sin \alpha_4 = 300 \cdot \sin 160° = 102.606043 \left[\mathrm{N}\right]\\ \sum F_y &=& 25 + 98.4807 + 196.9615 + 102.606043 = 423.048243 \left[\mathrm{N}\right] \end{eqnarray} \begin{eqnarray} F_R &=& \sqrt{F_x^2 + F_y^2}\\ F_R &=& \sqrt{\left(-221.2423\right)^2 + \left(423.048243\right)^2}\\ F_R &=& \sqrt{48948.155 + 178969.8159} = \sqrt{227917.97}\\ F_R &=& 477.4075 \left[\mathrm{N}\right] \end{eqnarray} Graphical Solution To solve this problem graphically the plan of forces must be created and after drawing all forces the resultant force can be determined. However, the scale must be defined. In this problem 40 N will be equal to 1 cm, so 100 N is equal to 2.5 cm which is indicated in the Figure 4.1.
The length of the resultnat force\(\vec{F}_R\) is 11.935 cm. By multiplying the length with the defined scale in Figure 4 the magnitude of the force \(\vec{F}_R\) can be calculated. $$F_R = 11.935 \left[\mathrm{cm}\right] \cdot \frac{100 \left[\mathrm{N}\right]}{2.5\left[\mathrm{cm}\right]} = 477.5 \left[\mathrm{N}\right]$$ Problem 2 For a given system of colinear forces shown in Figure (xx) calculate resultant force using analythical and graphical method. \(F_1 = 150 \left[\mathrm{N}\right]\), \(F_2 = 200 \left[\mathrm{N}\right]\), \(F_3 = 300 \left[\mathrm{N}\right]\).
Analythical solution For a given system of colinear forces the resultant force can be easily calculated.
\begin{eqnarray} F_R &=& \sum F_x + \sum F_y = ? \\ \sum F_x &=& 0\\ \sum F_y &=& 0 \\ \sum F_x &=& -F_1 + F_2 + F_3 \\ \sum F_x &=& -150 + 200 + 300 = 350 \left[\mathrm{N}\right]\\ F_R &=& 350 \left[\mathrm{N}\right] \end{eqnarray} Graphical solution To determine the length of resultant force the plan of forces must be developed first. The plan of forces is given in Figure 4.7.
\(M = \frac{100 \left[\mathrm{N}\right]}{0.5\left[\mathrm{cm}\right]}\)
Analythical solution