In order to find the total force, we need to sum the forces F1 and F2.
To do so, let's calculate the horizontal and vertical components of each force.
Before calculating these components, let's use the following angles, considering as reference the positive x-axis:
So we have:
[tex]\begin{gathered} F_{1x}=F_1\cdot\cos (\theta_1) \\ F_{1x}=210\cdot\cos (72\degree) \\ F_{1x}=210\cdot0.309=64.89\text{ N} \\ \\ F_{1y}=F_1\cdot\sin (\theta_1) \\ F_{1y}=210\cdot\sin (72\degree) \\ F_{1y}=210\cdot0.951=199.71\text{ N} \\ \\ F_{2x}=F_2\cdot\cos (\theta_2) \\ F_{2x}=210\cdot\cos (108\degree) \\ F_{2x}=210\cdot(-0.309)=-64.89\text{ N} \\ \\ F_{2y}=F_2\cdot\sin (\theta_2) \\ F_{2y}=210\cdot\sin (108\degree) \\ F_{2y}=210\cdot0.951=199.71\text{ N} \end{gathered}[/tex]
Now, adding the components of each direction, we have:
[tex]\begin{gathered} F_x=F_{1x}+F_{2x}=64.89-64.89=0 \\ F_y=F_{1y}+F_{2y}=199.71+199.71=399.42\text{ N} \end{gathered}[/tex]
The resulting force is given by:
[tex]\begin{gathered} F=\sqrt[]{F^2_x+F^2_y} \\ F=\sqrt[]{0^2+399.42^2} \\ F=399.42\text{ N} \end{gathered}[/tex]
Since the resulting force has only vertical component, its direction is upwards (positive y direction), and the type of movement caused is a vertical movement.
