We can find the resultant forces as the sum of the forces as vectors.
We can find the component in the x-axis as:
[tex]\begin{gathered} R_x=|F_1|\cdot\cos (30\degree)+|F_2|\cdot\cos (-45\degree) \\ R_x=50\cdot\frac{\sqrt[]{3}}{2}+10\cdot\frac{\sqrt[]{2}}{2} \\ R_x=25\sqrt[]{3}+5\sqrt[]{2} \end{gathered}[/tex]We now can find the component in the y-axis as:
[tex]\begin{gathered} R_y=|F_1|\cdot\sin (30\degree)+|F_2|\cdot\sin (-45\degree) \\ R_y=50\cdot\frac{1}{2}+10\cdot(-\frac{\sqrt[]{2}}{2}) \\ R_y=25-5\sqrt[]{2} \end{gathered}[/tex]We can express the resultant force as:
[tex]\bar{F_1}+\bar{F_2}=(25\sqrt[]{3}+5\sqrt[]{2})\hat{i}+(25-5\sqrt[]{2})\hat{j}[/tex]We can find the magnitude and direction by converting the resultant force in component form into polar form:
[tex]\begin{gathered} |R|=\sqrt[]{R^2_x+R^2_y} \\ |R|=\sqrt[]{(25\sqrt[]{3}+5\sqrt[]{2})^2+(25-5\sqrt[]{2})^2} \\ |R|=\sqrt[]{(25\sqrt[]{3})^2+2\cdot25\sqrt[]{3}\cdot5\sqrt[]{2}+(5\sqrt[]{2})^2+25^2-2\cdot25\cdot5\sqrt[]{2}+(5\sqrt[]{2})^2} \\ |R|=\sqrt[]{625\cdot3+250\sqrt[]{6}+25\cdot2+625-250\sqrt[]{2}+25\cdot2} \\ |R|=\sqrt[]{1875+250\sqrt[]{6}+50+625-250\sqrt[]{2}+50} \\ |R|=\sqrt[]{2600+250\sqrt[]{2}(\sqrt[]{3}-1)} \\ |R|\approx53.47 \end{gathered}[/tex]And the direction can be calculated as:
[tex]\begin{gathered} \tan \theta=\frac{R_y}{R_x} \\ \theta=\arctan (\frac{25-5\sqrt[]{2}}{25\sqrt[]{3}+5\sqrt[]{2}}) \\ \theta=\arctan (\frac{5-\sqrt[]{2}}{5\sqrt[]{3}+\sqrt[]{2}}) \\ \theta\approx19.59\degree \end{gathered}[/tex]Answer:
We can express the resultan force as (25√3+5√2)i + (25-5√2)j.
The resultant magnitude is approximately 53.47 N and the direction is approximately 19.59°.
