Answer
The magnitude of forces ${{\bf{F}}_1}$ and ${{\bf{F}}_2}$:
${F_1} \simeq 27.32$
${F_2} \simeq 24.5$
Work Step by Step
From Figure 29, we have the vectors:
${{\bf{F}}_1} = {F_1}{\bf{j}}$,
${{\bf{F}}_2} = {F_2}\cos 45^\circ {\bf{i}} - {F_2}\sin 45^\circ {\bf{j}}$,
where ${F_1} = ||{{\bf{F}}_1}||$ and ${F_2} = ||{{\bf{F}}_2}||$.
Let ${{\bf{F}}_3}$ denote the vector on the right. So,
${{\bf{F}}_3} = - 20\cos 30^\circ {\bf{i}} - 20\sin 30^\circ {\bf{j}}$
Since there is no net force on the object, ${{\bf{F}}_1} + {{\bf{F}}_2} + {{\bf{F}}_3} = {\bf{0}}$.
So, the components of the total net forces are
(1) ${\ \ }$ ${F_2}\cos 45^\circ - 20\cos 30^\circ = 0$
(2) ${\ \ }$ ${F_1} - {F_2}\sin 45^\circ - 20\sin 30^\circ = 0$
From equation (1) we get
${F_2} = 20\frac{{\cos 30^\circ }}{{\cos 45^\circ }} = 10\sqrt 6 \simeq 24.5$
Substituting ${F_2} = 10\sqrt 6 $ in equation (2) gives
${F_1} - 10\sqrt 6 \sin 45^\circ - 20\sin 30^\circ = 0$
So, ${F_1} = 10 + 10\sqrt 3 \simeq 27.32$.
