Answer
${F_1} = 71.74$ N, ${\ \ }$ ${F_2} = 87.86$ N
Work Step by Step
Let $g$ denote the gravitational acceleration. So, the gravitational force on the mass is ${\bf{W}} = - 10g{\bf{j}}$ N.
Let the forces that act on the mass by Rope 1 and Rope 2 be denoted by ${{\bf{F}}_1}$ and ${{\bf{F}}_2}$, respectively. By Newton's third law, the forces on Rope 1 and Rope 2 are the same magnitude but in opposite directions. So, we have
${{\bf{F}}_1} = - {F_1}\cos 30^\circ {\bf{i}} + {F_1}\sin 30^\circ {\bf{j}}$,
${{\bf{F}}_2} = {F_2}\cos 45^\circ {\bf{i}} + {F_2}\sin 45^\circ {\bf{j}}$,
where ${F_1} = ||{{\bf{F}}_1}||$ and ${F_2} = ||{{\bf{F}}_2}||$.
Since the system is in equilibrium, by Newton's first law, the total net force is zero. Thus,
$\sum {\bf{F}} = {\bf{W}} + {{\bf{F}}_1} + {{\bf{F}}_2} = {\bf{0}}$
So, the components of the total net forces are
$ - {F_1}\cos 30^\circ + {F_2}\cos 45^\circ = 0$
$ - 10g + {F_1}\sin 30^\circ + {F_2}\sin 45^\circ = 0$
From the first equation we obtain ${F_2} = {F_1}\frac{{\cos 30^\circ }}{{\cos 45^\circ }}$. Substituting it in the second equation gives
$ - 10g + {F_1}\sin 30^\circ + {F_1}\frac{{\cos 30^\circ }}{{\cos 45^\circ }}\sin 45^\circ = 0$
$ - 10g + \frac{1}{2}{F_1} + {F_1}\left( {\frac{{\frac{1}{2}\sqrt 3 }}{{\frac{1}{2}\sqrt 2 }}} \right)\left( {\frac{1}{2}\sqrt 2 } \right) = 0$
Solving this equation gives ${F_1} = \frac{{20}}{{1 + \sqrt 3 }}g \simeq 7.32g$ N and
${F_2} = {F_1}\frac{{\cos 30^\circ }}{{\cos 45^\circ }} = 7.32g\left( {\frac{{\frac{1}{2}\sqrt 3 }}{{\frac{1}{2}\sqrt 2 }}} \right) \simeq 8.97g$ N
Let $g=9.8$ $m/{s^2}$. So,
${F_1} = 7.32g = 71.74$ N, ${\ \ }$ ${F_2} = 8.97g = 87.86$ N
