Answer
Magnitude of the force on cable 1 and cable 2 (respectively):
${F_1} \simeq 453.154$ N
${F_2} \simeq 211.309$ N
Work Step by Step
Let the gravitational acceleration be $g=10$ $m/s^2$. So, the gravitational force on the mass is ${\bf{W}} = - 500{\bf{j}}$ N.
Let the forces that act on the mass be denoted by ${{\bf{F}}_1}$ and ${{\bf{F}}_2}$, respectively. By Newton's third law, the forces on cables 1 and 2 are the same magnitude but in opposite directions. So, we have
${{\bf{F}}_1} = - {F_1}\cos 65^\circ {\bf{i}} + {F_1}\sin 65^\circ {\bf{j}}$,
${{\bf{F}}_2} = {F_2}\cos 25^\circ {\bf{i}} + {F_2}\sin 25^\circ {\bf{j}}$,
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,
$\sum {\bf{F}} = {\bf{0}} = {\bf{W}} + {{\bf{F}}_1} + {{\bf{F}}_2} = {\bf{0}}$
So, the components of the total net forces are
$ - {F_1}\cos 65^\circ + {F_2}\cos 25^\circ = 0$
$ - 500 + {F_1}\sin 65^\circ + {F_2}\sin 25^\circ = 0$
So, ${F_2} = {F_1}\frac{{\cos 65^\circ }}{{\cos 25^\circ }}$. Substituting this in the last equation above gives
$ - 500 + {F_1}\sin 65^\circ + {F_1}\frac{{\cos 65^\circ }}{{\cos 25^\circ }}\sin 25^\circ = 0$
Solving this equation gives ${F_1} \simeq 453.154$ N and
${F_2} = {F_1}\frac{{\cos 65^\circ }}{{\cos 25^\circ }} = 376.244\frac{{\cos 65^\circ }}{{\cos 25^\circ }} \simeq 211.309$ N.
