Since the block is at rest, i.e., equilibrium, the net force on the system must be zero: That means:
[tex]\begin{gathered} F_{1x}=F_{2x} \\ F_1\sin 30^o=F_2\sin 60^o \\ F_1=\frac{F_2\sin 60^o}{\sin 30^o}^{}\rightarrow eq(1) \end{gathered}[/tex]
Also we have:
[tex]\begin{gathered} F_{1y}+F_{2y}=W \\ F_1\cos 30^o+F_2\cos 60^o=mg \\ \frac{F_2\sin60^o}{\sin30^o}\cos 30^o+F_2\cos 60^o=mg \\ F_2\sin 60^o\cot 30^o+F_2\cos 60^o=mg \\ F_2(\sin 60^o\cot 30^o+\cos 60^o)=mg \\ F_2=\frac{mg}{\sin 60^o\cot 30^o+\cos 60^o} \\ \end{gathered}[/tex]
Lets say that g = 9.8 m/s², then if we plug the data we get:
[tex]\begin{gathered} F_2=\frac{16\times9.8}{(0.87)\times(1.73)+(0.50)} \\ \\ F_2=\frac{156.8}{2.01} \\ \\ F_2=78N \end{gathered}[/tex]
Answer: the tension in string 2 is 78 N.
