30.1 N
Explanation:
Given:
[tex]W_1 = 16\:\text{N}[/tex]
[tex]W_2 = 8\:\text{N}[/tex]
Let's write the components of the net forces at the intersections. Note that the system is equilibrium so all the net forces are zero.
Forces involving W1:
[tex]x:\:\:\:-T_1 + T_3\cos \alpha = 0\:\: \\ \text{or}\:\:T_2 = T_3\cos \alpha\:\:\:\:\:(1)[/tex]
[tex]y:\:\:\:T_3\sin \alpha - W_1 = 0\:\:\: \\ \text{or}\:\:\:T_3\sin \alpha = W_1\:\:\:\:\:\:(2)[/tex]
Forces involving W2:
[tex]x:\:\:\:T_1\sin 53 - T_3\sin \alpha = W_2\:\:\:\:\:\:\:(3)[/tex]
[tex]y:\:\:\:T_4 - T_1\cos 53 - T_3\cos \alpha = 0\:\:\:\;(4)[/tex]
Substitute (2) into (3) and we get
[tex]T_1\sin 53 - W_1 = W_2[/tex]
Solving for [tex]T_1[/tex],
[tex]T_1 = \dfrac{W_1 + W_2}{\sin 53} = 30.1\:\text{N}[/tex]
