Solution: \(\displaystyle\sum{T}={0}\)

condition of rotational equilibrium, we take the left end as a pivot point

\(\displaystyle-{700}{\left({0.500}\right)}-{\left({9.8}\cdot{30}\right)}{\left({1.00}\right)}+{T}_{{1}}{\sin{{\left({40}\right)}}}{\left({2.00}\right)}={0}\)

Solving for \(T_1\)

\(T_1=501\ N\)

Then we apply the translational equilibrium condition, In the horizontal

\(\displaystyle\sum{F}_{{x}}={0}\)

\(\displaystyle{T}_{{1}}{\cos{{\left({40}\right)}}}-{T}_{{3}}={0}\)

\(\displaystyle{T}_{{3}}={501}{\cos{{\left({40}\right)}}}={384}\ {N}\)

In the vertical axis:

\(\displaystyle\sum{F}_{{y}}={0}\)

\(\displaystyle{T}_{{2}}-{30}\cdot{9.8}-{700}+{T}_{{1}}{\sin{{\left({40}\right)}}}={0}\)

\(\displaystyle{T}_{{2}}-{30}\cdot{9.8}-{700}+{501}{\sin{{\left({40}\right)}}}={0}\)

solving for \(T_2:\)

\(\displaystyle{T}_{{2}}={672}\ {N}\)