The uniform $120\mathrm{N}$ board shown in figure is supported by two ropes. A $400\mathrm{N}$ weight is suspended one-fourth of the way from the left end. Choose the correct options

 

Taking moments about the left edge and resolving $T_1$ into $x$ and $y$ components,

$\sum _{ }^{ }\zeta =0$ yields $L{T}_1\cos 30°-(0.25L)\left(400\right)-(0.5L)\left(120\right)=0$

Dividing throughout by $L$ and solving, we get

$T_1=185N$

Also,

Force balance in horizontal direction gives,  $T_2\sin \theta =T_1\sin {30}^0$

Force balance in vertical direction gives, $T_2\cos \theta +T_1\cos {30}^0=400+120$

Solving these equations, we get

$T_2\sin \theta =92.5N$ and $T_2\cos \theta =360N$

Dividing the equation yields

$\tan \theta =0.257$, or $\theta =14.4°$

and $T_2=371N$

One can always check moment problem results by taking moments about another point, such as the right end of the bar for this problem