A $200\ \text{lb}$ weight and a $100\ \text{lb}$ weight are held in static equilibrium by cables $OA$ and $OB$ as shown below. The pulleys are frictionless and you may neglect the pulley weights. Determine the tension in cables $OA$ and $OB$.
This is how I did it. I know how it is wrong.
Could anyone show me the correct way step by step? I am a novice here.
There are four forces acting at $O$, the tensions in the three cables and the weight of the 200 lb object. Let's call $T_A$ and $T_B$ the tensions in the cables toward $A$ and $B$. The tension in the third cable is equal in magnitude to the weight of the 100 lb object. So, using $5^2+12^2=13^2$ and $3^2+4^2=5^2$, writing forces by $x$ and $y$ components look like $$T_A\frac5{13}-100\frac 45-200=0\\T_A\frac{12}{13}+100\frac35-T_B=0$$ You get $T_A$ from the first equation, then plug it into the second equation to get $T_B$.