In this motionless situation, find the mass of the hanging block on the right side of the system. The mass of the left block is 3.5 kg, the masses of both the carts are 1.0 kg, and the left angle is 12 degrees and the right angle is 36 degrees.
I would guess to start by making Ft equal to the Fw of the left hanging block, but I don't know what to do then. (https://i.stack.imgur.com/DhRRW.jpg)
\begin{equation} \begin{cases} T_1=3,5g\\T_2=T_1+1g\sin12^°\\T_2=T_3+1g\sin36^°\\T_3=mg \end{cases} \end{equation}
In the first equation the weight of the mass on the left must be equal to the string tension (it is motionless);
on the left cart we have a component of the weight force ($1g\sin12^°$) downward along the inclined plane and also the tension $T_1$ (downward along the inclined plane), this two forces (summed) must be equal to $T_2$ (the tension of the string upward), so I wrote the second equation.
Now we go to the right side:
on $m$ we have the weight force downward and the string tension ($T_3$) upward, so we have the last equation;
on the right cart we have a component of the weight force ($1g\sin36^°$) downward along the inclined plane and also the tension $T_3$ (downward along the inclined plane), this two forces (summed) must be equal to $T_2$ (the tension of the string upward), so I wrote the third equation.