# [Physics] How does the net torque have to be 0 at all points for static equilibrium?

staticsstructural-beamtorque

I know that for a mass to be in static equilibrium two things have to be satisfied: $\sum F=0$ and $\sum T=0$, where $T$ represents torque. However, I am not sure how the latter can apply in a situation such as this

Here, a beam is is static equilibrium with only two forces acting on it (let us only consider the y-axis for simplicity's sake), $F_g$ and $F_t$. Clearly the rule of $\sum F=0$ holds and this can be proven with a simple $F_{net}$ statement. However, if the pivot is placed at the very end on the right hand side $\sum T = F_t \sin(55)\cdot0 – F_g\cdot x$, where $x$ is some arbitrary non-zero distance. How can this be if we need $\sum T = 0$?

There is also a force from the wall acting on the left end of the beam. This force provides a positive $F_{y_{wall}}$ that helps hold the beam up and a positive $F_{x_{wall}}$ that prevents the beam from swinging to the left.
The torque is then: $T = -F_g x - 2x F_{y_{wall}}$. Setting this equation to zero gives us $F_g = 2 F_{y_{_wall}}$. Note that none of the forces depend on x.