I have a member that is at a 15 degree angle from the horizontal. I have a purely vertical load acting halfway along it, and an applied torque at one end. I am solving for the torque at the other end.
The Attempt at a Solution
My question is this: Because I am working in a non-angled reference frame, I have naturally used Lcos(15)*vertical load in my moment equilibrium equation. But how do I account for the applied torque? I don't think it needs to be multiplied by cos(15) as torques in an equilibrium equation are independent of their distance from the equilibrium point, but I would like to verify this. Thanks!