FBD with Applied Torques in Angled Member

[Fyrefly]

Homework Statement

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.

Homework Equations

Equilibrium equations.

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!

 

[anathema]

it is important to always consider all the variables and factors involved in a problem. In this case, the load and torque are acting at different points along the member, so it is important to account for their distances from the equilibrium point. The torque should be multiplied by its distance from the equilibrium point, just like the load is multiplied by its distance. This will ensure that both the load and torque are balanced in the equilibrium equation. Therefore, the correct equation would be Lcos(15)*vertical load + torque*distance from equilibrium point = 0. I hope this helps clarify your question.