It is understood that F is the perpendicular component of the force. Rotational equilibrium is established when the clockwise torques balance the sum of the counter-clockwise torques1. We have ignored, until now, the weight of the beam itself. Fortunately, this is easily included. We imagine the entire weight of the beam to be concentrated at the precise center of the beam.
All calculations then proceed normally.
A 25 pound beam is supported by a string as shown. In each case the bar weighs 320 nt and is 6 meters long. Find the tension in the supporting strings. Now here is one with an angled beam of 6 m length and 320 nt weight. Remember, the weight still pulls straight down, so its perpendicular component will have to be calculated.