A uniform vertical beam of mass 40 kg is acted on by a horizontal force of 520 N at its top and is held, in the vertical position, by a cable as shown. a) Draw a free-body diagram for the beam, clearly labeling all of the forces acting on it. b)Calculate the tension in the cable? c)Determine the reaction forces acting on the beam by the ground? Sorry I could not post the picture but its a vertical beam that is 5 m long and the force is coming out at the horizontal at the top of the beam. The cable is connected 3 m up the beam with a theta of 28 degrees. Formulas: Fx=0 Fy=0 Torque=0 So what I have done is broken the components of tension into its x and y components. Horizontally it is Tcos theta and vertically it is Tsin tetha so: Fy=0 Fn-mg-Tsin theta= 0 Tsin theta=0 (because Fn and mg as equal and opposite) Fx=0 F- Tcos theta=0 F= Tcos theta T= F / cos 28 T= 588.93 is this correct for the part b? Not using the torque equation is kind of throwing me off because of how it is connected. Any suggestions on if I did it right or not?