A uniform steel beam of length 5.50 m has a weight of 4.50 103 N. One end of the beam is bolted to a vertical wall. The beam is held in a horizontal position by a cable attached between the other end of the beam and a point on the wall. The cable makes an angle of 25.0° above the horizontal. A load whose weight is 12.0 103 N is hung from the beam at a point that is 4.60 m from the wall.
(a) Find the magnitude of the tension in the supporting cable.
(b) Find the magnitude of the force exerted on the end of the beam by the bolt that attaches the beam to the wall.
torque = Force * distance
sum the vertical forces and set to zero
sum the horizontal forces and set to zero.
sum the moments about any point and set to zero.
ok i seriously am confused how to solve this...cause we just learned about torques..but
EFx = 0
Tcos25 = 0..but this doesnt make sense cause this means that T= 0???
Ffy = 0
Tsin25 - 4.5E3N = 0
then once you get T somehow... and then magically F appears..hmm im soo confused HELP PLEASE!!!!