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. N (b) Find the magnitude of the force exerted on the end of the beam by the bolt that attaches the beam to the wall. N my attempt: ETorque = 0 Fl(D) + Wb (.5 D) - Ty(D) = 0 12.0E3N(4.60 m) + 4.50E3N(2.75) - Ty(6.069m) = 0 Ty = 11134.457 N Find Tx : tan25 = 11134.457 / x Tx = 23877.913 N T = sqrt(23877.913^2 + 11134.457^2) T = 26346.36 N Efx = 0 23877.913 N - Fx = 0 Fx = 23877.913 N EFy = 0 Fy + Ty-Fl _fb = 0 11134.457 N - 12E3 N - 4.5 E3 N + Fy = 0 Fy = 5365.543 N F = sqrt(23877.913^2 + 5365.543^2) F = 24473.327 N When i tried the T and F in the online submission, there were wrong both times, and now im on my last try. Can someone please help me??