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.

Question

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??

drwls · Answer

You start out doing (a) correctly by setting the net torque about the bolt equal to zero. Ty is the vertical component of the cable tension force. However, why are you multiplying it by 6.069 m instead of 5.50 m, the length of the beam? Once you have Ty, get T by setting Ty/T = sin 25