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??

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

Sure, I can help you with that. Let's go through the problem step by step.

(a) To find the magnitude of the tension in the supporting cable, let's consider the equilibrium of forces in the vertical direction. Since the beam is held in a horizontal position, the vertical components of the tension in the cable and the weight of the beam should balance the vertical component of the load.

Starting the equation:
∑Fy = Ty + Wb*sin(25°) - Load = 0

We know that Wb = 4.50 * 10^3 N and Load = 12.0 * 10^3 N. Now, we can solve for Ty:
Ty = Load - Wb*sin(25°)
Ty = (12.0 * 10^3 N) - (4.50 * 10^3 N * sin(25°))
Ty = 8.71 * 10^3 N

Therefore, the magnitude of the tension in the supporting cable is 8.71 * 10^3 N.

(b) To find the magnitude of the force exerted on the end of the beam by the bolt, let's consider the equilibrium of forces in the horizontal direction. Since the beam is in a horizontal position, the horizontal components of the tension in the cable and the weight of the beam should balance out.

Starting the equation:
∑Fx = Tx - Wb*cos(25°) = 0

We already have the expression for Wb and the value of Tx can be found using trigonometry:
Tx = Ty / sin(25°)
Tx = (8.71 * 10^3 N) / sin(25°)
Tx = 2.07 * 10^4 N

Therefore, the magnitude of the force exerted on the end of the beam by the bolt is 2.07 * 10^4 N.

Please note that the values you calculated for Ty and Tx are incorrect because of some errors made along the way. I hope this explanation helps you understand how to solve the problem correctly.