In Fig. 13-35, one end of a uniform beam that weighs 928. N is attached to a wal
ID: 2211899 • Letter: I
Question
In Fig. 13-35, one end of a uniform beam that weighs 928. N is attached to a wall with a hinge. The other end is supported by a wire. Fig. 13-35 (a) Find the tension in the wire. N (b) What is the horizontal component of the force of the hinge on the beam? N (c) What is the vertical component of the force of the hinge on the beam? N
Explanation / Answer
We have a "Fixed Pin" connection at the wall ===> Moment at the wall = 0. Pin can support horozontal and vertical forces Sum of the vertical forces = 0 Sum of the horizontal forces = 0 Sum of the moments about any point = 0 1) Sum moments at the wall ===> We need to calculate the lever arm from the loads to the pin. Assume beam is 20 meters long Arm to CG of Beam cos 30 = Acg/10 Acg = 8.66 m Mocg = 8.66m * 1240 N = 10738.72 N-m CW Arm to crate cos 30 = Ac/20 Ac = 17.32 m Moc = 17.32 m * 1930 N = 33428.58 N-m CW Total CW Moment 33438.58 + 10738.72 = 44167.30 N-m CW Now we translate the coordinate system to the end of the beam and to the trig to get the lever arm for the cable. The cable is 50 degrees from the horizontal on the left ====> it is 90 - 50 = 40 degrees from the vertical. We figure the needed vertical force applied to the same lever arm as the crate needed to balance the CW moment at the wall Cable Lever arm = 17.32 m Mocb = 44167.30 44167.30 = y * 17.32 y = 2550.08 N Now the trig to find the horizontal component of that. Tan 40 = X/2550.08 X = 2139.77 N That is the horizontal thrust on the pin.