Question

A 1260-N uniform beam is attached to a vertical wall at one end and is supported by a cable at the other end. A W = 1815-N crate hangs from the far end of the beam. Using the data shown in the following drawing, find the magnitude of the tension in the wire and the magnitudes of the horizontal and vertical components of the force that the wall exerts on the left end of the beam.

(See the drawing in the comments.)

Answer 1

TORQUE
i would calculate the torque about the beam touches the wall, which should be **zero**. call the length of the beam L, & tension in string T.

with clockwise torques treated as positive, Torque = 0 = ∑ r x F =

L/2<cos 30, sin 30> x < 0, 1260> + [the weight of the beam]
L <cos 30, sin 30> x < 0, 1815> + [the crate]
L <cos 30, sin 30> x T< -cos50, -sin50> [the wire]


giving:
L/2 cos30 1260 + L cos30 1815 + LT ( -cos30 sin 50 + sin30 cos50) = 0
L's cancel: T = (cos30 •1260/2 +cos30•1815) / ( cos30 sin 50 - sin30 cos50)

cos30 sin 50 - sin30 cos50 = sin (50 - 30) = sin 20 [double angle formula]

T = cos30 (630 +1815) / sin20

FORCES AT WALL.
horizontal: F = T cos50
vertical: F = 1260 + 1815 - T sin 50