A uniform steel beam of length 5.00 m has a weight of 4.50 ✕ 10^3 N....

A uniform steel beam of length 5.00 m has a weight of 4.50 ✕ 10^3 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 ✕ 10^3 N is hung from the beam at a point that is 3.70 m from the wall.

(a) Find the magnitude of the tension in the supporting cable.

(b) Find the magnitude of the force exerted on the end of the beam by the bolt that attaches the beam to the wall.

Expert Solution

(a) Normally when you say "bolted" here it is implied that the bolt creates a moment between the wall and the beam, which makes the problem indeterminate. I'll assume you meant "pinned."
Assuming the tension creates a positive torque and the other forces create negative torques (about the pin), then
ΣM = 0 = T*5.00m x sin25.0º - 4500N x 5.00m/2 - 12000Nx3.70m
tension T = 26336 N approx.

(b) Fx = Tcos25º = 23868 N
Fy = 4500N + 12000N - Tsin25º = 5370 N
|F| = √(23868² + 5370²) N = 24464 N

genius_generous answered 2 years ago

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