# A uniform steel beam of length

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 beamis held in a horizontal position by a cable attached between theother end of the beam and a point on the wall. The cable makes anangle 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 thewall.

(a) Find the magnitude of the tension in the supportingcable.

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

(a) Normally when you say “bolted” here it is implied that thebolt creates a moment between the wall and the beam, which makesthe problem indeterminate. I’ll assume you meant “pinned.”Assuming the tension creates a positive torque and the other forcescreate negative torques (about the pin), thenΣM = 0 = T*5.00m x sin25.0º – 4500N x 5.00m/2 – 12000Nx3.70mtension T = 26336 N approx.(b) Fx = Tcos25º = 23868 NFy = 4500N + 12000N – Tsin25º = 5370 N|F| = √(23868² + 5370²) N = 24464 N

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