520-N uniform rectangular sign 4.00 m wide and 3.00 m high is suspended from a h

Question

520-N uniform rectangular sign 4.00 m wide and 3.00 m high is suspended from a horizontal, 6.00-m-long, uniform, 150-N rod as indicated in the figure below. The left end of the rod is supported by a hinge and the right end is supported by a thin cable making a 30.0

Explanation / Answer

(a) balancing moment of weight of sign, weight of rod and tension, T, in the cable, (440) * (4) + (130) * (3) = T * (6cos30°) => T = 310.32N (b) Balancing all forces in the vertical direction, vertical component of force of the rod by the hinge, F(v) = 440 + 130 - Tcos30° = 570 - 268.74 = 301.25 N Balancing forces in the horizontal direction, horizontal component of force of the rod by the hinge, F(h) = Tcos60° = 310.32 * (1/2) = 155.16 N

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