HW 11 solution_pdf

# HW 11 solution_pdf - ghei(mg42382 HW 11 opyrchal(11130 This...

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ghei (mg42382) – HW 11 – opyrchal – (11130) 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points A 1096 N uniform boom of length is sup- ported by a cable, as shown. The boom is pivoted at the bottom, the cable is attached a distance 3 4 from the pivot, and a 3138 N weight hangs from the boom’s top. F T 3138 N 24 66 Find the force F T applied by the supporting cable. Correct answer: 1998 . 98 N. Explanation: Let : W b = 1096 N , W m = 3138 N , θ b = 66 , and θ t = 24 . Applying the second (rotational) condition of equilibrium (with axis of rotation at the base of the beam and beam length ), τ net = F T parenleftbigg 3 4 parenrightbigg - W b parenleftbigg 2 parenrightbigg cos θ b - W m cos θ b = 0 . 3 F T - 2 W b cos θ b - 4 W m cos θ b = 0 F T = (2 W b + 4 W m ) cos θ b 3 = [2 (1096 N) + 4 (3138 N)] cos 66 3 = 1998 . 98 N . 002 (part 2 of 3) 10.0 points Find the horizontal component of the reaction force on the bottom of the boom. Correct answer: 1826 . 15 N. Explanation: Applying the first (translational) condition of equilibrium horizontally, F x = R x,base - F T cos θ t = 0 R x,base = F T cos θ t = (1998 . 98 N) cos 24 = 1826 . 15 N . 003 (part 3 of 3) 10.0 points Find the vertical component of the reaction force on the bottom of the boom. Correct answer: 3420 . 94 N. Explanation: Applying the first (translational) condition of equilibrium vertically, F y = R y,base + F T sin θ t - W m - W b = 0 R y,base = W m + W b - F T sin θ t = 1096 N + 3138 N - (1998 . 98 N) sin 24 = 3420 . 94 N . 004 10.0 points A light string has its ends tied to two walls separated by a distance equal to five-eighths the length of the string as shown in the figure.

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