oldhw 23 - alexander(jra2623 – oldhomework 23 – Turner...

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Unformatted text preview: alexander (jra2623) – oldhomework 23 – Turner – (92510) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A uniform bar of length L and weight W is attached to a wall with a hinge that exerts a horizontal force H x and a vertical force H y on the bar. The bar is held by a cord that makes a 90 ◦ angle with respect to bar and angle θ with respect to wall. L W θ What is the magnitude of the horizontal force H x on the pivot? 1. H x = 1 2 W cos 2 θ 2. H x = 1 2 W sin 2 θ 3. H x = 1 2 W sin θ 4. H x = 1 2 W tan θ 5. H x = 1 2 W cos θ 6. H x = 1 2 W sin θ cos θ correct Explanation: Analyzing the torques on the bar, with the hinge at the axis of rotation, summationdisplay τ = LT − parenleftbigg L 2 sin θ parenrightbigg W = 0 T = 1 2 W sin θ . Analyzing horizontal equilibrium on the bar, summationdisplay F x = H x − T cos θ = 0 H x − 1 2 W sin θ cos θ = 0 H x = 1 2 W sin θ cos θ . 002 (part 1 of 3) 10.0 points A 1347 N uniform boom of length ℓ is sup- ported by a cable, as shown. The boom is pivoted at the bottom, the cable is attached a distance 3 4 ℓ from the pivot, and a 4922 N weight hangs from the boom’s top. F T 4922 N 32 ◦ 58 ◦ Find the force F T applied by the supporting cable. Correct answer: 3953 . 55 N. Explanation: Let : W b = 1347 N , W m = 4922 N , θ b = 58 ◦ , and θ t = 32 ◦ . Applying the second (rotational) condition of equilibrium (with axis of rotation at the base of the beam and beam length ℓ ), alexander (jra2623) – oldhomework 23 – Turner – (92510) 2 τ net = F T parenleftbigg 3 4 ℓ parenrightbigg −W b parenleftbigg ℓ 2 parenrightbigg cos θ b −W m ℓ 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 (1347 N) + 4 (4922 N)]cos 58 ◦ 3 = 3953 . 55 N ....
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oldhw 23 - alexander(jra2623 – oldhomework 23 – Turner...

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