Homework 8, torque 18-19-solutions.pdf

# Homework 8, torque 18-19-solutions.pdf - litzner(bnl645 –...

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litzner (bnl645) – Homework 8, torque 18-19 – marder – (OlsonM302K1819 6) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001(part1of2)10.0points The arm of a crane at a construction site is 15.0 m long, and it makes an angle of 10 . 7 with the horizontal. Assume that the max- imum load the crane can handle is limited by the amount of torque the load produces around the base of the arm. What maximum torque can the crane with- stand if the maximum load the crane can handle is 456 N? Correct answer: 6721 . 07 N · m. Explanation: Let : d = 15 . 0 m and W max = 456 N . θ = 90 . 0 - 10 . 7 = 79 . 3 , so τ max = F d sin θ = W max d sin θ = (456 N)(15 m)(sin 79 . 3 ) = 6721 . 07 N · m . 002(part2of2)10.0points What is the maximum load for this crane at an angle of 31 . 2 with the horizontal? Correct answer: 523 . 837 N. Explanation: Let : θ = 90 . 0 - 31 . 2 = 58 . 8 We have the same maximum torque, so W = τ max d sin θ = 6721 . 07 N · m (15 m) sin 58 . 8 = 523 . 837 N . 003 10.0points If the torque required to loosen a nut that is holding a flat tire in place on a car has a magnitude of 35 N · m, what minimum force must be exerted by the mechanic at the end of a 27 cm-long wrench to loosen the nut? Correct answer: 129 . 63 N. Explanation: Let : τ = 35 N · m and d = 27 cm = 0 . 27 m . For a given torque, the minimum force must be applied perpendicular to the lever arm so that sin θ = sin 90 = 1, and τ = F d F = τ d = 35 N · m 0 . 27 m = 129 . 63 N . 004(part1of2)10.0points The figure shows a claw hammer as it pulls a nail out of a horizontal board. Single point of contact 5 . 05 cm 29 F 29 cm

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litzner (bnl645) – Homework 8, torque 18-19 – marder – (OlsonM302K1819 6) 2 If a force of magnitude 184 N is exerted horizontally as shown, find the force exerted by the hammer claws on the nail. (Assume that the force the hammer exerts on the nail is parallel to the nail). Correct answer: 1208 . 11 N. Explanation: Let : h = 29 cm , = 5 . 05 cm , α = 29 , and F = 184 N . F @ h α R @ cos α Let R be the reaction force of the nail on the hammer. Taking the sum of the torques about the point of contact of the hammer with the table, summationdisplay τ = R ℓ cos θ - F h = 0 R = F h cos α = (184 N) (29 cm) (5 . 05 cm) cos 29 parenleftbigg 1 kN 1000 N parenrightbigg = 1208 . 11 N . 005(part2of2)10.0points Find the force exerted by the surface on the point of contact with the hammer head. As- sume that the force the hammer exerts on the nail is parallel to the nail. Correct answer: 1130 . 42 N. Explanation: Let f be the horizontal force exerted by the surface at the point of contact on the hammer. From Newton’s second law, horizontally, summationdisplay F x = f + F - R sin α = 0 f = R sin α - F = (1208 . 11 N) parenleftbigg 1000 N 1 kN parenrightbigg sin 29 - 184 N = 401 . 702 N and vertically, summationdisplay F y = n - R cos α = 0 n = R cos α = (1208 . 11 N) parenleftbigg 1000 N 1 kN parenrightbigg cos 29 = 1056 . 63 N , so F = radicalbig f 2 + n 2 = radicalBig (401 . 702 N) 2 + (1056 . 63 N) 2 = 1130 . 42 N .
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