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Unformatted text preview: yang (ey942) – HW10 – TSOI – (58160) 1 This printout should have 20 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The figure shows a claw hammer as it pulls a nail out of a horizontal board. 5 . 75 cm Single point of contact 34 . 2 ◦ F 28 . 1 cm If a force of magnitude 90 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: 0 . 531781 kN. Explanation: Let : h = 28 . 1 cm , ℓ = 5 . 75 cm , α = 34 . 2 ◦ , and F = 90 N . F @ h α R @ ℓ 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 of the table, summationdisplay τ = Rℓ cos θ F h = 0 R = F h ℓ cos α = (90 N) (28 . 1 cm) (5 . 75 cm) cos 34 . 2 ◦ 1 kN 1000 N = . 531781 kN . keywords: 002 (part 1 of 3) 10.0 points A 1810 N uniform boom is supported by a cable as shown. The boom is pivoted at the bottom, and a 1130 N object hangs from its end. The boom has a length of 14 m and is at an angle of 53 ◦ above the horizontal. A support cable is attached to the boom at a distance of 0 . 72 L from the foot of the boom and its tension is perpendicular to the boom. 90 ◦ . 7 2 L 1130 N 5 3 ◦ T Find the tension in the cable holding up the boom. Correct answer: 1700 . 96 N. Explanation: Let : W 1 = 1810 N , yang (ey942) – HW10 – TSOI – (58160) 2 W 2 = 1130 N , α = 53 ◦ , L = 14 m , and L 1 = 0 . 72 L = 10 . 08 m . L T . 7 2 L W 1 W 2 α β In equilibrium summationdisplay vector τ = 0 . Taking torques about the pivot point, T (0 . 72 L ) = W 2 ( L cos α ) + W 1 parenleftbigg L 2 cos α parenrightbigg T = W 2 cos α . 72 + W 1 cos α 2 (0 . 72) = (1130 N) cos 53 ◦ . 72 + (1810 N) cos 53 ◦ 2 (0 . 72) = 1700 . 96 N . 003 (part 2 of 3) 10.0 points Find the horizontal component of the reaction force on the boom by the floor. Correct answer: 1358 . 45 N. Explanation: The horizontal component of the tension is T sin α . From summationdisplay F x = 0, F h = T sin α = (1700 . 96 N) sin53 ◦ = 1358 . 45 N . 004 (part 3 of 3) 10.0 points Find the vertical components of the reaction force on the boom by the floor. Correct answer: 1916 . 33 N. Explanation: The vertical component of the tension is T cos α . From summationdisplay F y = 0 F v = W 1 + W 2 T cos α = 1810 N + 1130 N (1700 . 96 N) cos53 ◦ = 1916 . 33 N . 005 (part 1 of 3) 10.0 points A crane of mass 310 kg supports a load of 50 kg. The crane’s boom is 6 m long and the angle it makes with the horizontal is 49 ◦ . The distance between the front and rear wheels is 5 m....
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This note was uploaded on 11/21/2010 for the course PHY 58160 taught by Professor Tsio during the Fall '10 term at University of Texas at Austin.
 Fall '10
 TSIO

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