This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: cerutti (cpc566) 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. 4 . 89 cm Single point of contact 28 . 2 F 29 . 2 cm If a force of magnitude 175 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: 1 . 18573 kN. Explanation: Let : h = 29 . 2 cm , = 4 . 89 cm , = 28 . 2 , and F = 175 N . F @ h R @ cos 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 = R cos  F h = 0 R = F h cos = (175 N) (29 . 2 cm) (4 . 89 cm) cos 28 . 2 1 kN 1000 N = 1 . 18573 kN . keywords: 002 (part 1 of 3) 10.0 points A 1600 N uniform boom is supported by a cable as shown. The boom is pivoted at the bottom, and a 2070 N object hangs from its end. The boom has a length of 25 m and is at an angle of 50 above the horizontal. A support cable is attached to the boom at a distance of 0 . 78 L from the foot of the boom and its tension is perpendicular to the boom. 90 . 7 8 L 2070 N 5 T Find the tension in the cable holding up the boom. Correct answer: 2365 . 13 N. Explanation: Let : W 1 = 1600 N , W 2 = 2070 N , = 50 , L = 25 m , and L 1 = 0 . 78 L = 19 . 5 m . cerutti (cpc566) HW10 TSOI (58160) 2 L T . 7 8 L W 1 W 2 In equilibrium summationdisplay vector = 0 . Taking torques about the pivot point, T (0 . 78 L ) = W 2 ( L cos ) + W 1 parenleftbigg L 2 cos parenrightbigg T = W 2 cos . 78 + W 1 cos 2 (0 . 78) = (2070 N) cos 50 . 78 + (1600 N) cos 50 2 (0 . 78) = 2365 . 13 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: 1811 . 79 N. Explanation: The horizontal component of the tension is T sin . From summationdisplay F x = 0, F h = T sin = (2365 . 13 N) sin50 = 1811 . 79 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: 2149 . 72 N. Explanation: The vertical component of the tension is T cos . From summationdisplay F y = 0 F v = W 1 + W 2 T cos = 1600 N + 2070 N (2365 . 13 N) cos50 = 2149 . 72 N . 005 (part 1 of 3) 10.0 points A crane of mass 550 kg supports a load of 90 kg. The cranes boom is 8 m long and the angle it makes with the horizontal is 69 . The distance between the front and rear wheels is 6 m....
View Full
Document
 Spring '08
 Turner
 Physics

Click to edit the document details