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Unformatted text preview: hanna (brh687) hk 12 Opyrchal (11105) 1 This printout should have 14 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 3) 10.0 points A 1448.0 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 3506.0 N weight hangs from the booms top. F T 3506 N 24 66 Note: Figure is not drawn to scale a) Find the force F T applied by the sup porting cable. Correct answer: 2293 . 99 N. Explanation: Basic Concept: The second (rotational) condition of equi librium (axis of rotation at the base of the beam and beam length ): F T parenleftbigg 3 4 parenrightbigg F g,b parenleftbigg 2 parenrightbigg (cos b ) F g,m (cos b ) = 0 Given: F g,b = 1448 . 0 N F g,m = 3506 . 0 N b = 66 t = 24 Solution: 3 F T 2 F g,b (cos b ) 4 F g,m (cos b ) = 0 F T = (2 F g,b + 4 F g,m )(cos b ) 3 = [2(1448 N) + 4(3506 N)](cos 66 ) 3 = 2293 . 99 N . 002 (part 2 of 3) 10.0 points b) Find the horizontal component of the reac tion force on the bottom of the boom. Correct answer: 2095 . 67 N. Explanation: Basic Concept: The first (translational) condition of equi librium: Horizontally, F x = R x,base F T (cos t ) = 0 Solution: R x,base = F T (cos t ) = (2293 . 99 N)(cos24 ) = 2095 . 67 N . 003 (part 3 of 3) 10.0 points c) Find the vertical component of the reaction force on the bottom of the boom. Correct answer: 4020 . 95 N. Explanation: Basic Concept: The first (translational) condition of equi librium: Vertically, F y = R y,base + F T (sin t ) F g,m F g,b = 0 Solution: R y,base = F g,m + F g,b F T (sin t ) = 1448 N + 3506 N (2293 . 99 N)(sin 24 ) = 4020 . 95 N . 004 10.0 points The system shown in the figure is in equilib rium. A 16 kg mass is on the table. A string hanna (brh687) hk 12 Opyrchal (11105) 2 attached to the knot and the ceiling makes an angle of 63 with the horizontal. The coeffi cient of the static friction between the 16 kg mass and the surface on which it rests is 0 . 36. The acceleration of gravity is 9 . 8 m / s 2 . 16 kg m 6 3 What is the largest mass m can have and still preserve the equilibrium? Correct answer: 11 . 3046 kg....
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This note was uploaded on 04/07/2012 for the course PHYSICS 121 taught by Professor James during the Spring '12 term at University of Texas at Austin.
 Spring '12
 James
 Physics

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