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Unformatted text preview: hanna (brh687) hk 12 Opyrchal (11105) 1 This print-out should have 14 questions. Multiple-choice 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