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5nsi30 n sin50 800nsn50 i q 35292 n 353 or 73 dc217

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Unformatted text preview: cable ACB, (b) the magnitude of load Q. SOLUTION Free­Body Diagram: Pulley C Σ Fx = 0: (a) − T ACB (cos 30 ° − cos 50 ° ) ( 800 N c) s 50 o ° = 0 T ACB = 2303.5 N Hence T ACB = 2.30 kN Σ F y = 0: (b) + T ACB (sin 30 ° + sin 50 ° ) ( 800 N s)n 50 i (2303.5 N si)(30 n ° + sin 50 ° ) ( 800 N s)n 50 + i Q = 3529.2 N 3.53 or 73 dc217. 4shared. com/ doc/ AOPLVdUr/ prev iew . html ° − Q = 0 ° − Q = 0 Q= kN 3/ 11 2/ 6/ 14 / usr/ hosting_files4/ main/ prev iew s/ doc431728704/ prev iew P R O B L E M 2 . 7 2 A 2000­N load Q is applied to the pulley C, which can roll on the cable ACB. The pulley is held in the position shown by a sec...
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This test prep was uploaded on 02/21/2014 for the course PHYS 1120 taught by Professor Dick during the Spring '10 term at WPI.

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