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2 c or p sin35 350lb 0 p 14597 lb 1460 p lb 72 dc217

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Unformatted text preview: SOLUTION Free­Body Diagram: Pulley A Σ = 0: 2 sin P Fx β − P cos 25 ° = 0 Hence: (a) (b) 1 sin β = cos 25 ° or 2 Σ = 0: 2 coP s Fy β = 24.2 ° β + P sin 35 ° − 350 lb = 0 Hence: 2 Pos 24.2 c or ° + P sin 35 ° − 350 lb = 0 P = 145.97 lb 146.0 P= lb 72 dc217. 4shared. com/ doc/ AOPLVdUr/ prev iew . html 2/ 11 2/ 6/ 14 / usr/ hosting_files4/ main/ prev iew s/ doc431728704/ prev iew P R O B L E M 2 . 7 1 A 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 second cable CAD, which passes = 800 N, over the pulley A and supports a load P. Knowing that P determine (a) the tension in...
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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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