HW 3 Solution - SOLUTION Cable BC Force; Fr = —(1451b)fl...

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Unformatted text preview: SOLUTION Cable BC Force; Fr = —(1451b)fl = —105 1b :16 72x: ‘2' “9 Fy = (1451102 = 1001b 116 | , I 0( IOU-lb Force: I Fx = 410011;)? = ~601b I 1 4 I Fy = —(1001b)§ = —30 1b I — _ ' 121’ 40“: 156-11) Force: E Fx = (1561b)% = 1441b F = —(1561b)i = —60 lb y 13 and x =2Fx =—211b, Ry :EFy =—401b R : (—211b)2 + (410 11:.)2 = 45.177 lb Further: 4o tana = — 21 a = tan"4—0 = 623° 21 Thus: R 2 45.2 lb 762.3°4 SOLUTION Free-Body Diagram 60011: a From the geometry, we calculate the distances: AC = (16 in.)2 + (12 in.)2 = 20 in. BC = (20 inf + (21 in.)2 = 29 in. Then, from the Free Body Diagram of point C: 1;. 2px = 0: —-;—ETAC +:—;-TBC = 0 BC = % x gT/{C +1213 = 0: £23m + grim — 6001b = 0 %TAC + %E% x gig”) — 600 lb = o TAG = 440.56 lb TAC = 441 lb 4 TBC = 487 lb < SOLUTION The components of the forces were determined in 2.23. m» =(71.9lb)i—(43.861b)j 43.86 tana = — 71.9 a = 31.380 R = (71.9113)2 + (—43.861b)2 = 84.23 1b R 2 84.2 lb ‘q 31.4°‘ ...
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HW 3 Solution - SOLUTION Cable BC Force; Fr = —(1451b)fl...

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