HW 1 Solution - SOLUTION Using the force triangie and the...

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Unformatted text preview: SOLUTION Using the force triangie and the Law of Cosines 7 . r, 5 .I, . R2=00bf+Q0bY—ZQMMQOEfimHW =[1004—400——400(—0342[pb2 ~ = 636.81?)Z R=fiflm Using now the Law of Sines 1011:. _ 25.231b sinfl sin¥10° mfi=(£:;Jmnm° = 0.3724 So: 7 fl = 21.87° Angle of inclination of R, 4} is then such that: ¢+fi=3m ¢ = 8J3" Hence: R : 25.21b b; 8.13° 4 SOLUTION Using the force triangle, the Law of Cosines and the Law of Sines We have: ' ’ a = 180° — (50° + 25°) 2105? Then: " R2 = (4.5 1<N)2 + (6 kN)2 — 2(4.5 kN)(6 kN)cosl()5° = 70.226 m2 or R = 8.3801 kN NOW: 8.3801kN : 6kN sin105° sin ,6 sinfi =[ 6 kN sin105° 8.3801 2 0.6916 5 = 43.756° R = 8.38 kN T18.76°{ 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 1 Solution - SOLUTION Using the force triangie and the...

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