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Unformatted text preview: F, : +500 Ib, F. =125.0Ib .. PROBLEM 2.93 Knowing that the teMion is 425 Ib in cable AD and 510 Ib in cable A C. determine the magnitude and direction ofthe resultant ofthe forces exerted atA by the two cables. , SOLUTION AD = (40 in.);  (45 in.)j +(60 in.)k AD = J(40inl +(45 ini +(60 inl = 85 in. AC = (100 in.); (45 in.)j + (60 in.)k AC = J(IOO ini +(45 ini + (60 in.)2 = 125 in. T =T l. =T AD =(425 Ib)[(40in.)I(45 in.)i + (60 in.1k] AB AB AB Ail AD 85 in. T" = (200 Ib)i  (225Ib)j + (300 Ib)k T =T l. =T AC =(510 Ib)[(IOO in.)I(45 in.)i + (60 in.)k] AC AC AC AC AC 125in. T,c = (408 Ib)I(183.6 Ib)j + (244.8Ib)k R = T" + T,c = (608)1 (408.6Ib)j +(544.8Ib)k Then: R=912.92Ib R=913lb ... and oos e = 608 Ib 0.66599 • , 912.921b cosO = 408.61b 0.44757 e, =116.6 0 ... J' 912.92lb cosO = 544.8Ib = 0.59677 , 912.921b...
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This note was uploaded on 10/05/2009 for the course CE statics taught by Professor Epstein during the Spring '09 term at UConn.
 Spring '09
 Epstein

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