Hw 3 - F, : +500 Ib, F. =-125.0Ib .. PROBLEM 2.93 Knowing...

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, PROBLEM 2.71 \-)IX) K Determine (a) the x, y, and z components of the 750-N force, (b) the angles 8.", ~, and B.. that the force forms with the coordinate axes. "', , SOLUTION F h =Fsin 35 0 = (750 N)sin 35° F h =430.2 N ./ (aJ F , =Fcos35° = (430.2 N) cos 25 0 = (750 N)cos 35° :::: (430.2 N)sin 25° =+390 N, F =+614 N, F =+181.8 N F z y z cosO =~ = +390N (b) x F 750N F -+{;14 N cos e :::: --. .L == --=e:-:c- y F 750 N cosO =F z :::: +181.8N z F 750 N
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PROBLEM 2.85 A ttansmission tower is held by three guy wires anchored by bolts at B, C, and D. If the tension in wire AD is 525 lb, detennine the components ofthe force exerted by the wire on the bolt atB. SOLUTION BA = (20 ft)1 + (100 ftlj - (25 ft)k BA : ~(20 ft)' + (100 ft)' + (-25 ft)' : 105 ft F = Fl BA BA :F BA = 5251b [(20 ft)1 + (100 ftlj- (25 ft)k] 105 ft F:(100.0 Ib)i +(500 Ib)J-(125.0 Ib)k F. :+loo.Olb,
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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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Hw 3 - F, : +500 Ib, F. =-125.0Ib .. PROBLEM 2.93 Knowing...

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