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# HW_6 - PROBLEM 3.4 A foot valve for a pneumatic system is...

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Unformatted text preview: PROBLEM 3.4 A foot valve for a pneumatic system is hinged at 8. Knowing that =I'1‘ 6,516, a 2 28°, determine the moment of the 4-lb force about point B by _-1 " r1 resolving the force into horizontal and vertical components. 20“ \q.“_ _ . 311i. 7 “a B ' I 313%. \mié" - .635' SOLUTION r - -- -x Note that 19 z a — 20° : 28° — 20° : 8° F). ’“ 2V" and Fx z {410)0038° : 3,961] lb F}. = (41b)sin8° = 0.556691}: . Also x = (6.5111.)cos20° = 6.1080 in. y z (6.5 in_)sin20“ : 2.2231111. Noting that the direction of the moment of each force component about B is counterclockwise, ME : 2F}: +ny = (6.1080 in.)(0.55669 1b) + (2.2231 in.)(3.9611lb) = 12.2062 lb -in. or M3 = 12.21 115-111.) 4 PROBLEM 3.12 , It is known that a force with a moment of 7840 lein. about D is required _ to straighten the fence post CD‘ If a 2 8 in, b = 35 in., and d = 1 l2 in., determine the tension that must be developed in the cable of winch puller AB to create the required moment about point D. SOLUTION Slope ofline EC = : 1 112m. +8 in, 24 24 Then T431. 2 ET” 7 and T , = —T AB} 25 AB Have MD =T4Bx(y)+TABJ 24 7' . 7840lb-in. : *2" G +——T 112 in. 25 AB() 25 AB( ) TAB = 250 lb or Tm : 250 lb 4 PROBLEM 3.17 A plane contains the vectors A and B. Determine the unit vector normal to the plane when A and B are equal to, respectively, (a) 4i — 2j + 3k and —2i+6'—5k, b 7i+'74kand76i—3k+2k. J ( ) J SOLUTION ((1) Have 1 = A x B lA x B| where A :4i72j+3k B = —2i + 6j » 5k i j k Then Ax B = 4 —2 3 : (10718)i+(76+20)j+[24—4)k = 2(—4i+7j+10k) —2 6 —5 and M x 3| = 21/(—4)2 + (7}2 + (10)2 : 2J165 2(4i + 7j+10k) 1 . . _ 7t: _— 7L = 4 7 10k 4 2\/165 or J155( H H ) (1)) Have A : A x B \A x Bl where A = 7i + j — 4k B : 761 ~ 3j + 2k i j k Then A><B = 7 1—4 = (2—12)i+(24—14)j+(—21+6)k = 5(—2i +2j—3k} ~6 —3 2 and |A x B] : 5 {—2)2 +(2}1 + (—3)2 = sJﬁ - 17W m1“;(_2i+2'_3k)4 ' 7 5J5 v57 J ...
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