hw2-ed11

hw2-ed11 - Problem 2—?9 Given 1‘1 r and rs determine...

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Unformatted text preview: Problem 2—?9 Given 1‘1, r:, and rs, determine ihe magmmde and direction Df 1' = 2r] — I: + 313. Given: M |’ 4 ‘| f3 ‘1 1~l:|—4}m P}: [I‘m 132—2131 '~ 3 -' k—jz' x 5 , Solution: 1‘ = 211 — r: + 3|} “ ll 1‘ = —14]m |.-| = 31.5 m a 26 J {3‘11 ‘5 {at} (59.5‘ ﬂ = acosf' i| ﬂ = 116.4 deg k lrl ,. | J a J"; 3r £34.14 a Problem 2—32 Express the position vector 1‘ in Cartesian voctor form: then detonﬂioe its mang and coordinate direction angle-5. Given: a = 4 to E:- = Sm r: = 3 m d = 4m _ - Solution: , —c "I —3 r 1‘ = —a'—b r: —12 m (I 4 Problem 2—92 Determine the magnitude and coordinate direction angles of the resultant force acting at point A. Given: F; =15UN u ,1“ "a F2=2ﬂﬂN ‘ F: azljm " x {1 524m , x c = 3111 d = 2m 2 = 3m 6' = 613 deg Solution: Deﬁne the position vectors and then the forces ecosliﬁ'ii r 33 . "AB r LAB = a + esinigi = Ir l = h L —E:- .1 AB R —m1.4 ,: C x r 119.4 ‘w "AC LAC = a — d F}; = F2 Ir l F21. = —19_9 N x —b . AC \$159.2 , Add the forces and find the magnitude of the resultant / 1514 I FR = Fh. + F21. FR = 33.9 N IFRI = 315 N k—26ﬂﬁ ,4 Find the direction cosine angles for“- rr F “a few [5:11 _. I? i = aces? R ] e = ":45 ' I t IFRI “J L J Liaise I . xi"; deg 3—..___—r" Problem Z—lﬂﬂ Determine the position (I, y, [I] for ﬁxing cable BA so that the resnitant of the forces exerted on the pole is directed along its axis, ﬂour: 3 toward 0, andhas magnitude FE. Also, what is the magnitude offorce F3? IGiven: F} = EDD N F2 = 4130 N FR = 1900 N azlm Solution: Initial Guesses F3=1N x=11n y=ln1 Given Problem 2—103 Each of the four forces acting at E has magnitude F. Express each force as a Cartesian vector and determine the resultant force. Units used: W = m3 N Given: F=23kN Solution: Find the position vectors and then the forces a E] 3' FEB l‘EB = 9 FEB = F |rEB| x —r:' ,I .r _b“. 1H: = a ‘ FEE” = F lrEcil kﬂu rEc. .1 _5 W. IE” 2 | —::I | FED = F IIEDI x _C J TED Find the resultant sum FR = FEA + FEB + FEC + FED K12 ‘ FEB: 3 EN .1‘ —24 ., (—12 1. lx—24, K—12 FED: —3 EN k—E‘i ' U FR = 0 kN .k —9ﬁ ,1 ...
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