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HW2 - Problem 2-51 Determine the magnitude of force F so...

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Unformatted text preview: Problem 2-51 Determine the magnitude of force F so that the resultant FRof the three forces is as small as possible. What is the minimum magnitude of FR? Units Used: v kN = 1000 N Given: ——————X F 1 = 5 kN | I F2 = 4 kN ‘3 i E M x a t. 1 t9 = 30 deg F? _ x‘: | - 1. Solution: Scalar Notation: Suming the force components algebrically, we have ‘ + _—> FRx=ZFx; FRx = F] —Fsin(t9) +T FRy= 2F); FR), = Fcos(6) — F2 The magnitude of the resultant force FR is FR = ,/ 1:sz + FR; = (F; _ Fsin(6))2 + (“03(0) _ F2)2 56 FR2 = F12 + F22 + F2 — 2FF1 sin(€) _ 2F; Fcos(l9) dF‘ ZFR—R = 2F — 2F] sin(0) - 2172 cos(t9) dF dF IfF is aminimum, then ((1—15 = ] F = F] sin(¢9) + F2 005(0) FR = (F1 _ mm)? + (“05(0) _ F2)2 Problem 2-58 Express each force in Cartesian vector form. Units Used: kN = 103 N Given: F] = 5 kN F2 = 2 kN 91 = 60 deg 62 = 60 deg 63 = 45 deg‘ Solution: cos(62) Flv = F1 cos(03) cos(61) 0 sz = F2 -1 X iiiii Problem 2-61 The stock S mounted on' the lathe is subjected to a force F, which is caused by the die D. Determine the coordinate direction angle ﬂ and express the force as a Cartesian vector. - Given: F = 60 N a = 60 deg y = 30 deg 65 Solution: cos2(a) + cosz(ﬂ) + cos2(y) = 1 ﬂ = acos(1 — cos(a)2 —- cos(y)2) ,6 = 90deg Loos(a) ~30 n = F mo) F 0 N —cos(y) 5 _‘ ~52 Problem 2-68 Determine the magnitude and coordinate direction angles of the resultant force. , z Given: F1=350N a=60deg F2 = 250N ,6 = 60 deg 0:3 7:45 deg mi 61 = 4 t9 = 30 deg ﬂ 3 1’ Solution: 1: Fa cos(a) 175 Flv = F] cos(ﬂ) Flv = 175 N —cos(7) —247.5 F F d F F C 2h = 2 —--- 2y = 2 —-- ~ \/c2+d2 J3”? F2hCOS(6') 173.2 FZV : —F2hsin(0) F2v = —100 N F2y 150 [FR = F1v+F2v i won.) ...
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