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# hw05 - 5—40 The magnitude ofthe force F in Fig P5-4U is...

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Unformatted text preview: 5—40 The magnitude ofthe force F in Fig. P5-4U is 976 N. Determine (a) The component of the moment at point C parallel to line CE. (1)) The component of the moment at point C perpendicular to line CE and the direction angles associated with this moment vector. SOLUTION 7350i7 300j+160k J3502+3002+1602 MC : (0141+ 0.4 j)>< (7700.06i7 600.06 j +320.03 k) = (128.01i—44.80j+196.02 k) N-m c2 : M : (70.46483i+0.88540j) 5/2102 +4002 MCE = MCDBCE = (128.01) (—0.46483)+ (—44.80)(0.88540) = —99.169 N-m (a) F 976 (700.061 600.06j+320.03k)N B MCE = Mme” = (46.101—87.80j) N -m Ans (b) Mi : MC 7M”. : (128.01i744.80j+196.02k)7 (46.10i78780j) : (81.91i+43.00j+196.02k) N- m Ans 120 STATICS AND MECHANICS OF MATERIALS, 2".1 Edition RILEY, STURGES AND MORRIS Mi = (01.91)2 + (43.00)2 + (196.02)2 = 216.75 N-m 0x : cos'1 81'” : 67.80° Ans 216.75 0? 2mg;1 43'00 278.560 Ans 216.75 92 = cos—119602 = 25.26021.“ 216.75 5742 Determine the moment of the couple shown in Fig. P542 and the perpendicular dismnce between the two forces. SOLUTION UM = (76060535°)(0.2)+ (7605111350) (0.1) = 168.103 N-m O ................... Ans. 163.103 N-m:760d d : 0.221 m : 221 mm Ans 121 STATICS AND MECHANICS OF MATERIALS, 2"” Edition RILEY, STURGES AND MORRIS 5743 A lug wrench is being used to tighten a lug nut on an automobile Wheel as shown in Fig. P5-43. Two equal magnitude, parallel forces of opposite sense are applied to the wrench. If the magnitude of each force is 2.5 lb, determine the couple applied to 3 lug nut and express the result in Cartesian vector form. SOLUTION M = —(25)(12)i = —3001 1b- in. Ans 5—77 Determine the y—cnordinate of the centroid of ﬁle shaded area shown in Fig. P5-77. SOLUTION dAV=(x2/Zb)dx dAH : [b’JZEde 130 STATICS AND MECHANICS OF MATERIALS, 2mi Edition RILEY, STURGES AND MORRIS M _ 8333.33 X A _ 833.33 5—30 Determine the y—coordinate of the centroid of the shaded area shown in Fig. PS-SO. SOLUTION dAH : (“f/52)“)? = 10.00 in. Ans yo: 031/4 : g 06/3 4 5-94 Locate the centroid of the shaded area shown in Fi_ P5-94. SOLUTION A = A, +A2 +4, =(20x60)+(20x80)+(20x60)=4000 mm2 E!xCA! 30(1200!+10!1600!+30!1200! x0 = = A 4000 = 22.0 mm .................................................................................. Ans. _ 2(ycA) 10(1200)+60(1600)+l [0(1200) ‘ TZ—m— = 60.0 mm .................................................................................. Ans. Yo SOLUTION 5,000 0 57,800 68,000 29,000 159,300 2 xCL) Ic- HR 57108 The loads acting on a beam are distributed in a triangular manner as shown in Fig. PS-IOS. Determine and locate the resultant with respect to the left end ofthe beam. SOLUTION I R = LX275” = 750 N “W" (4)(750) “F _‘I®lllllllllllilll.- I . : — : 1500 N - - 2 2 LEM—17:..." In-—J R : 750+1500 : 2250 N l... ...............Ans. OM 7 2250d: 750[(2/3)(2 ]+1500[2+ 4/5)] (1 = 2.67 m A115 144 5—109 Determine the resultant of the distributed loads acting on the beam shown in Fig. P5-109, and locate its line of action with respect to the support at A. S OLUTION R,:(200) (3): 600 lb (150) (3): 450 1b 237000“ )730011) R=600+450+300=13501b l... OM =1350d= 600(1.5)+450(4. 5)+5.00(7 5) 57111 A distributed load acts on a beam as shown in Fig. PS-l ] 1. Determine and locate the resultant of the distributed load with respect to the support at A. S OLUTION R1 = jwdx = [010528 dx = [x3]: = 1000 lb 4 10 0Ma : R1071 :wadx : [:03x3dx : [3%] : 75001b-ﬂ 0 R2 :(300)(5):15001b R :1000+1500 : 25001b l, Ans 0M2 : 2500d: 7500 +1500 (10+2.5) d =10.50 ﬁt Ans ...
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