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Unformatted text preview: 5—40 The magnitude ofthe force F in Fig. P54U 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) Nm c2 : M : (70.46483i+0.88540j)
5/2102 +4002 MCE = MCDBCE = (128.01) (—0.46483)+ (—44.80)(0.88540) = —99.169 Nm (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 Nm 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 Nm O ................... Ans.
163.103 Nm: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. P543. 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. P577.
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. PSSO.
SOLUTION dAH : (“f/52)“)? = 10.00 in. Ans yo: 031/4 : g
06/3 4 594 Locate the centroid of the shaded area shown in Fi_ P594.
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. PSIOS. 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. P5109, 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. PSl ] 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%] : 75001bﬂ
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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 Fall '08
 Brody

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