hw8_sol

# hw8_sol - ENGINEERING MECHANICS STATICS 2nd Ed W F RELEY...

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Unformatted text preview: ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RELEY AND L. D. STURGES 5-1* Locate the center of gravity for the three particles shown in Fig. P5—1 if WA = 25 lb, WE = 30 lb, and NC = 45 1b. SOLUTION 25 + 30 + 45 = 100 lb 25:0; + 30(—10 cos 40°} + 45(10 cos 50°) = —4.813 in.'lb 7 ' w v + * “AKA + BEE hcyc 25(10! + 30(“10 sin 40°} + 45(—10 Sin 600} = -332.55 in.'lb Zw.x. I. 1 EN x _ i i _ —4.813 n _ . _ - ——i—~ — -T56- — 0.04813 1n. 2 0.481 1n. ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 5-4 Locate the center of mass for the four particles shown in Fig. P5-4 if mA = 16 kg, = 24 kg, mC = 14 kg, and = 36 kg. SOLUTION 16 + 24 + 14 + 36 = 90 k2 + [BAXA meB + mcxc + meD 16(300) + 24(0) + 14(0) + 36(0) = 4800 kg-mm + + m y mByB + m V mDy A A C‘C D 16(0) + 24(0) + 14(500) + 36(0) 7000 kg~mm + + mAzA mBzB + chC szD = 16(0) + 24(0) + 14(0) + 36(400) 14,400 kg'mm 273 ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES Locate the center of gravity for the five particles shown in Fig. P5—5 if WA = 15 lb, WE = 24 1b, NC = 35 1b. WD = 18 lb, and WE = 26 lb. SOLUTION A + h?B WAxA + W + W x BXB C 15110) + 24(10) + 35(0) + 18(10) + 26(0) WAyA + hBYE 15(0} + 24(11} + 35116) + 18(16) + 25(0} N z + W A A + Wcz Z 8 B 15(0) + 24f0) + 35(0) + 18(111 + 28(111 E = 15 + 24 + 35 + 18 + . C wDXD c + wDZD M XG — .li W 12‘?<S + W + W EXE + hcyc + HDyD + WEyE 2 E E 26 = 118 lb 570 in.°lb 1112 in.-1b 484 in.°lb ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 5-11* Locate the centroid of the ’ shaded triangular area showu in Fig. P5—11 if b = 12 in. and h = 8 ih. SOLUTION For the differential element of area shown in the sketch at the right: y=%(b-Xi dA = y dx = E{b - x} dx UK I x III'Jx U h2 2 de = E dA = "—E(b - x} dx 2b -1_1_ y — b(b x] x dx Q, Z I >< D. Sb- 1 ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 5—15* Locate the centroid of the shaded area shown in Fig. P5—15. Fig. P5-15 SOLUTION For the differential element of area shown in the sketch at the right: VG y dx lllllilll'l'll'. v 2 dA ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 5—17* Locate the centroid of the shaded area shown in Fig. P5—17. SOLUTION For the differential element of area shown in the sketch at the right: 2 2X. b2 x dy ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 5-19 Locate the centroid of the shaded area shown in Fig. P5-19. SOLUTION For the differential element of area shown in the sketch at the right: ~ n u I I I V I I I I n I I I I I u a 5 2. _ I2 2x dx ‘ 5/2 5 3/2 - x I2 /5 x dx — /3 [ 5/2 5 x3/Z 5 I2 /5; dx = /5 [ 3/2]2 ...
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## This note was uploaded on 03/26/2008 for the course CVEN 221 taught by Professor - during the Spring '08 term at Texas A&M.

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hw8_sol - ENGINEERING MECHANICS STATICS 2nd Ed W F RELEY...

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