Chapter5

# Chapter5 - ENGINEERING MECHANICS STATICS 2nd Ed W F RILEY...

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Unformatted text preview: ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 5-2* Locate the center of mass for the three particles shown in Fig. P5—2 if mA = 26 kg, m8 = 21 kg, and mC = 36 kg. SOLUTION + mC = 26 + 21 + 36 = 83 kg IIIAXA + meB + 111ch = 26(-200 cos 60°) + 21(-200) cos 30°) + 36(200) = 962.69 kg-mm ‘+ mAyA mByB + mcyc 26(200 sin 60°) + 21(—200 sin 30°) + 36(0) = 2403.33 kg-mm giggégg = 28.956 mm 3 29.0 mm '29] 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 mass for the five particles shown in Fig. P5—6 if mA = 2 kg, mB = 3 kg, and :11E 2 kg. #SOLUTION : 9 c + mD + mE 2 + 3 + 4 + 3 + _ = + + ’ + + . EAXA meB mch meD mExE 2(300) + 3(150) + 4(300) + 3(300) + 2(0) = 3150 kg'mm + + + + mAyA mByB mcyc mDyD mEyE 2(240) + 3(400) + 4(400) + 3(0) + 2(200) 3680 kg'mm + + + mAZA mBzB mczc + szD mEzE 2(0) + 3(0) + 4(270) + 3(270)+ 2(270) = 2430 kg'mm ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 5—12* Locate the centroid of the shaded triangular area shown in Fig. P5—12 if b = 200 mm and h = 300 mm. Fig. P5-12 SOLUTION ’hor the differential element 3 of area shown in the sketch at the right: _ h y—gx _ -2 d dA — y dx — b x x ux dMY : % dA = h—; x2 dx 2b dM = x dA = h x2 dx y b _ b h2 2 _ h2 b 2 h2 x3 b hzb MY— —§‘(dX—-—2 xdxz—ZB— =_6— ' 0 2b 2b 0 2b 0 b b 3 b 2 - 22 -2 2 _2x_ Lb My - J0 b x dx — b [0 x dx — b[ 3]0 — 3 -l A — 2 bh M 2 - -_v-m-z -2 - Axc - My xc - A - bh/Z - 3 b - 3(200) — 133.3 mm Ans. M 2 AyC = Mx yC = 31 = EB??? - g = Egg = 100 0 mm Ans. ENGINEERING MECHANICS ~ STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 5-14 Locate the centroid of the shaded area shown in Fig. P5_14o F%.P544 SOLUTION For the differential element of area shown in the sketch at the right: 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 ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 5—35 Locate the centroid of the shaded area shown in Fig. P5—35. Flg. P5-3S SOLUTION The shaded area can be divided into a rectangle with two circles removed. The centroid for the composite area is determined by listing the area, the centroid location, and the first moment for the individual parts in a table and applying Eqs. 5-13. Thus, 1440 -37.71 -113.13 1289.16 M _ Y _ 1289.16 _ . C j Ci y C — 3- — -§TZT§E — 6.00 in. Ans. > X ll [‘1 > >< U 2 X -—_—.._._.—_—_.— ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 5-37 Locate the centroid of the shaded area shown in Fig. P5~37. SOLUTION The shaded area can be divided into three rectangles. The centroid for the composite area is determined by listing the area, the centroid location, and the first moment for the individual parts in a table and applying Eqs. 5—13. Thus, ___§£_ - YC - A — 48 - 4.00 in. (Note symmetry about y axis) 326. ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 5—40 Locate the centroid of the shaded area shown in Fig. P5—40. “SOLUTION The shaded area can be divided into a rectangle, and a quarter circle, with a square removed. The centroid for the composite area is determined by listing the area, the centroid location, and the first moment for the individual parts in a table and applying Eqs. 5-13. Thus, 2 2 = %n(150)2 = 17,671 mm 4r _ 4(150) 3n = 63.66 mm, 113.66 mm My yCi Mx (mmZ) (mm) (mm3) (mm) (mma) __ 7500 75 . 562,500 25 187,500 17,671 63.66 1,124,936 113.66 2,008,486 -5625 37.5 -210,938 87.5 —492,188 19,546 1,476,498 1,703,798 = 1,476,498 : 19,546 = 1,703,798 : 19,546 ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 5-44 Locate the centroid of the shaded area shown in Fig. P5—44. SOLUTION The shaded area can be divided into a square with a circle and a quarter circle removed. The centroid for the composite area is determined by listing the area, the centroid location, and the first moment for the individual parts in a table and applying Eqs. 5—13. Thus, = an: n(60)2 = 11,310 mm 2 2: ln(1zo)2 11,310 x _ _ 4(120) _ - 240 3“ - 189.07 mm 4(120) 240 3" ' 189.07 mm Ai XCi My yCi Mx 3 (mmz) (mm) (mma) (mm! (mm ) 1 57,600 120 6,912,000 120 6,912,000 2 —11,310 100 -1,131,000 80 -904,800 3 -11,310 189.07 -2,138,382 189.07 -2,138,382 2 34,980 3,642,618 3,868,818 = 3,642,618 _ 34,980 = 3,868,818 = 34,980 ENGINEERING MECHANICS — STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 5-69 Determine the resultant g of the system of distributed loads and locate its line of action with respect to the left support for the beam shown in Fig. P5—69. SOLUTION 200(3) 1.5 ft. 150(3) 4.5 ft. 100(3) 7.5 ft. 2F = A1 + A2 + A3 = 600 + 450 + 300 = 1350 lb = 1350 lb 1 A + A x + A x 1XC‘1 2 C2 3 C3 = 600(1.5) + 4.50(4.5) + 300(7.5) = 5175 ft'lb '"' "'"UL ----- ---. 3.56 ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 5-72 Determine the resultant ﬁ of the system of distributed loads and locate its line of action 25kNm1 with respect to the left support for the beam shown in Fig. 95—72. WE ‘ ‘ ‘ L—nglv‘im Fig. P5-72 SOLUTION 2.5(4) = 10 RN 4 m %(2.5)(4) = 5 MI 2 + gm = 4.667 In SF = A1 + A2 = 10 + = 15.00 kN ¢ + Azxc2= 10(4) + 5(4.667) = 63.335 kN-m 357 ENGINEERING MECHANICS - STATICS. 2nd. Ed. W. F. RILEY AND L. D. STURGES 5—85* The width of the rectangular gate .shown in Fig. P5—85 is 8 ft. Determine the magnitude of the resultant force R exerted on the 3 gate by the water (7 = 62.4 lb/ft ) pressure and the location of the center of pressure with respect to the hinge at the bottom of the gate. SOLUTION p,, = 7d = 62.4(7) = 436.8 lb/ftz max 1 Vps = Epmax %(436.8)(7)(8) R hw 12230.4 lb 2 12.23 kip Ans. %(7) = 2.33 ft Ans. The width of the rectangular gate shown in Fig. P5-86 is 2 m. Determine the magnitude of the resultant force ﬂ exerted on the gate 9y the water (0 = 1000 kg/m ) pressure and the location of the center of pressure with respect to the hinge at the top of the gate. SOLUTION p x: pgh 1000(9.807)(3) = 29,421 N/m2 - -l R — V — 2p hw ps max - §<29.421)(3)(2) 88,263 N a 88.3 RN 2 _ 1 + 3(h) — 1 + ...
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Chapter5 - ENGINEERING MECHANICS STATICS 2nd Ed W F RILEY...

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