Chapter2 - ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F....

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Unformatted text preview: ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES Determine the magnitude of the resultant fl and the angle 6 between the x axis and the line of action of the resultant for the two forces shown in Fig. P2-3. SOLUTION 0 0 ¢ = 90° + 29° - 21 = 98 From Eqs. 2—1 and 2-2: 2 2 - F1 + F2 + 2F1F2 cos ¢ a’isoz + 4002 + 2(480)(400) cos 989 = 580.48 a 580 lb sin ¢ . o _ . -1 400 Sin 98 _ o R — Sln -—§§673§——- - 43.03 B + 21° = 43.03° + 21° = 64.030 e 64.0° § = 580 lb A 64.00 Ans. Determine the magnitude of the resultant R and the angle 9 between the x axis and the line of action of the resultant for the two forces shown in Fig. P2—4. SOLUTION From Eqs. 2—1 and 2-2: 2 2 2 R — F1 + F2 + 2F1F2 cos ¢ = 2502 + 2002 + 2(250)(200) cos 50° 408.39 a 408 N 51“ "’ - sin-1 - 22 03° - 408.39 “ ' 36.030 a 36.00 fl = 408 N a 36.0° Ans. ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 2-5* Determine the magnitude of the resultant fl and the angle 6 between the x axis and the line of action of the resultant for the two forces shown in Fig. P2—5. SOLUTION From Eqs. 2-1 and 2-2: 2 2 , F1 + F2 + ZFiF2 cos ¢ 2502 + 6002 + 2(250)(600) cos 30° 826.02 lb 3 826 lb 1 F2 51“ ¢ _ sin-1 600 sin 30° _ R ‘ 826.02 ' 0 B + 30° = 21.30° + 30 = Determine the magnitude of the resultant fl and the angle 6 between the x axis and the line of action of the resultant for the two forces shown in Fig. P2—6. SOLUTION ¢ = 36 + From Eqs. 2—1 and 2—2: 2 2 — F1 + F2 + 2F1F2 cos ¢ 102 + 252 + 2(10)(25) cos 60° 31.22 kN a 31.2 kN 51“ ¢ _ sin-1 25 sin 60° R “ 31.22 B - 24 = 43.91 - 24 = 19.91° fl = 31.2 RN 2 19.91° Ans. = 43.91° ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 2—7 Determine the magnitude of the resultant R and the angle 9 between the x axis and the line of action of the resultant for the two forces shown in Fig. P2-7. SOLUTION O = 45° + 60 2 2 - F1 + F2 + 2F1F2 cos ¢ 5 6002 + 8002 + 2(600)(800) cos 105° 866.91 lb 2 867 lb 31“ ¢ _ . -1 800 sin 105° R “ 51“ 866.91 B + 30 = 63.046 + 30 = 93.046° e 93.0° fi = 867 lb 5 87.00 = 63.0460 Determine the magnitude of the resultant fl and the angle 9 between the x axis and the line of action of the resultant for the two forces shown in Fig. PZ—8. SOLUTION = 107° 2 2 , F1 + B“,a + 8}?sz cos (p 7802 + 6502 + 2(780)(650) cos 107° 856.99 N K 857 N 81“ (1) Sin" w - 46 ‘0° 856.99 ' '3 R = 857 N s 12.5o° Ans. ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 2-13 Determine the magnitude of the resultant E and the angle 9 between the x axis and the line of action of the resultant for the two forces shown in Fig. P2-13. SOLUTION From Eqs. 2 2 , — F1 + F2 + ZFlF2 cos ¢ 6002 + 5002 + 2(600)(500) cos 54.1s° 980.47 lb 3 980 lb 51“ ¢ _ . -1 500 sin 54.160 R ’ $1" 980.47 ' E + 21.80 = 24.42 + 21.80 = 46.22 g 46.‘ Ans. Determine the magnitude of the resultant § and the angle 9 between the x axis and the line of action of the resultant for the two forces shown in Fig. P2—14. SOLUTION 1 2 4 2~22 ¢ = 180° - tan- From Eqs. 2—1 and 2 2 — F1 + F2 + 2F1F2 cos ¢ 2402 + 1802 + 2(240)(180) cos 79.700 324.73 N a 325 N Si“ ¢ _ sin-1 130 sin 79.70° _ 33 R ‘ 324.73 ' ' O 8 + 36.87 = 33.05 + 36.87 = 69.92° a 69.9 ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 2—15* Determine the magnitude of the resultant R and the angle 6 between the x axis and the line of action of the resultant for the two forces shown in Fig. P2-15. SOLUTION -1 = tan tan—1 = 32.01° 180° - 20.850 - 32.01° = 127.14° Eq. 2—1: 2 2 F1 + F2 + 2F1F2 cos ¢ 8002 + 10002 + 2(800)(1000) cos 127.140 820.96 lb a 821 lb Eq. 2-2: -1 F2 31" ¢ _ . -1 1000 sin 127.14° _, o R - 5m - 16.17 8 + 20.35° = 76.17 + 20.85 = 97.02° e 97.o° fi = 821 lb s 83.00 ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 2-29* Determine the magnitude of the resultant § and the angle 9 between the x axis and the line of action of the resultant for the four forces shown in Fig. P2-29. SOLUTION From Eq. 2-1: 2 2 R12 — F1 + F2 + 2F1F2 cos ¢1 2502 + 3502 + 2(250)(350) cos 50° R12 545.42 lb 2 545 lb From Eq. 2-2: ' _1 F2 sin $1 1 R -1 350 sin 50° 545.42 B sin = 29.44° Similarly: = 130° — 20° - 30° = 130° 3 2 + F4 + ZFaF‘1 cos $2 3002 + 6002 + 2(300)(600) cos 130° 467.54 lb a 468 lb F4 Sln ¢2 R34 -1 000 sin 130° 0 467.54 — 100.56 ENGINEERING MECHANICS — STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 2-29* (Continued) 360° - 80° - 31 — B 2 360° - 80° — 29.44° - 100.560 = 150.oo° 2 2 R12 + R3“ + 2R12R34 cos ¢3 545.422 + 467.542 + 2(545.42)(467.54] cos 150° 272.75 lb 2 273 lb Sln $3 R -1 467.54 sin 150° 272.75 34 = 58.99° 0 81 + 83 — 30 0 29.44° + 58.990 - 30 fi = 273 lb & 58.4° Ans. ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 2—37 Two ropes are used to tow a boat upstream as shown in Fig. P2-37. The resultant E of the rope forces Fu and Ev has a magnitude of 400 lb and its line of action is directed along the axis of the boat. Determine the magnitudes of forces and Fv. U SOLUTION sin 30° sin 110° —: sin 40° 274 lb sin 110 400 . 0 Sln 110 sin 30° 213 lb Two cables are used to support a stoplight as shown in Fig. P2~38. The resultant R of the cable forces Eu and Ev has a magnitude of 1350 N and its line of action is vertical. Determine the magnitudes of forces E and E . u V SOLUTION F U sin 36.870 sin 45° sin 98.130 1350 sin 93.13° 1350 sin 98.13° sin 36.37° = 818 N sin 45° 964 N ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 2-54* Determine the x and y scalar components of the force shown in Fig. P2-54. Fig. P2-54 SOLUTION F cos 6x‘ 675(-5//§§) F sin ex 675(—2//§§> 2—55* For the force shown in Fig. P2-55 (a) Determine the x, y, and z scalar components of the force. (b) Express the force in Cartesian vector form. F = 100011) SOLUTION (a) F‘y 1000 cos 25° 906.3 lb 1000 sin 25° = 422.6 lb 2 423 lb 906.3 cos 130° = —582.6 lb 2 -583 lb 906.3 sin 130° = 694.2 lb 2 694 lb + 694 3 + 423 E ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 2-60* For the force shown in Fig. P2-60 (a) Determine the x, y, and z scalar components of the force. (b) Express the force in Cartesian vector form. SOLUTION (ar'd = x2 + y2 F cos 9x ' . = ~5.612 kN a —5.61 kN F cos 9y —11.224 kN a ~11.22 kN = F cos 92 16.836 kN a 16.84 kN (b) ? = — 5.61 2-61 For the force shown in Fig. P2-61 F=1mow (8) Determine the x, y, and ' z scalar components of the force. (b) Express the force in Cartesian vector form. SOLUTION (a) d = /(-4)2 + (12)2 + (7)2 = /209 = 14.457 F cos 9x = 1500(-4//209) = -415.03 1b % -415 1b r cos 6y 1500(12//209) = 1245.09 lb 2 1245 lb = F cos 62 = 1500(7//209) = 726.30 1b a 726 lb = -415 i + 1245 j + 726 R 1b ENGINEERING MECHANICS - STATICS, 2nd. Ed; W. F. RILEY AND L. D. STURGES 2—63* Two forces are applied to an eyebolt as shown in Fig. P2—63. (a) Determine the x, y, and z scalar components of force F1. (b) Express force F1 in Cartesian vector form. (c) Determine the angle a between forces F1 and F2. Fig. P2-63 SOLUTION (—6)2 + (3)2 + (7)2 = 9.695 ft = -556.99 lb 2 -557 lb 278.49 lb a 278 lb 649.82 lb a 650 lb + 273 3 + 650 R lb (—6)2 + (6)2 + (3)2 = 9.00 ft 9.695 E 6 _ g + 3 9.695 3 9.695 30.73 2 30.7° ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 2-83* Determine the magnitude R of the resultant and the angles 6 , 9 , and 6 between the x y 2 line of action of the resultant and the positive x—, y~, and z- coordinate axes for the three forces shown in Fig. P2-83. SOLUTION a1 2 0 1 + 2 + 2 E = 0 i + 0.7071 3 + 0.7071 2 (0)2 + (2)2 + (2)2 = 4 1 + 4 + 0 fi = 0.7071 3 + 0.7071 3 + 0 E /(4)2 +14)2 + (0)2 = 3 1 + 0 + 2 E = 0.7071 1 + 0 j + 0.7071 R (0)2 + (2)2 + (2)2 = 600(0 1 + 0.7071 3 + 0.7071 2) 424.26 3 + 424.26 E lb A 800(O.7071 1 + 0.7071 3 + 0 R) 565.68 1 + 565.68 3 lb 70010.7071 i + 0 j + 0.7071 R) 494.97 E + 494.97 E lb + F3 = 1060.65 1 + 989.94 3 + 919.23 E lb ,/R: + R: + a: = /(1060.65)2 + (989.94)2 + (919.23)2 = 1717.54 lb E 1718 lb Rx _ -1 1060.65 “"‘ COS R 1717.54 0 51.860 a 51.9 -1 939.94 ’ COS 1717.54 -1 919.23 0 0 1717.54 57.64 a 57.6 — COS ...
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This note was uploaded on 03/25/2008 for the course MEEN 221 taught by Professor Mcvay during the Spring '08 term at Texas A&M.

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Chapter2 - ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F....

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