hw2_sol - Determine the magnitude of the resultant fl and...

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Unformatted text preview: Determine the magnitude of the resultant fl and the angle 8 between the x axis and the line of action of the resultant for the two forces shown in Fig. PZ—Z SOLUTION From Eq. 2—1: 2 .2 I . £1 + E2 + JF1F2 cos ¢ 602 + 542 + 2i60}(541 cos 60° 98.?? 3 98.8 N Eq. 2-2: sin ¢ R _ . 0 sinui 3i+§i3:§9— = = 98.8 N z 23.3° 98.x! . -1 2 311} 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 ¢ = 90° + 29° - 21 = 98° From Eqs. 2—1 and 2-2: 2 2 - F1 + F2 + 2F1F2 cos ¢ a’4802 + 4002 + 2(480)(400) cos 989 = 580.48 a 580 lb 51“ ¢ _ . -1 400 sin 98° T—Sln W 0 580.48 ' 43°03 B + 21° = 43.03° + 21° = 64.030 g 64.0° § = 580 lb A 64.00 Ans. 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“ ¢ _ S. -1 300 sin 105° R “ 1“ 866.91 B + 30 = 63.046 + 30 = 93.046° e 93.0° fi = 867 lb 5 87.00 Ans. = 63.0460 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“ ¢ _ 811.1 500 sin 54.160 R ’ ‘ 980.47 ' E + 21.80 = 24.42 + 21.80 = 46.22 g 46.‘ 980 lb & 46.20 Ans. 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. Two cables are used to support a stoplight as shown in Fig. P2~38. The resultant R of the cable forces F" and Fv has a magnitude of 1350 N and its line of action is vertical. Determine the magnitudes of forces F and F . 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 ...
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hw2_sol - Determine the magnitude of the resultant fl and...

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