hw5_sol - 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 4-1* Determine the moment of the 250—1b force shown in Fig. P4—1 about point A. SOLUTION IFId = 250(3) = 750 ft-lb A = 750 ft-lb D Ans. ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 4—9* Three forces are applied to a circular plate as shown in Fig. P4—9. Determine (a) The moment of force F1 about point 0. (b) The moment of force F3 about point 0. The moment of force F2 about point A. QOLUTION (a) Mo = IF1Id1 = 50(12) 600 in.'lb = 600 in.-lb D Ans. . o _ . ' (b) MO IF3Id3 - 100(12 s1n 30 ) - 600 1n. lb 600 in.'lb 5 Ans. (0) MA |F2|d2 80(24 sin 45°) = 1358 in.-lb 1358 in.-lb 3 Ans. ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 4-11 Determine the moment of the 350—lb force shown in Fig. P4—11 about points A and B. Fig. P4-11 SOLUTION MA”: |F|dA 350(10 sin 40° + 12 cos 40°) 5467 in.°lb a 5.47 in.-kip ME = IFIdB 350(10 sin 40°) = 2250 in.'lb a 2.25 in.-kip ENGENEERING MECHANICS — STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 4-13* Two forces are applied F1:3251h to an angle bracket as shown in Fig. P4—13. Determine the moments of forces F1 and F2 about points A and B. SOLUTION o .H . ’ . MM _ {F1|dM _ 325(12 cos 30 } _ 33:? 1n. lb fl 3330 in.°lb D Ans. A1 . o __ I- I . MAE z |F2|dA2 _ 425(12 cos 50 1 — 2350 1n. lb HA2 2550 in.'lb D Ans. _ o , o _ r. , . M31 = |F1|dBi = 323(12 cos 30 + 12 Bin 30 } _ 532: 1n. lb 831 z 5330 in.-lb D Ans. ,_ , O O _ ‘ . MHZ = |F2|dB2 = 420(12 Sln 60 - 12 cos 60 ) _ 1866.? 1n. lb ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 4-14* Two forces are applied to an eye bracket as shown in Fig. P4-14. Determine the moments of forces ?1 and F2 about points A and B. SOEUTION M = IFlldAl = 250 (0.150 cos 20° + 0.325 sin 20°) = 63.03 N-m flAl a 63.0 N-m 5 Ans. 400(o.325 sin 65° — 0.150 cos 65°) = 92.46 N-m Weld” fihz a 92.5 N-m D Ans. 250(0.325 sin 20°) 81.25 N-m H II (Fild81 331 a 81.3 N-m 5 Ans. 400(0.325 sin 65°) 117.82 N'm |F2|d82 R82 2 117.8 N-m D Ans. M2. 4-20 A 160-N force is applied to the handle of a door as shown in Fig. P4-20. Determine the moments of the force about hinges A and B. SOLUTION + C MA 160 cos 45° (1.500 — 0.900) + 160 sin 45° (0.800) = 158.39 N'm a 158.4 N'm ' EA = 158.4 N'm j + C MB = -150 cos 45° (0.900 — 0.400) + 160 sin 45° (0.800) = 33.94 N-m a 33.9 N-m fl = 33.9 N'm 5 MA ENGINEERING MECHANICS - STATICS, 2nd. F. RILEY AND L. D. STURGES _‘.' 5011:: 4—23* A 50-1b force is applied to the handle of a lug wrench which is being used to tighten the nuts on the rim of an automobile tire as Shown in Fig. P4—23. The diameter of the bolt circle is 5 1/2 in. Determine the moments of the force about the axle for the wheel {point 0} and about the point of contact of the wheel with the pavement (point A}. SOLUTION + C M0 = -50(20 + 2.75 sin 20°) = —1o47.03 in.-ib a —1047 in.-ib E 1047 in.‘lb D Ans. + C MA = —50{20 + 16.?5 sin 20°} = —1235.44 in.-1b a —1286 in.'lb E II ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 4-25 Three forces ?A. FE, and ?c are applied to a beam . . F,:3001b as shown in Fig. P4—25. Determine {a} The moments of forces FA and FC about point D. (b) The moment of force FB about point D. SOLUTION ° (91 + FA sin 50° (30) «200 cos 60° {91 + 200 sin 60° {30} 4296 in.'lb a 4.30 in.‘kip H OH F cos 30° (9} — Fc sin 30° {60} C 250 cos 30° (9) - 250 Sin 30° (60) —5551 in.-lb = —5.55 in.-kip F! 0C 0 (b) + C M = —FB cos 45 {9 + 4] + FB sin 45° (601 —300 005 45° {13) + 300 sin 45° (60] 9970 in.-lb = 9.97 in.-kip ...
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hw5_sol - ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F....

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