hw 2 - W384. Hudsa Juli-f‘xan (i) ENGINEERING MECHANICS -...

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Unformatted text preview: W384. Hudsa Juli-f‘xan (i) ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES For the force shown in Fig. P2—58 (a) Determine the x, y, and z scalar components of the force. (b) Express the force in Cartesian vector form. SOLUTION (a) ny F cos ¢ 475 cos 60° 237.5 F sin ¢ 475 sin 60° = 411.4 a 411 N va cos = 237.5 cos (-127°) -142.93 N a —142.9 N ny sin 9 = 237.5 sin (-1270) -189.68 N 3 -189.7 N = -142.9 E - 189.7 3 + 411 E N Determine the magnitude R of the resultant and the angles 6 , 6 , and 9 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. P2-84. SOLUTION E = 2 1 + 5 + 0 E = 0.3714 1 + 0.9285 3 + 0 E ‘1 {(2)2 + (5)2 + (o)2 a = 4 1 + 5 + 4 E = 0.5298 1 + 0.6623 3 + 0.5298 R 2 /(4)2 + (5)2 +14)2 6 0 1 + 2 + 4 E - 0 i + 0.4472 3 + 0.8944 2 3 /(o)2 + (2)2 + (4)2 10(o.3714 i + 0.9285 3 0 E) = 3.714 1 + 9.285 3 kN = 12(0.5298 i + 0.6623 3 + 0.5298 i) = 6.358 1 + 7.948 3 + 6.358 E RN 5333 = 8(0 1 + 0.4472 3 + 0.8944 2) = 3.578 3 + 7.155 2 kN F1 + F2 + F3 = 10.072 1 + 20.811 J + 13.513 2 am yle /R: + R: + R: /(1o.072)2 + (20.811)z + (13.513)2 26.780 RN 3 26.8 kN 67.91° a 67.9° 39.00° 59.70° A homogeneous cylinder weighing 500 lb rests against two smooth planes that form a trough as shown in Fig. P3-3. Determine the forces exerted on the cylinder by the planes at contact points A and B. SOLUTION _From a free-body diagram for the cylinder: + T 2F = F cos 30o - W y B FB cos 30° - 500 = 0 577.35 lb s 577 lb 0 577 lb 5 60 . 0 FA - FB Sln 30 FA - 577.35 sin 30° = 0 288.68 lb 3 289 lb FA = 239 lb -e Ans. An automobile stuck in a muddy field is being moved by using a cable fastened to a tree as shown in Fig. P3-8. When the cable is in the position shown and force F = 500 N, determine the x- and y—components of the cable force being applied to the automobile. SOLUTION From a free-body diagram of a point on the cable where load P is applied: 0 + T 2Fy = P - 2T sin 5 - 2T sin 5° = T = 2868 N 2868 cos 5° 2357 N s 2.86 kN 2868 sin 5° 249.93 N s 250 N 3—10* Determine the magnitude and direction angle 9 of force F4 so that the particle shown in Fig. P3—10 is in equilibrium. SOLUTION From the free—body diagram for the particle: + -+ EFX = P4 cos 9 — 650 cos 70o - 700 cos 26° + 300 cos 35° F4 cos 6 = 605.72 N + T'zgy = F4 sin 6 + 650 sin 70° — 700 sin 26° - 300 sin 35° F4 sin 6 = -l31.87 N Solving for force F4 and angle 9 yields: -1 -131.87 0 0 6 - tan -§6§T7§ — ‘12.282 3 -12.28 F4 = 34233 = -——:l§lL§1———— = 619.91N s 520 N sin 6 sin (-12.2820) F4 = 620 N s 12.28° liiii ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 3-23* Determine the forces in legs AB, AC, and AD of the tripod shown in Fig. P3—23 when force F = 75 lb. The legs can transmit only axial forces. SOLUTION "it n (0.3982 3 - 0.3186 3 — 0.8602 E)FB (0.3437 i + 0.5579 3 --0.7531 2’5 25 T + 40 i — 54 E F F c '11! n H (-0.4180 i + 0.1003 3 — 0.9029 E)FD F = -75 2 lb fl = FE + FC + FD + F = (0.3932FB + 0.3437FC — 0.4180FD) 1 + (—0.3136FB + 0.5579FC + 0.1003FD) + (-0.8602FB - 0.75311?C — 0.90ngD — 75) E = 0 Thus: 0.3982FB + 0.3487FC - 0.4180FD = 0 —0.3186FB + 0.5579Fc + 0.1003FD = O -0.8602FB — 0.7531FC - 0.9029FD = 75 Solving yields: FB = —33.42 lb 2 33.4 lb (C) Ans. FC = -11.62 lb = 11.62 lb (C) Ans. F = -41.53 lb 3 41.5 lb (C) Ans. 136 ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 3—37 The hot-air balloon shown in Fig. P3-37 is tethered with three mooring cables. If the net lift of the balloon is 900 lb, determine the force exerted on the balloon by each of the cables. SOLUTION = (0.3244 1 + 0.4867 3 — 0.8111 ErrA = (0.2752 1 - 0.4299 3 — 0.3599 ErrB 900 E EF=TA+TB+TC+E = (0.3244TA + 0.2752TB — 0.4319Tc) i + (0.4867TA — 0.4299TB — 0.2592TC) + (-0.,81111‘A - 0.35991"B - 0.8639TC 0.3244TA 0.2752TB 9.4319TC 0.4867TA 0.4299TB 0.2592TC -O.8111TA 0.8599TB 0.84539TC Solving yields: 418.2 lb 205.2 lb 444.9 lb ...
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hw 2 - W384. Hudsa Juli-f‘xan (i) ENGINEERING MECHANICS -...

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