HWS3 - Haw Mew .2771 Foo; oz" ENGINEERING...

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Unformatted text preview: Haw Mew .2771 Foo; oz" ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 2—58 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 ¢ 475 005 60° 237.5 N = F sin ¢ = 475 sin 60° = 411.4 N 2 411 N = ny cos 6 = 237.5 cos (—127°) = -142.93 N = ny sin 6 = 237.5 sin (-127°) —189.68 N (b) F = —142.9 f — 139.7 3 + 411 E N 2—59* For the force shown in Fig. P2—59 (a) Determine the x, y, and z scalar components of the force. (b) Express the force in Cartesian vector form. 'SOLUTION (a) d = / x2 +'y2 + z2 = /(,6)2 o (10)2 + (8)2 = 14.142 = 254.6 lb 3 255 lb — 424.3 1b a 424 lb 339.4,1b ,339 lb 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 componehts of force F1. (b) Express force F1 in F1=9mnb Cartesian vector form. ' (c) Determine the angle a between forces F1 and F2. SOLUTION '9.595 ft = —556.99 lb'g —557 lb 278.49 lb 278 lb 649.82 lb 650 1b -557 i.+ 278 3 + 650 E lb ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 2-84 Determine the magnitude R of the resultant and the angles 6 , 6 , 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—84. SOLUTION é : 2 1 + 5 + 0 fl = 0.3714 1 + 0.9285 3 + 0 E 1 1/12.)2 + (5)2 +10)2 62 = ——3—l>1¥31414LJLJ§—— = 0.5298 1 + 0.6623 3 + 0.5298 E 1/1412 +1512 +14)2 3 = 0 1 + 2 + 4 E — 0 i + 0.4472 3 + 0.8944 E 3 - N0)2 + (2)2 + (4)2 1 1e1 10(0.3714 i + 0.9285 3 '11! H q > II + 0 E) = 3.714 T + 9.285 3 kN F -= F a = 12(0.5298 i + 0.6623 3 + 0.5298 2) = 6.358 3 + 7.948 3 + 6.358 E kN 8(0 1 + 0.4472 3 + 0.8944 E) = 3.578 3 + 7.155 E kN fi = F1 + F2 + Pa = 10.072 1 + 20.811 J + 13.513 2 gay yle ' R = ,/R: + R: + a: = /(10.072)2 + (20.811)2 + (13.513)2 = 26.780 kN E 26.8 kN I Ans. 10.072 0 0 x R 26.780 — 67.91 a 67.9 Ans. Q) n o o m u o o m -1 20.811 ' _ 0 y 00 26.780 ' 39'00 II (A {D O Ans. 13.513 _ 0 26.780 7 59.70 . Ans. 03 II 0 O m n o o m H o: :0 Q 77 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 p = 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: + T 28y = p — 2T sin 5° 2T sin 5° T = 2868 N 2868 cos 5° 2857 N e 2.86 RN 2868 sin 5° 249.96 N a 250 N ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES ’ 3-10* Determine the magnitude and direction angle 6 of force F4 so that the particle shown in Fig. 93-10 is in equilibrium. ‘ SOLUTION From the free—body diagram for the particle: + —+ ZFX 2 F4 cos 6 ~ 650 cos 70o - 700 cos 26° + 300 cos 35° cos 9 = 605.72 N F4 + T 29y = F4 sin 6 + 650 sin 70° — 700 sin 26° ~ 300 sin 35° F4 sin 9 = —131.87 N . Solving for force F4 and angle 6 yields: '1 0 : 0 ' — “ ... - 0 tan ’EEET7E 12.282 12.28 3‘944 = ———ll§li§i———— = 619.91N a 620 N ' Afis. sin a sin (—12.2820) mg ll 4 F = 620 N E 12.280 »/ Ans. .IIZJ? ...
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HWS3 - Haw Mew .2771 Foo; oz" ENGINEERING...

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