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Class 4 otes

# Class 4 otes - PROBLEM 2.75 Determine(a the x y and 2...

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Unformatted text preview: PROBLEM 2.75 Determine (a) the x, y, and 2 components of the 900-N force, (b) the angles 0x, 0),, and 02 that the force forms with the coordinate axes. MOO ,V SOLUTION (a) Fx = (900 N)cos30°cos25° = 706.40 N F; = 706 N 4 Fy = (900 N)sin30° = 450.00 N Fy = 450 N 4 F2 = —(900 N)cos30°si1125° = —329.04 N Fz = —329 N 4 ([7) COSQX = W 900 N or ex = 383° 4 450.00 N 005% = ————- 900 N or By = 60.0° 4 —329.40 N cosBz = —~—— . 900 N or 02 =111.5°4 PROBLEM 2.90 A transmission tower is held by three guy wires anchored by bolts at B, C, and D. If the tension in wire AD is 1260 N, determine the components of the force exerted by the wire on the bolt at D. SOLUTION m = (4 m)i + (20 m)j + (14.8 m)k DA = (4 m)2 + (20 m)2 + (14.8 m)2 = 25.2 In F = FKDA = FE =1260N DA 25.2m [(4 m)i + (20 m)j + (14.8 m)k] F = (200 N)i + (1000 N)j + (740 N)k Fx = +200 N, F) = +1000N, FZ = +740N‘ PROBLEM 2.95 The boom 0A carries a load P and is supported by two cables as shown. Knowing that the tension is 510 N in cable AB and 765 N in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables. SOLUTION TB = —(600 mm)i + (360 mm)j + (270 mm)k AB = (—600 mm)2 + (360 mm)2 + (270 mm)2 AB = 750 m I: = —(600 mm)i + (320 mm)j — (510 mm)k AC = (—600 mm)2 + (320 mm)2 + (—510 mm)2 AC = 850 mm TB 510 N . . TAB = TABE _ 750 mm[—(600 mm)! + (360 mm)] + (270 mm)k] TAB = —(408 N)i + (244.8 N)j + (183.6 N)k A—C 765 N . . TAC — TACAC — 850 mm[—(600 mm)1 + (320 mm)] — (510 mm)k] TAC = —(540 N)i + (288 N) j — (459 N)k R = TAB + TAC = —(948 N)i + (532.8 N) j — (275.4 N)k , Then R=1121.80N R=1122N< and cos0x = ﬂ X = 147.7° 4 1121.80 N cos0 = i32—8—N— 0 = 61.6° 4 y 1121.80 N y osez = 12.7315 02 = 104.2° 4 1121.80 N ...
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Class 4 otes - PROBLEM 2.75 Determine(a the x y and 2...

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