HW_4 - SOLUTION (a) (b) PROBLEM 2.71 A horizontal circular...

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Unformatted text preview: SOLUTION (a) (b) PROBLEM 2.71 A horizontal circular plate is suspended as shown from three wires which are attached to a support at D and form 30° angles with the vertical. Knowing that the x component of the force exerted by wire AD on the plate is 220.6 N, determine (a) the tension in wire AD, (b) the angles 19x, By, and 62 that the force exerted at A forms with the coordinate axes. @536“ E vs wine Fx = F sin30°sin50° = 220.6 N (Given) A D F - “r—*—.‘—" \ sm3 0° sm50° ,. ~ 220'6N = 575.95 N . cosfix = i = = 0.3830 F 575.95 0x = 67.5°4 Fy = Fcos30° = 498.79N F 0030}, = —y = 498'79 = 0.86605 F 575.95 Q, = 300° 4 F2 = —Fsin30°cos50° = —(575.95 N)sin30°cos50° = —185.107 N c0502 = E— = = —0.32139 F 575.95 Oz = 108.7° 4 PROBLEM 2.85 A force F of magnitude 400 N acts at the origin of a coordinate system. Knowing that 6x = 28.5°, Fy = —80 N, and F1 > 0, determine (a) the components Fx and F2, (b) the angles fly and 02. SOLUTION (a) Have Fx = Fcosflx = (400 N)cos28.5° Fx = 351.5 N 4 Then: fl=fi+fi+fi So: (400 N)2 = (352.5 N)2 + (—80 N)2 + F} Hence: F2 = + (400 N)2 — (351.5 N)2 — (—80 N)2 Fz = 173.3 N4 (19) cosfiy = % = 2T? = —0.20 0y =101.5° 4 c050 = g = = 0.43325 (92 = 64.3° { 2 F 400 PROBLEM 2.39 n . A 150W Plateismsupportedpy three cables as shown. Knowing that the tension in cable 4:3 is. 204 1b,:detem1ine the components Of the force exerted on thepIatg gt 3, . V ' SOLUTION 37 e (32 my. .+ (4,8 huh: (36 in.)k (32 in)? 13(48'in32 2+ (33$,m.)? .= 68 ‘ 1 __... BA 2043-11) 3’ ' 3" BA 68m; j ,— (36 in.)k‘]':»; 1333(961b)i§+;(14421b)j;1—-(108x1b)k g .e .» I t ., Fx =+96.0 lb, Fy =+144.o lb, F, =—108.01b< ...
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This note was uploaded on 02/12/2010 for the course EGM 2511 taught by Professor Jenkins during the Spring '08 term at University of Florida.

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HW_4 - SOLUTION (a) (b) PROBLEM 2.71 A horizontal circular...

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