4-84, 4-90, 4-111

# 4-84, 4-90, 4-111 - 4—34 Two Parallel forces of opposite...

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Unformatted text preview: 4—34 Two Parallel forces of opposite sense = 125 i + 200 3 + 250 E N and = -125 i — 200 3 — 250 E N act at points A and B of a body as shown in Fig. P4-84. Determine the moment of the couple and the perpendicular distance between the two forces. Fm.P4£4 SOLUTION P1 = 125 I + 200 3 + 250 E N F1 = {(125}2 + (200)2 + {250)2 = 343.09 N I.AfB H = F x F B A/E 1 A = 0.200 1 - 0.030 3 + 0.000 E m = {0.200 E - 0.030 3 + 0.000 E} x (125 i + 200 3 i 3 E 0.200 —0.030 0.000 125 200 250 -19.50 i — 42.50 3 + 43.75 E N-m M = /(~19.5012 + (-42.50)2 + {43.7512 = 04.04 N'm s 04.0 N-m B M _ a _ 04.04 _ d — E: _ 535735 - 0.10033 m a 130.3 mm Three couples are applied to a rectangular block as shown in Fig. P4-90. Determine the magnitude of the resultant couple C and the direction angles associated with the resultant couple vector. NON rm.P¢90 SOLUTION 80 1 N rB’A -0.375 + 0.250 E m —75 E N Fain -0.200 - 0.375 3 + 0.250 E m A = 100 J i-‘mA = —0.200 1 + 0.250 E m Summing moments about point A yields: {rB/A x F3) + {r CIA x Flcl + {rD/A x F l D {—0.375 3 + 0.250 ﬁ) x {30 1) + {—0.200 1 — 0.375 3 + 0.250 E} x {—75 E) a. + [—0.200 1 + 0.250 E] x [100 3) 3.125 1 + 5.000 3 + 10.000 2 N-m C = {13.12512 + {5.000)2 + (10.000)2 = 11.609 N-m a 11.61 N-m O 74.4 0 ' 64.5 0 a 30.5 4-111 Replace the GEO-lb force shown in Fig. P4-111 by a forca at point B and a couple. Express your answer in Cartesian vector form. Fig. 94-111 SOLUTION F = F = 660 7 i - 20 i — 11 E 1712 + (-2012 + (—1112 = 193.51 T — 552.89 3 — 304.09 E 5 193.5 I — 553 j — 304 E 15 Ans. A n FMB x F = 115 j + 17 E} x (193.51 1 — 552.39.1 J - 304.09 E} i j E 0 15 17 193.51 -552.39 -304.09 4838 i + 3290 3 - 2903 E in.-lb A 4.34 i + 3.29 — 2.90 E in.-kip ...
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## This homework help was uploaded on 04/16/2008 for the course ENGR 1100 taught by Professor Anderson during the Fall '06 term at Rensselaer Polytechnic Institute.

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4-84, 4-90, 4-111 - 4—34 Two Parallel forces of opposite...

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