Question

+ -/30 points NorEngStatics 1 4.P.031. Force F = 1,545 N acts from A towards B, as shown. 275 mm FV 550 mm 220 mm Determine the magnitudes of the moments of force F about the following lines in N. m. (a) oc Nom (b) OD Nom (c) CD (d) BC N·m (e) DE DNm (f) DF N·m (g) FG DN.m

Answer 1

p, Co, C, 0.235) (О. 22, oo, 255) В; - (0, 0,55,0.235) (О. 22,0 55, 0.25) (0.22,90) 10,0-55,0) (О. 22,0-55,0)

Moment about oc uz 7 OC = 0.55j +0.275€ u= 0.55j +0.2757 - 2. ht Ik V5 T5 V0.55 +0.275 r. a position vector from any pointon line to any point on the line of action of Force. the fore wewill take OA OA = 0.227+0.559 r= 0.227 +0.555 ² = 0.559 -0.275 ² = F = AB = 0.55 9 – 0.275² No.55²+0.2752 - 2 ] _ i Vo V5 F = 1545 (21 - eto , 15450 )

F = 0.55 ] _ 0.275 k 10.55 +0.295 E 1545x - 138 1.89 F - 690-94k Moment aboutoc - & u. rx = 271.91N-m. 10.22 0.55 o lo 1381.29 -690.194

A souTop из о-245E оп», Ү- Yo3 - 0.221. с. 13 81, 84 Т – 6qo -44 le M = UerXF 3o4.016 N- 0.22 оо в о 138 1. 34 - Geo.44 Ароит ср . о. 55 г. - 5 yo-95” "ps = 0.22 Y = E (38 (-940 — Gao.S4 E M = Urr AF = - 152.007 N-my

ABOUT BC Moment is zero since we can take it between line and line as any of action points of force % = 0 so is one vector M= v. r XF 0 ABOUT DE 037137 +0.9284 F un 0.229 +0.553 10, 22² +0.550 r=ar DB = 0.227 F = 1381.899-690.947 Moment z vorXF - 141. 123 N-m.

ABOUT DE U= 0.559 – 0.295k 1.0.55 +0.255 = 2 5 J-1 15 r= DB = 0.221 F = 1381.893 - 690-94R. - v.rtf ABOUT FG U ) = 0.373 î - 0.9284 ] 0.22 - 055 No.22 +0.55 r=FA = 0.221 F = 1381.89] - Gav.94 if = +1 +23 - 141. 123 M =

Summary if u= ai+by+ck radit éjt fk F = gî that ke Then Mau. r XF detaminant of d e fl