Force F acts on the frame such that its component acting along member AB is 690

Question

Force F acts on the frame such that its component acting along member AB is 690lb , directed from Btowards A, and the component acting along member BCis 530lb , directed from B towards C. Determine the direction

? of force I found the magnitude of F to be 970 N

Force F acts on the frame such that its component acting along member AB is 690lb , directed from Btowards A, and the component acting along member BCis 530lb , directed from B towards C. Determine the direction? of force I found the magnitude of F to be 970 N

Explanation / Answer

xx: Fx=-F1cos(60)+F2cos(45) = -690cos60 + 530 cos45 = 29.77 lb
yy: Fy=-F1sin60-F2sin45 = -972.32 lb

F1=FAB=690, F2=FBC=530

Fx=Fcos(Theta+60)
Fy=Fsen(Theta+60)
Fy/Fx=tan(theta+60)-->theta=tan-1(Fy/Fx) - 60->theta=-148.25

theta along xx = 180-148.25 = 31.75 degrees

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