Determine the components of F that act along rod AC and perp

Determine the components of F that act along rod AC and perpendicular to it. Point B is located at the midpoint of the rod. Given: F = 600 N i = 4m b = 6 m c = 4 m d = 3m e = 4 m

Solution

In 3-D space co-ordinates of A, C, D are

A(0,0,4), C(3,-4,0), D(-4,-6,0)

CD² = 7² + 2² , CD= 7.28

OC² = 4² +3² , OC =5

OA = 4

AC² = 4² +5²   , AC = 6.4

BC = 3.2 , B is midpoint of AC

Drop a perpendicular BM from B to the xy plane then

M= (1.5, -2, 0) , BM = 2

DM² = 5.5² + 4² , DM = 6.8

BD² = BM² + DM²   = 2² +6.8² , BD = 7.08

Consider the triangle BDC

BD=c = 7.08, BC =d = 3.2 , CD =b= 7.28

Use the triangle formula

b² = c² + d² -2cdCos(B)

Cos(B) = (7.08² + 3.2² – 7.28² )/(2*7.08*3.2) = 0.163

Sin(B) = 0.99

Component of F along AC   = FCos(B) = 600*0.163 = 97.56 N

                          ^_r to AC   = FSin(B) = 600*0.99 = 594 N