Thread: Impulsive force
View Single Post
tomrule007
tomrule007 is offline
#5
Apr30-07, 08:17 PM
P: 18
So this is the same thing i did before but updated with the new signs
B)
I_x=.2(8.48528-9.02723)= -.10839

I_y=.2(8.48528-(-11.9795))= 4.092996


and to solve for the F_average i just do divide by .05
so
F_x=-.10839/.05 = -2.1678 N

F_y=4.092996/.05 = 81.8591 N