20. If we write Grxyz=++###ijk, then (using Eq. 3-30) we find GGrF×is equal to yFzFzFxFxFyFzyxzy x−+−dibg###.ijk(a) In the above expression, we set (with SI units understood) x= –2.0, y= 0, z= 4.0, Fx= 6.0, Fy= 0 and Fz= 0. Then we obtain ˆ(24N m)j.rFτ=× =⋅GGG(b) The values are just as in part (a) with the exception that now Fx= –6.0. We find ˆ
This is the end of the preview. Sign up
access the rest of the document.