Originally Posted by **ChipB** I think you're on the right track. If you know the velocity of the forced air (call it v_a) and the velocity of the cylinder (v_c) on its track, and assuming the air in front of the cylinder has zero velocity, then the net force of the air acting behind and in front of the cylinder is:
F_net = (1/2) rho C_d A ((v_a-v_c)^2 - v_c^2)
= (1/2) rho C_d A (v_a^2 - 2 v_a v_c)
This net force must equal friction between the cylinder and the supporting structure. This assumes that the track is horizontal (i.e. we can ignore any force needed to go up or down hill). Note that if there is no friction F_net is zero if v_c = (1/2)v_a. |

I am happy that I am on the right track. However, I have to admit I am a Little lost on the difference of sq(va-vc) and vc square. Which two forces are you considering? Drag and maybe the pushing force?

I only need to know the force air is applying to the bottles to move them.

Thank you so much for your help.