Rod is 1. pulled, 2. pushed horizontally. Force required to sustain constant vel.??? 1. The problem statement, all variables and given/known data There are 2 scenarios. 1. A rod of mass 5kg is first pulled at constant velocity by a force at 45° to the horizontal. 2. Then the same rod is pushed at constant velocity by a force at 45° to the horizontal. Assuming that in both cases the frictional force is horizontal and equal to 0.4 times the normal reaction force on the rod, find the force F in each case. Why is the force different in 1 and 2? (Especially important to me, because I dont understand why it should be different) 2. Relevant equations weight = mass x gravity normal reaction force N = cos(x) x weight friction force f = 0.4 x N f = x-component of F. therefore.. F = f/ cos (x) 3. The attempt at a solution In scenario 1: I found the weight to be 50N. Then the reaction force must be cos (45) x 50N = 35N. Thus the friction must be 0.4 x 35 N = 14 N Thus F = 14 / cos (45) =19.71 N This answer is wrong (F = 20.2 N according to answer key) and I don't understand why. In scenario 2, I don't know why it makes a difference if the rod is pushed or pulled. I again i get F = 19.71 N, even though the answer is supposed to be F = 47.1 N Thank you for your support!