A 204-kg log is pulled up a ramp by means of a rope that is parallel to the surface of the ramp. The ramp is inclined at 29.6 ° with respect to the horizontal. The coefficient of kinetic friction between the log and the ramp is 0.825, and the log has an acceleration of 0.734 m/s2. Find the tension in the rope. I drew a beautiful free body diagram like the one in my book that is kind of like this problem, but I can't show you that. Anyway; The sum of the forces in the x direction = ma in the x direction and The sum of the forces in the y direction = 0 So, sigmaFx = (204kg)(0.734m/s^2) = mg(sin(29.6)) - 0.825 + T T = 149.736/(204 x 9.8 x sin(29.6) - 0.825) T = 0.1518 N Keep in mind when reading this, my teacher is a total douche and won't explain what "sum of the forces" even means so I am getting what I can out of the book but obviously I am not doing it right because the homework thing is telling me the answer is wrong.