I posted this question once and am still confused as to how to solve, I have spent much time going at it from different directions....It can't be as hard as I am making it out to be. The question is: "A 60kg skier is in a tuck and moving straight down a 30 degree slope. Air resistance pushes backward on the skier with a force of 10 Newtons. The coefficient of dynamic friction between the skis and the snow is 0.08. What is the resultant force that actson the skier?" I get very confused trying to set it up and I am confused as to how to factor in the normal friction. I am not looking for the ANSWERS, but I am looking for the steps to take to get towards them. This is what I've set up, I'm just not sure if it is rigt or if I'm heading in the right direction....: 1) Drew a right angle triangle with the hypoteneuse at angle of 30 degrees (so it looks like the slope). 2a) Placed skier on slope (hypoteneuse) drew a line straight through skier with the force of gravity acting on him (60)(9.8) = 588N. 2b) Drew a vector acting parallel to slope against the skier with a force of 10 N 3) Then I solved for the vertical and horizontal side of the triangle and got: V = sin 30*10 = 5 N H = cos 30*10 = 8.7 N 4) Then I calculated normal force: Fn = (.08)(588) = 47.04 N 5) I don't know what to do from here. I am very much new to this and am being rushed through the class, but I can pick up things very easily. Please help!