See sketch. If Fcosq is the only horizontal force, then the acceleration is
a = (Fcosq )/m. If friction (magnitude f) is significant, then
a = (Fcosq - f)/m. Note that if the wagon has constant velocity, a = 0 and Fcosq = f. If the coefficient of friction m is given, the force of friction is m times the normal force, so you must consider all vertical forces: Fsinq and the normal are up and mg is down. So
F_N + Fsinq = mg, or F_N = mg - Fsinq . Now the force of friction is
f = m F_N, and it opposes the motion.

Coast back to the main physics page or the Newton's 2nd law stuff.