A farm tractor tows a 4400-kg trailer up a 21 degree incline at a steady speed of 3.0 m/s. What force does the tractor exert on the trailer? (Ignore friction.) How would you get the answer 15000 N.
equilibrium Draw yourself a diagram of the trailer showing all the forces acting on it. Then realize that since it's moving at a steady speed, it must be in equilibrium: the net force must be zero. In particular, that means that the components of the forces along the incline must add to zero.
Hint: if you are allowed to ignore friction, does the steady speed of the tractor even need to enter into your calculations? ADDED NOTE: As usual, Doc Al got there first.
Ok! I made a coordination where x-axis goes through this inclination. Sigma Fx = max = 0 because ax = 0 Fx = N + W + "Pull" Fx = 0 + mgsin21 + "Pull" = 0 Pull = -15452 N am I right?
Your notations are rather cryptic, but if I understand you right, you balance the cable tension (from the tractor) with the component of gravity along the incline. In addition, it seems you set zero acceleration, which also is right (steady speed). So it looks OK to me, your answer is in magnitude equal to 15000 if you round off.
Right! But be careful interpreting that minus sign: it just means that the pull is opposite to the weight. The weight pulls down the incline, so the "pull" must pull up the incline. (The only reason it turned out negative is because you chose down to be positive.)