A child pulls a 10lb sled up a 15 degree incline at a constant speed. The child is pulling on a rope attached to the sled. If the rope is inclined at 37 degrees to the horizontal and there is an 8 pound tension in the rope, what is the coefficient of friction?
3. The Attempt at a Solution
First what I did was draw a diagram break up the forces using a free body (see attached). Since the object is not floating up or going downwards through the ground the normal force and the force of gravity must add up to zero so the normal force and gravitational force are equal. With that done, I realized that since it is going up the incline with an 8lb the force opposing it must be mgsin(15). Thus I concluded that Fk=mgsin(15) and since the object is in motion kinetic friction is at play here Fk=μN and with some algebra i came up with tan(15) which was incorrect.
Next attempt I reread the problem and it said at "a constant speed" from my calc classes I know that when the velocity is not changing the acceleration is zero so i figured Fnet=ma a is zero so I came up with 8lb-mgsin15=0 but that made no sense. Is my diagram correct? Where am i going wrong here?