What is the coefficient of friction in a child pulling a sled up an incline?

[Sprockets]

Homework Statement

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?

Homework Equations

F=ma
Fk=μN
Fs≤μN
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?

 

[Simon Bridge]

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 ...

No ... there are three forces perpendicular to the slope - one of them is the y component of the force of gravity. But you are correct that the sum of the forces is zero.

Is my diagram correct? Where am i going wrong here?

You have drawn the tension parallel to the slope - check the angles again.
Write down the equation for the forces parallel to the slope.

 

[AdityaDev]

What are the forces acting along the normal to the incline?
What are the forces acting along the incline?

 

[Sprockets]

Sorry Simon I meant to say mgcos(15) is equal to the normal force, its on my diagram.

AdityaDev, I have the force along the incline as mgsin(15). I think what's really confusing me is that 37 degree angle where is it? "rope inclined at 37 degrees to the horizontal," to me says that there is a rope that a kid is pulling on, on top of the 15 degree incline the angle between the rope and the horizontal is 37 and since the kid is pulling on this block that's on top of this 15 degree incline the tension must be parallel to the slope of the 15 degree incline. Like what i tried to draw.