Calculating Frictional Force: Tension in Sleigh Rope & Mass of 47kg

[F.B]

I am really stuck on a few questions but i'll only post one.

An adult is pulling two small children in a sleigh over level snow. The sleigh and children have a total mass of 47 kg. The sleigh rope makes an angle of 23 degrees with the horizontal. The coefficient of kinetic friction between the sleigh and the snow is 0.11. Calculate the magnitude of the tension in the rope needed to keep the sleigh moving at a constant velocity. (Hint. The normal force is not equal to the force of gravity.)

I don't know what to figure out first. See i need Fn but i can't figure it out.
I think Fn=Fg - Fapp(in the y direction). But i don't have a force to help me out. So can anyone please help me.

 

[Tide]

HINT: Now look at the horizontal forces and realize what "constant velocity" means.

 

[F.B]

a=0, so does it equal:
0=Fapp(in the x direction) - Ff
But i still don't know what Fapp is because i don't have a force there.

 

[Tide]

You now have two equations each of which contains Fapp. Use one equation to eliminate it in the other!

 

[F.B]

If i do that then i have:
Fapp=Fg - Fn
Fapp=Fnet + Ff

I can't use those because i still don't know Fn. It won't work.

 

[Tide]

You need to separate the applied force into horizontal and vertical components:

$$F_{app, y} = F_{app} \times \sin 23^o$$

and

$$F_{app, x} = F_{app} \times \cos 20^o$$

The normal force is therefore

$$F_n = -F_{app} \times \sin 20^o + mg$$

and the horizontal force is

$$0 = F_{app} \times \cos 20^o - F_f$$

Now use the fact that

$$F_f = \mu F_n = -\mu(- F_{app} \times \sin 20^o + mg)$$

and you should be able to determine the magnitude of the applied force.