Finding coefficient of static friction with applied force and normal force

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A Man dragged a 51 Kg bag at a 45 degree angle with a force of 400 N. The bag moved along hte pavement with a constant speed. What is the coefficient of static friction?

OK SO, you can make a triangle and figure out that 283 N are pulling horrizontal and 283 pulling vertically. But I am not sure how to plug any of that into the

F=(Friction coef)N

where F is force of friction, and N is normal force...

How do i find the coeff. of static fricton?

Another one that is similar is the following:

In grass-skiing, a man goes down at 92 Km/hr. It takes him 6.6 to reach that speed from rest at a 35 degree incline. His mass is 75 Kg. What is the coeffcient of friction between the grass and his skis that day?

I'm totally lost...help please!?
 

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1) Draw a free-body diagram and resolve the forces parallel and perpendicular to the inclined plane.

Since there is no acceleration along both these directions, the forces must be balanced. One equation will tell you what N is, and the other will tell you what (mu)N is.

From these two numbers you can find (mu).

2) What is the use of the knowing the time and velocity ? Can you find something useful from that ?
 
Since the bag is moving at constant velocity, the net force acting on it is zero.
Given the force which is pulling the bag, you can get the force of friction.
 
fs, max = 400 N : Fs, max/fn=U sub s : Fn= mass times gravity times the cosine of 45
 
In grass-skiing, a man goes down at 92 Km/hr. It takes him 6.6 to reach that speed from rest at a 35 degree incline. His mass is 75 Kg. What is the coeffcient of friction between the grass and his skis that day?

Is areodynamic drag part of this problem? If so, you need to use terminal velocity equations (do a web search for this). Also, you don't need to know his mass to solve this problem, since gravity and friction will both be relative to the mass. Treat the friction as an acceleration factor instead of a force factor and then you can ignore the mass.