Finding coefficient of friction given slope

[Carriebun]

Homework Statement

A skier coasts down a 3.5 degree slope at a constant speed. Find the coefficient of kinetic friction between the skis and the snow covering the slope.

Homework Equations

coefficient of friction x Normal force = force friction

therefore: coefficeint of friction = force friction / normal force

The Attempt at a Solution

I'm totally stuck. I drew the Free Body Diagram, but i have no idea where to go from there.

 

[rock.freak667]

Well what is the force parallel to the slope and what is the force normal to the slope?

 

[Carriebun]

i wasnt given any of that information, the only other thing i know is that Fn=-Fg, since there is no accel, so the Ffr= (9.81)μm... but where do i go from there?

 

[rock.freak667]

i wasnt given any of that information, the only other thing i know is that Fn=-Fg, since there is no accel, so the Ffr= (9.81)μm... but where do i go from there?

You need to split the weight into two components, one parallel to the slope and one normal to it.

The one normal to it will be your normal force which will give you the frictional force.

 

[rwisz]

I would first start by defining the variables in the problem. The slope angle of 3.5 degrees can be converted to radians (0.0611 rad) and the constant speed of the skier can be represented by the velocity (v). The force of friction (Ff) can be calculated using the equation Ff = μN, where μ is the coefficient of friction and N is the normal force.

Next, I would analyze the forces acting on the skier. The only forces acting on the skier are the force of gravity (mg) and the force of friction (Ff). Since the skier is moving at a constant speed, the net force on the skier must be zero. This means that the force of friction must be equal and opposite to the force of gravity.

Using this information, we can set up the equation μN = mg. We know that the normal force is equal to the component of the force of gravity perpendicular to the slope, which can be calculated using N = mgcosθ, where θ is the slope angle.

Substituting this into our equation, we get μmgcosθ = mg. The mass of the skier (m) cancels out, leaving us with μ = 1/cosθ. Plugging in the slope angle of 3.5 degrees, we get μ = 1/cos(0.0611) = 1.002.

Therefore, the coefficient of kinetic friction between the skis and the snow is approximately 1.002. This means that the force of friction is slightly greater than the force of gravity, which is why the skier is able to maintain a constant speed down the slope.