Downhill Skier: Calculating Resultant Forces

[ruzoo]

Please Help ASAP: Downhill Skier Question

Here is the question:

"A 60kg skier is in a tuck and moving straight down a 30 degree slope. Air resistance pushes backward on the skier with a force of 10 Newtons. The coefficient of dynamic friction between the skis and the snow is 0.08. What is the resultant of the forces that act on the skier?"

This is what I have done:

Made a right angle triangle with the hypoteneuse at a 30 degree angle and the skier coming down the hypoteneuse. Converted skier's weight into mass: 588 N. Calculated the vertical component of the triangle (5 N, which is the Fn?) and the horizontal component (8.7 N, which is the parallel force?). I drew an arrow with a force of 10 N acting parallel against the slope (hypot) acting against the skier. I am stuck as of here and how to calcultate the resulting forces.

 

[ruzoo]

In Addition...

I was also told to try and redo this problem by making my hill or slope be the x axis...I guess draw a tilted axis diagram to make the forces opposing the skier one value...but that confused me a little.

 

[Diane_]

First off, you converted the skier's mass into weight.

Second - when you talk about "vertical" and "horizontal", do you mean that literally or are you speaking with respect to the plane? Normally in the latter case we talk about "parallel to the plane" and "perpendicular to the plane".

I'll assume the latter is what you meant. You have the skier's weight component perpendicular to the plane - this means you know the normal force, or should. From the normal force and the coefficient of friction you can get the frictional force. You know that the normal force will cancel the perpendicular component of the weight, so you can reduce this to a one-dimensional problem, parallel to the hill. You can get the net force, then, by doing a vector sum on the forces you know - which is the answer for which you're looking.

As to your addition - I will confess I'm puzzled. If I'm reading the question correctly, then that's exactly what you'd do to answer the first one. So either I'm not reading it right or I'm missing something about the first one.

Anyway - does this help any?