Calculating Drag & Friction on 80kg Skier Down 40° Slope

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SUMMARY

The discussion focuses on calculating the forces acting on an 80 kg skier descending a 40° slope using the drag equation and incorporating friction. The skier's dimensions are specified as 1.8m tall and 0.4m wide, with a friction coefficient of 0.06. The calculated drag force results in a speed of 53 m/s, but the user seeks guidance on integrating friction into the overall force calculations. The relevant friction formula is provided as F_{\mu}=\mu N, indicating the need to determine the normal force for accurate calculations.

PREREQUISITES
  • Understanding of basic physics concepts, specifically forces and motion.
  • Familiarity with the drag equation and its application in fluid dynamics.
  • Knowledge of friction coefficients and their role in motion on inclined surfaces.
  • Ability to calculate normal forces on inclined planes.
NEXT STEPS
  • Research the application of the drag equation in real-world scenarios.
  • Learn how to calculate normal force on an inclined plane.
  • Explore the relationship between friction and motion on slopes.
  • Study the dynamics of skiing and the impact of various forces on performance.
USEFUL FOR

This discussion is beneficial for physics students, sports scientists, and engineers interested in the dynamics of skiing and the forces involved in downhill motion.

Gott_ist_tot
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I have a question where an 80 kg skier goes down a 40 degree snow slope on wooden skis. The skier is 1.8m tall and .4m wide. I use the drag equation and get 53 m/s. However I am not sure how to add friction into this. Any guidance is appreciated. Oh, friction is .06.

I am not looking for an answer only direction. The numbers were put down just in case.

Thank you for your help.
 
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[tex]F_{\mu}=\mu N[/tex]
So the total force downwards (along the slope) must then be - what?
 

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