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

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An 80 kg skier is descending a 40-degree slope on wooden skis, with a drag calculation yielding a speed of 53 m/s. The skier's dimensions are provided for context, but the focus is on incorporating friction into the analysis. The coefficient of friction is noted as 0.06, and the user seeks guidance on how to calculate the total force acting down the slope. The discussion emphasizes the need to understand the relationship between drag and friction forces in this scenario. Clarification on the total force calculation along the slope is requested.
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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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F_{\mu}=\mu N
So the total force downwards (along the slope) must then be - what?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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