What is the acceleration of a snowboarder when turning on a slope?

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The discussion centers on calculating the acceleration of a snowboarder on a slope while turning. Initially, a 70kg snowboarder moving straight down a 30-degree incline at 20 mi/h experiences an acceleration of 4.9 m/s² down the slope, derived from gravitational force calculations. When the snowboarder turns 35 degrees to the right, the dynamics change, complicating the calculation due to the altered angle of the board relative to the slope and the forces involved. The participant seeks clarification on how to account for the slope of the board and the effects of friction at different angles. The conversation highlights the challenges of applying physics concepts to real-world scenarios in snowboarding.
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Say a 70kg snowboarder is cruising straight down a hill at 20 mi/h. The slope of the hill is due south and at a 30 degree angle. The snowboarder sees a big tree ahead and turns his board 35 degrees to the right.

What is the snowboarders acceleration while going straight?
What is the snowboarders acceleration after having turned 35 degrees?


Thanks
 
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Are you including air resistance and friction of the snow?
PS: If this is your HW feel free to post it in the HW forum.

Tachyon.
 
This is not for homework, I finished uni a few years ago. I'm just studying physics for myself and I have had a bit of difficulty conceptualising this particular problem. I'm not including air resistance, but friction, yes and no. No in the sense of friction slows you down depending on the friction coefficient. But sort of yes, because when the board is at an angle it seems to me the friction will vary along the edge of the board and this variance will of course affect the direction of the acceleration.

If the boarder is going straight down the incline(ignoring friction) here's what I'm getting:

Fgrav = m * g = (70 kg) * (9.8 m/s/s) = 686 N
Fparallel = 686 N * sin (30 degrees) = 343 N
acceleration = Fparallel / m = (343 N) / (70 kg) = 4.9 m/s/s, down the incline

Hopefully that's right. The problem is when the boarder turns 35 degrees to the right. Because although the slope is 30 degrees, the boarder is now facing an angle less than 30 degrees in terms of slope, and now there is a 35 degree angle between the board and Fparallel as well.

So here I have problems, how do you calculate the "slope" of the board? How do you calculate the force that gravity drags the board along the diagonal? What kinds of friction are there going to be at the front of the board and the forward facing side since they are now at an angle? If you know any online resources that would be great as well because all I find is stuff http://www.glenbrook.k12.il.us/gbssci/phys/Class/vectors/u3l3e.html" which deal with the 2d case and not 3d slopes.

Thanks
 
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Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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