Dynamics question

lussi

I'm studying for my exam and I don't understand the solution to a problem. I hope someone could help. It's actually for Dynamics.

problem: A skier coasts down a hill without initial velocity. The slope is at angle α = 30° with respect to the horizon from p. A to p. B. From p. B to p. C the track is with a reverse angle β = 10° with respect to the horizon. The skier stops at p. C. The motion time for section AB is t1 = 6s, and for section BC - t2 = 15s. Determine the friction coefficient μ between the skies and the snow, as well as the skier's velocity at p. B.

A part of the given solution is: m*y1 = 0 = N1 - m*g*cosα. And I thought that it should be the sin of the angle.

If you could actually explain the solution to the problem, I would really appreciate it.

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collinsmark

Homework Helper
Gold Member
Hello lussi,

Welcome to Physics Forums!

I'm studying for my exam and I don't understand the solution to a problem. I hope someone could help. It's actually for Dynamics.

problem: A skier coasts down a hill without initial velocity. The slope is at angle α = 30° with respect to the horizon from p. A to p. B. From p. B to p. C the track is with a reverse angle β = 10° with respect to the horizon. The skier stops at p. C. The motion time for section AB is t1 = 6s, and for section BC - t2 = 15s. Determine the friction coefficient μ between the skies and the snow, as well as the skier's velocity at p. B.

A part of the given solution is: m*y1 = 0 = N1 - m*g*cosα. And I thought that it should be the sin of the angle.
There's some dimensionality issues with the equation, the way it written above. The left side of the equation is written in mass times distance, while the right side is force. Something about that isn't consistent.

Anyway, let's just take the right side of the equation:

0 = N1 - mgcosα.​
Rearranging,
N1 = mgcosα.​

That's the magnitude of the normal force. Finding the magnitude of the normal force is one step along the way to solving the problem.

bobc2

I'm studying for my exam and I don't understand the solution to a problem. I hope someone could help. It's actually for Dynamics.

problem: A skier coasts down a hill without initial velocity. The slope is at angle α = 30° with respect to the horizon from p. A to p. B. From p. B to p. C the track is with a reverse angle β = 10° with respect to the horizon. The skier stops at p. C. The motion time for section AB is t1 = 6s, and for section BC - t2 = 15s. Determine the friction coefficient μ between the skies and the snow, as well as the skier's velocity at p. B.

A part of the given solution is: m*y1 = 0 = N1 - m*g*cosα. And I thought that it should be the sin of the angle.

If you could actually explain the solution to the problem, I would really appreciate it.
When you used the cos(a), maybe you were thinking of the component of the gravitational force along the direction of the downhill path? Concentrate on just one force component at a time, and take collinsmark's suggestion.

lussi

Well, thank you for the quick answers. I've forgotten to write that y1 in the equation is a second derivative, i.e. the acceleration. In the book, it was written with 2 dots on top of y1.

collinsmark

Homework Helper
Gold Member
Well, thank you for the quick answers. I've forgotten to write that y1 in the equation is a second derivative, i.e. the acceleration. In the book, it was written with 2 dots on top of y1.
Okay, so that's saying that the component of acceleration in the direction of the normal is 0, which makes sense.

However, the component of the acceleration parallel to the surface is not. So, you'll need to sum the force components along this direction, set that equal to $ma_{\|}$. Then kinematics can be used.*

*(If I'm not mistaken, there is an alternate way to solve the problem using conservation of energy instead kinematics. Either way though, you'll still need to calculate the force of friction, and that involves the magnitude of the normal force.)

[Edit: nevermind, about the conservation of energy. I think kinematics is the best approach.]

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