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Skiers and their frictions

  1. Sep 18, 2014 #1
    1. The problem statement, all variables and given/known data

    So, I'm studying for the MCAT and I can't for the life of me understand why I'm wrong here.
    Here is the question:

    A skier is given a strong push so that he slides up the hill in Figure 2 (it’s a non-descript right triangle with a stick man skier on it, the angle of incline is irrelevant) for a certain distance with a μk =.1. When he gets to the highest point, he slides back down. How does the acceleration of the skier on his ascent compare to the acceleration on his descent? Do not consider the acceleration of the push.

    A. The acceleration on the descent is smaller in magnitude than on the ascent.
    B. The acceleration on the ascent is smaller in magnitude than on the descent.
    C. Both accelerations are the same.
    D. The accelerations have the same magnitude but different directions.



    2. Relevant equations

    Fk=FNμk
    FN= mgcosΘ.
    Fk= mgsinΘ.


    3. Attempts at explanation

    So, these questions tend to be broadly theoretical, and the answer to this particular one is B. The book reasons that friction is additive as the skier is moving up the hill and subtractive as the skier moves down, i.e. the forces on the skier moving up the hill (gravity + friction) add, and the forces as she moves down the hill (gravity - friction) subtract. My issue here is that friction opposes movement/force, not slope. There would still be some friction as the skier went up the hill, right? Right? Or am I losing my mind?
     
    Last edited: Sep 18, 2014
  2. jcsd
  3. Sep 18, 2014 #2

    PhanthomJay

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    you are right regarding friction in either direction, and the book answer is wrong, although it's reasoning is correct. Gravity force components act down the slope no matter which direction the skier is traveling, but friction forces always oppose the relative motion between the contact surfaces. But you don't say what the correct answer is. The correct answer is ?????. Note also you have a typo error in your 3rd relevant equation.
     
  4. Sep 19, 2014 #3
    The correct answer was B, and I can logically reach that conclusion, I think, but I can't really sort the math out. I don't understand why I'd subtract friction one way and not the other. But I also don't have a very strong grasp on force summation, and I'd love any suggestions anyone can offer to help me in that regard, especially if it can be explained conceptually.
     
  5. Sep 19, 2014 #4

    BvU

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    As Jay explained, B is NOT the right answer.

    [edit edit] OOPSOOPS B is really wrong. A is right. Boy, what a mess!

    When in doubt, make a drawing.
    Top pic shows situation going UP the slope: There are only three forces working on the skier: gravity, normal force and friction. Normal force compensates component of gravtiy force perp to slope (red). The gravtiy force component along the slope (##mg \sin \theta##, red) plus the friction force (blue) add up and decelerate.
    Bottom pic almost the same, but now the movement is downwards, so the friction force is upwards along the slope. Same magnitude, but clearly the resultant has smaller magnitude than in the first picture.

    And (nitpicking): your remark "the angle of incline is irrelevant" is not completely correct: if the slope has ##\sin\theta < \mu## there won't be any sliding down !
     

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  6. Sep 19, 2014 #5
    My apologies! I hadn't had my cuppa yet; I'm not sure I count among the living until I do. Fair enough on the angle of incline, you made a small but important distinction that I had heretofore not considered. Your drawing was very helpful. My suspicion was correct --> the magnitude of the acceleration due to gravity doesn't change as it's just the parallel component of gravity. However, the summed magnitude of acceleration is affected by the kinetic friction, which is in the same direction as the skier ascends and opposes the skier as he descends, which means the magnitude of his acceleration is... magnitude implies absolute value, right? then, as you said, A must be right. Thank you both for your help!
     
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