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Forces on skier: slope friction and wind

  1. Feb 11, 2014 #1
    1. The problem statement:

    Slick is skiing down a 27 degree slope (above horizontal) into a headwind (parallel to the slope) of 8N. His mass is 78Kg. The coefficient of friction between the snow and the skis is 0.11. What is the resultant force? If the coefficient of friction increases to 0.13, how much additional force will Slick have to add to maintain the same resultant force, assuming that the added force is parallel to the slope? How much additional force will he have to add to maintain the same resultant force in the added force is 45 degrees to the ski slope (ie. he pushes backward with his pole)? How can he potentially reduce the force that is opposing his progress down the hill

    2. Relevant equations
    I don't really know what to put in this?


    My attempt:

    I found the resultant force, which I calculated to be 264.4N? I also came up with 13.5N, for the additional force he would have to add to maintain the same resultant force. Its so late, and I want to be done. But I am having a hard time trying to figure out the equation for how much force hed have to add is the angle changed to 45 degrees?

    Any help at all would be greatly appreciated!
     
  2. jcsd
  3. Feb 12, 2014 #2
    I just came up with this:

    264.4N = (347.4N + Cos(45)(x)N)-8N-(.13(681.8N - Sin(45)(x)N))

    The X's in parenthesis after the Cos and Sin represent numbers, but I cant figure out which numbers go into these slots...does anybody have any idea?
     
  4. Feb 12, 2014 #3

    Simon Bridge

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    Welcome to PF;

    You do everything from a free-body diagram... just add the vectors.
    We cannot comment unless you show your working though.

    When the angle of the added force changes, then not all the added force will be in the same direction as the previous resultant - since some of the added force is downwards it will affect the friction. Redraw the free-body diagram.

    The last question is a matter of looking at where the opposing force comes from.

    [edit] I don't know what that equation (post #2) is supposed to show so I cannot comment on what x is supposed to be.
     
  5. Feb 12, 2014 #4
    I have a free body diagram drawn in front of me. I dont know if there is a way for me to post a picture of it on here thhough
     
  6. Feb 12, 2014 #5

    Simon Bridge

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    It is possible to scan a picture then upload it to PF using the "manage attachments" dialog.
    Usually you can describe it - put the +x axis pointing down along the slope and the +y axis pointing up, perpendicular to the slope.

    The forces are gravity and friction with a normal force from the slope along with an applied force 45deg to the slope provided by the skier. All forces need to be resolved into x and y components. Remember that the friction force is proportional to how hard the skis press into the slope.

    Ten you can tell me what the forces are in each direction - use symbols rather than numbers: it is best practice to leave the number stuff to last.

    note: sin(45)=cos(45)=1/√2
     
  7. Feb 12, 2014 #6
    This was given to use by our teacher to help us with this last problem:

    Maintain Downhill F = Downhill F (comp. wgt + comp. F poles) – Wind – Friction (uD * (comp. wgt – comp. F poles))

    The X's in the equation that I posted in post two represent the component F of the poles. I have a free body diagram drawn with a right triangle, but all I have in the triangle at this point is the angle. I cant figure out what goes in the opposite side of the triangle. Im guessing it is the normal force of the skiier (765.18N)
     
  8. Feb 12, 2014 #7

    BvU

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    I just love the
    as if you aimed a telescope at your classmate at work across the street. Or was it genuine inspiration? In that case you know what goes in the slots...

    Anyway, can I help you guys with a picture from my old album ? Very dusty. I already drew in what I think you know anyway. My mg is and has been a little more than 78 kg x g, so I left in the expression instead of writing 1000. So much more meaningful too. The others I left a little vague on purpose. But I did merge two already.

    And now a commercial (well meant): by the time you can write the quoted chunk of algebra, you can also write down the relevant equation. Must be. So do that, please!

    To boot, a tip (well meant 2): The exercise comes in segments. Your problem solving skills by now should be such that you know how to sort out big problems into small problems. Collect what you need for a), b) etc. Edit (complete or reduce) on the go. Same for the equations in 2) and the attempt in 3). Learn to do this and benefit the rest of your life.

    And as a PS: if you are tired and in a hurry, don't go to PF but go to bed. Being fresh in the morning is more important than wasting energy fighting such an exercise.
     

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  9. Feb 12, 2014 #8

    BvU

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    Furthermore, if you are done with a) and b), you can save time (yours and ours) and stress (yours) by showing your work. If that goes smoothly, adequate help for c) is quite different than when a) and b) landed in a heap of snow!
     
  10. Feb 12, 2014 #9

    Simon Bridge

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    It only helps if you understand it.

    BvU has provided a starting point.
    If the red F is the applied force from the poles, you need to ask if it is in the right direction.
    He also forgot the friction force.

    I gave you a list of forces to go on the diagram in my last post and suggestions for how to go about getting the components. Please follow suggestions.
     
  11. Feb 12, 2014 #10

    BvU

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    @Simon:
    So friction and wind were supposed to go in the red arrow. I was still in a) and b) without realizing the numbers he gave under attempt: were the right solutions. Cougar was in the 45° pole stuff already. In the next picture in my album I use the poles too ...;-)
     
  12. Feb 12, 2014 #11

    Simon Bridge

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    Oh I think I left out the wind force in my list.
     
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