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Dynamics Question

  1. Jan 23, 2008 #1
    [SOLVED] Dynamics Question

    1. The problem statement, all variables and given/known data
    A skier on a slope inclined at 4.7(degrees) to the horizontal pushes on ski poles and starts down the slope. The initial speed is 2.7m/s. The coefficient of kinetic friction between skis and snow is 0.11. Determine how far the skier will slide before coming to rest.


    2. Relevant equations
    Fnet=ma
    Fg=mg
    v2=v1+2ad
    Ff(force of friction)=uk(coefficiant of kinetic friction)xFn(normal force)

    3. The attempt at a solution
    Well after drawing a free body diagram, I concluded that Fnet=fg+fappcos4.7-(force of friction) after subbing in:
    ma=mg+fappcos4.7-uk(Fn)
    mass cancels out so:
    a=g+fappcos4.7-uk(Fn)

    but now I have no fapp value. I also tried solving it without a Fapp and subbing my acceleration answer into the relevant equation to get the distance, but my answer was wayyyyy of the right answer, which is 13m

    I wish I could explain more or show that I tired more, but these type of questions always confuse me, I've yet to solve one.
     
  2. jcsd
  3. Jan 23, 2008 #2

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    Resolve the forces along and normal to the slope. Along the slope, what is the component of gravity?

    (What are you denoting by Fapp?)
     
  4. Jan 23, 2008 #3

    Doc Al

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    Staff: Mentor

    There's no additional "applied" force. The only forces acting on the skier are: gravity, friction, and the normal force.

    What's the normal force equal to? What's the component of gravity acting down the slope?
     
  5. Jan 23, 2008 #4
    Fapp is meant to denote the force applied.

    Normally, I'd say that the normal force would equal the force of gravity, but since its at an angle the normal force equal something else.

    Maybe this will help: [​IMG]
    This is the solution I found, but I looked at it and I don't understand much of anything they're doing. Maybe explaining the steps or how the formula is initially set up will be easier to help me with then figuring the whole question from scratch. Thanks for the time and help!
     
  6. Jan 23, 2008 #5

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    It says exactly the same thing as we do. By x, they mean along and down the plane, and y is normal to the plane.

    Which portion are you not able to understand ?
     
  7. Jan 23, 2008 #6
    ^^ Well, to start, I don't really understand how they set up the formula for x and the formula for y. What does it mean in English? I also don't understand why they put mgsin(angle), I was taught that its always fappcos(angle).
     
  8. Jan 23, 2008 #7

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    Do you know how to resolve, i.e., break up a force into two mutually perpendicular directions?
     
  9. Jan 23, 2008 #8
    I'm not sure what you mean by that, I think I know it by another name. What I have been taught is too draw a free body diagram, then draw all the forces acting on the object. So for this question, the FBD would look like this: [​IMG]
    Hope this is what you mean.
     
  10. Jan 23, 2008 #9

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    Right. Now notice that two of the forces are along the plane, and one is normal to the plane, but the mg is pointing down. We want that force of gravity to be broken up so that the components are either along or normal to the plane. Then we can add up all the forces directly as x or y forces.

    The inclination theta of the plane is given. Can you do what I said?
     
  11. Jan 23, 2008 #10
    Ok, I see what your saying. So mg needs to be broken up into x an y. Now how would you do this? Would you say that the x component is mgsin4(degrees) and the y is mgcos4(degrees). My reasoning for putting the sine and cos where they are is that I remember the teacher saying that cos always goes to the angle opposite of...something..I forgot the exact word. Is this what you're trying to show me? Am I on the right path?
     
  12. Jan 23, 2008 #11

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    That's good enough. (Stop writing the whole sin4(degrees); I'll use 'b' for the angle.)

    Now write down the x and y forces and sum them. Refer to the solution you had attached. If stuck, get back here.

    You have to know that force of friction is kN. Think for some time.
     
  13. Jan 23, 2008 #12
    ALright, thanks a lot for the patience and help. If I do have more questions, I'll post back. Thanks.
     
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