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Homework Help: Conceptual difficulty - Work

  1. Sep 13, 2010 #1
    1. The problem statement, all variables and given/known data

    I am worried that I don't understand a basic part of figuring out the component forces in the following problem. I have a full worked example but there is a few steps which I don't understand why we use sin for the x component and not cos (understand why I am really worried as it appears to be basic trig :/)

    A 58-kg skier is coasting down a 25° slope, as Figure 6.7a shows. Near the top of the slope, her speed is 3.6 m/s. She accelerates down the slope because of the gravitational force, even though a kinetic frictional force of magnitude 71 N opposes her motion. Ignoring air resistance, determine the speed at a point that is displaced 57 m downhill.


    2. Relevant equations
    vf = √2(KEf) / m
    = √(2(1/2 mv02 + Sigma F cos theta s)/m
    = √(2(1/2 mv02 + mg sin 25 - fk s) /m
    = √(2(1/2 mv02 + 170N cos 0 x 57) / 58
    = 19m/s

    3. The attempt at a solution

    This was in my textbook:

    a free-body diagram for the skier and shows the three external forces acting on her: the gravitational force , the kinetic frictional force , and the normal force . The net external force along the y axis is zero, because there is no acceleration in that direction (the normal force balances the component mg cos 25° of the weight perpendicular to the slope). Using the data from the table of knowns and unknowns, we find that the net external force along the x axis is:

    SigmaF = mg sin 25 - fk
    = (58)(9.8)(sin 25)(71)
    = 170N


    I look at sin and think we are looking at y component. I can't see why we would used sin for x component - can someone explain?

    Many thanks

    H
    x

    ps. please forgive formating - each time I use latex it just puts large gaps in the place of symbols :)
     
  2. jcsd
  3. Sep 13, 2010 #2
    AH! My free body diagram was incorrect - all good don't need a reply. thanks anyway those who read through my problem. Sorry for inconvenience.
     
  4. Sep 13, 2010 #3
    Friction is a nonconservative force, so you can't use conservation of energy.

    Sin is used for the x component because you have to draw a similar triangle for the normal force of the skiier. The angle between the force opposite the normal force and gravity is theta. Then trig will tell you that the x component is mgsin(theta).
     
  5. Sep 14, 2010 #4
    Thank you very much! This really helped me consolidate my understanding :)
     
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