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Homework Help: Find the frictional force on the skis

  1. Jun 4, 2010 #1
    1. The problem statement, all variables and given/known data
    A skier of mass 57 kg skis down a 10° slope @ constant velocity.
    The force of air resist. has a magnitude of 74 N. Find the frictional force on the skis.


    2. Relevant equations

    F=ma
    F(fr) = (mu)(Fn)


    3. The attempt at a solution

    I found Fn= ma
    where is 558N
    so far i tried, 74sin10- 558
    and i tried drawing a body diagram, where 558 is directly upwards, 9.8 down for gravity, the downward slope of 74 N at 10 deg
     
  2. jcsd
  3. Jun 4, 2010 #2

    berkeman

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    Staff: Mentor

    Hint -- sum the forces in the direction of the motion of the skier (along the direction of the slope). What do those forces need to sum to?
     
  4. Jun 4, 2010 #3
    it would be the sum of normal force, and air resistance?
    so it should be 74 +558 ?
     
  5. Jun 4, 2010 #4

    berkeman

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    Staff: Mentor

    The normal force is orthogonal to the slope, not parallel to it. Which way does the air resistance vector point? The friction force vector? What force vector opposes those two vectors?
     
  6. Jun 4, 2010 #5
    air resistance is opposite to the velocity, and friction force should be the same,

    gravitainal force opposes it but i dont think directly bc its is not parallel to the slope.
     
  7. Jun 4, 2010 #6

    berkeman

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    Staff: Mentor

    But one component of the gravitational force is parallel to the slope... Is it the sin or cos component? Remember that if the slope is zero degrees, the gravitational component of force paralle to the slope is zero...
     
  8. Jun 4, 2010 #7
    it is sin, (lookin for the Y component of the slope)

    which i had (74 sin 10),
     
  9. Jun 4, 2010 #8

    berkeman

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    Staff: Mentor

    74 N is the force from air resistance, not gravity.

    Write the sum of the forces in the direction parallel to the slope. What do you want to set that sum to equal, and why?
     
  10. Jun 4, 2010 #9
    umm im lost now.... maybe i drew it wrong...

    the air resistance is the magnitude isnt it?

    iono.. start over?
     
  11. Jun 4, 2010 #10

    berkeman

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    Staff: Mentor

    Describe (or post) your FBD. What force vectors act on the skier?
     
  12. Jun 4, 2010 #11
    what is FBD? , the forces affecting the skier is air resistance, *frictional force, gravity, velocity
     
  13. Jun 4, 2010 #12

    berkeman

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    Staff: Mentor

    FBD = free body diagram. You alluded to it in your first post.

    And velocity is not a force...
     
  14. Jun 4, 2010 #13
    down is 9.8 (gravity,) down slope is 74 in mag. the slope from the X axis is 10 deg from it.the Y of the triangle i believe is (74 sin 10) and perpendicular to the slope i hav upward is 558 N
     
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