1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: What is the horizontal component of the force pushing him forward?

  1. Jan 2, 2010 #1
    1. The problem statement, all variables and given/known data

    An 80 kilogram skier slides on waxed skis along a horizontal surface of snow at constant veloctiy while pushing with his poles. what is the horizontal component of the force pushing him forward?

    2. Relevant equations[/]

    is there an equation?

    3. The attempt at a solution

    i only know the mass how can i solve this problem?

    this was mutiple choice

    1) 0.05 N
    2) 0.04 N
    3) 40 N
    4) 4N
     
    Last edited: Jan 3, 2010
  2. jcsd
  3. Jan 2, 2010 #2
    I think it's a trick question, seeing as he is skating (so friction is negligible) and moving at a constant velocity (no acceleration) he isn't exerting any more force, which seems to work conceptually, but the problem states he is actively moving the poles so he obviously is exerting a force so maybe it is a trick question...
     
  4. Jan 3, 2010 #3
    oh but is there a way to solve this problem?
     
  5. Jan 3, 2010 #4

    Doc Al

    User Avatar

    Staff: Mentor

    I suspect that you are given the coefficient of friction for waxed skis on snow. You'll need it.
     
  6. Jan 3, 2010 #5
    Neglecting friction: F = m A. since acceleration is zero, the force is zero.

    With friction this is a far more complex question than we are given data for. We would need to know not only the coefficient of friction for the skis, but also data on the poles such as the angle at which they push off the ground since this would lighten the force on the skis. Ideally this would be related as a periodic function of time.

    I suspect the trivial answer is the one you're looking for.
     
  7. Jan 3, 2010 #6

    Doc Al

    User Avatar

    Staff: Mentor

    You are correct that a realistic answer would depend on the how the force of the poles is exerted and would be periodic. But this is meant as a simple problem, so just pretend that the poles only push backwards with no vertical component and that the force is continuously applied. Then the problem is easily solved (albeit naively).
     
  8. Jan 3, 2010 #7
    im confused so i can use the equation F = M A to solve it? how can i calculate the answer?
     
  9. Jan 3, 2010 #8

    Doc Al

    User Avatar

    Staff: Mentor

    What forces act on the skier? What's the net force on the skier?

    Do you understand how to calculate the friction force? (You need the coefficient of friction.)
     
  10. Jan 3, 2010 #9
    yeh u = ff/fn im not sure what the net force on the skier is
     
  11. Jan 3, 2010 #10

    Doc Al

    User Avatar

    Staff: Mentor

    What's the acceleration of the skier?
     
  12. Jan 3, 2010 #11
    the acceleration is at constant so its zero
     
  13. Jan 3, 2010 #12

    Doc Al

    User Avatar

    Staff: Mentor

    The velocity is constant thus the acceleration is zero. So what does that tell you about the net force?
     
  14. Jan 3, 2010 #13
    the net force is zero?
     
  15. Jan 3, 2010 #14
    The forces on the skier include a gravitational force pushing down and a normal force opposing it. This normal force causes friction.
     
  16. Jan 3, 2010 #15
    The friction force and the force of the poles net to zero.
     
  17. Jan 3, 2010 #16

    Doc Al

    User Avatar

    Staff: Mentor

    Good. Now answer my other question: What forces act on the skier?
     
  18. Jan 3, 2010 #17
    theres a normal force?
     
    Last edited: Jan 3, 2010
  19. Jan 3, 2010 #18
    The force of the snow pressing up on the skis. Otherwise the skier would fall to the center of the Earth.
     
  20. Jan 3, 2010 #19
    how can i find out the normal force of the skier?
     
  21. Jan 3, 2010 #20
    Unless something is being partially lifted (which doesn't apply to your problem) then normal force = force of gravity (mg)
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook