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REALLY Need Help Solving Inclined Plane Question

  1. Oct 14, 2005 #1
    I posted this question once and am still confused as to how to solve, I have spent much time going at it from different directions....It can't be as hard as I am making it out to be. The question is:

    "A 60kg skier is in a tuck and moving straight down a 30 degree slope. Air resistance pushes backward on the skier with a force of 10 Newtons. The coefficient of dynamic friction between the skis and the snow is 0.08. What is the resultant force that actson the skier?"

    I get very confused trying to set it up and I am confused as to how to factor in the normal friction. I am not looking for the ANSWERS, but I am looking for the steps to take to get towards them.

    This is what I've set up, I'm just not sure if it is rigt or if I'm heading in the right direction....:

    1) Drew a right angle triangle with the hypoteneuse at angle of 30 degrees (so it looks like the slope).

    2a) Placed skier on slope (hypoteneuse) drew a line straight through skier with the force of gravity acting on him (60)(9.8) = 588N.

    2b) Drew a vector acting parallel to slope against the skier with a force of 10 N

    3) Then I solved for the vertical and horizontal side of the triangle and got:

    V = sin 30*10 = 5 N
    H = cos 30*10 = 8.7 N

    4) Then I calculated normal force: Fn = (.08)(588) = 47.04 N

    5) I don't know what to do from here. I am very much new to this and am being rushed through the class, but I can pick up things very easily. Please help!
     
  2. jcsd
  3. Oct 14, 2005 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Start by identifying all the forces acting on the skier and drawing them on a diagram. (I count 4 forces.)
    OK. But that's only useful as a diagram of the situation.
    This is the skier's weight, which acts straight down. You'll find it useful to find its components parallel and perpendicuar to the slope.
    That represents the air resistance.
    It looks like you are trying to find the vertical and horizontal components of the 10 N air resistance force: but since that force is parallel to the slope, you'll be better off using components parallel and perpendicular to the slope.
    Do this more accurately! (Find the components of the weight parallel and perpendicular to the slope. To use your triangle method, you'll need to draw a right triangle with the weight as the hypotenuse.)
    Identify all the forces. You mention the weight, air resistance, and the normal force. There's one more you left out. Then find the components of those forces parallel and perpendicular to the slope.

    Hint: You know that the skier cannot accelerate in the direction perpendicular to the slope, so what does that tell you about the net force in that direction?
     
  4. Oct 14, 2005 #3
    Thank You

    Did the question again, got resultant force of 243.3 N moving the skier down the slope. Checked my answers...they're right. Thank you. I broke the force of gravity into its parallel and perpendicular parts (to the slope). From the perpendicular component I obtained my normal force, used that to calculate Force of friction, which I added to the wind resistance... etc. I think you get the idea, thanks for your help, your instructions and hint helped. :biggrin: :biggrin:
     
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