Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Friction on a slope

  1. Sep 15, 2010 #1
    The question is quite simple by i dnt seem to be getting it..
    When a body is placed in a slope.
    force mg is resolved into rectangular components..
    the one perpendicular to the slope is taken as cos component n the one parallel is taken as sine component.
    I don't understand whisch one to take as cos and which one as sine..generally x-axis is cos n y-axis is sine bt m unable to make out in a figure that which ione to take as x-axis n which one as y-axis??

    plzz help
  2. jcsd
  3. Sep 15, 2010 #2

    Doc Al

    User Avatar

    Staff: Mentor

    When you want the x & y components of a vector F and the vector's angle is given with respect to the x-axis, then the x-component will be Fcosθ.

    But when resolving weight into components parallel (usually called the x-axis) and perpendicular (y-axis) to the incline, realize that the angle of the incline is the angle that the weight makes with the y-axis, not the x-axis. So the trig functions end up getting reversed.

    Read this: http://www.physicsclassroom.com/Class/vectors/u3l3e.cfm" [Broken]
    Last edited by a moderator: May 4, 2017
  4. Sep 15, 2010 #3
    your F_g fully does not constrabute to friction... becasue the y- component is driving the object down.. or making it stationary
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook