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HELP: Friction on upward motion in an inclined plane

  1. May 26, 2009 #1
    Guys, I'm just bothered of my professor's answer in the problem we solved. Here are the given

    Given:
    Weight: 5000 N
    Coefficient: 0.4
    Angle of the plane: 20º

    Solve for the required FORCE to move the box (5,000N) upward to the plane.

    This is the formula I used:

    Fn=Ff

    Fn(sin theta) = (coefficient)(5000 N - normal force)(cos theta)

    Fn = (coefficient)(5000 N - normal force)(cos theta)/sin theta

    Fn = 5, 494.95 N

    Am I right or wrong?

    Thanks for the help.
     
  2. jcsd
  3. May 26, 2009 #2
    I may be wrong but i think of it this way;

    Force down the plane by gravity is 5000*sin(20)=1710.1N
    Force of friction is 5000*cos(20)*0.4=1879.4N

    A force pushing it up the plane needs to overcome both of these forces, so the force needs to be about 1710.1+1879.4=3589.5N

    I think thats right. The force you calculated is actually larger than the weight of the box, which i think is intuitively too large
     
  4. May 26, 2009 #3
    The angle of plane is given 20degree
    the force is resolved into two component that is
    1.Horizontal component
    formula
    fx=cos(angle)f1
    2.Vertical component
    fy=sin(angle)f1
    Determine the net force here is the formula
    total force=sqrt(fx^2+fy^2)
    this is the exact solution
     
  5. May 27, 2009 #4
    This is the correct answer, as per my professor. Thanks for the light.
     
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