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Math Help - Friction and coefficient of friction questions!

  1. #1
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    Friction and coefficient of friction questions!

    Whenever a numerical value of g is required take g=9.8 metres per second per second
    1)A body of mass 2kg is held in limiting equillibrium on a rough plane inclined at 20 degrees to the horizontal by a force of Y at an angle of 45 degrees to the horizontal. The coefficient of friction between the body and the plane is 0.2. Modelling the body as a particle find Y when the body is on the point of slipping.
    a) up the plane
    b) down the plane

    Any help would be much appreciated
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  2. #2
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    Quote Originally Posted by Confuzzled?
    Whenever a numerical value of g is required take g=9.8 metres per second per second
    1)A body of mass 2kg is held in limiting equillibrium on a rough plane inclined at 20 degrees to the horizontal by a force of Y at an angle of 45 degrees to the horizontal. The coefficient of friction between the body and the plane is 0.2. Modelling the body as a particle find Y when the body is on the point of slipping.
    a) up the plane
    b) down the plane

    Any help would be much appreciated
    Here is one way.

    Friction.
    It tends to go against the direction of motion. In the above problem, it is about static friction.
    --If the tendency of the body is to slide down the plane, the static friction will tend to go up the incline. ----------(i)
    ---If the tendency of the body is to slide up the plane, the static friction will tend to down the incline. ------------(ii)
    ---Friction depends on the forces normal to the plane. Force due to friction is (normal force) times (coefficient of friction of the plane). -----(iii)

    Weight of body is vertical force, it is m*g = 2(9.8) = 19.6 newtons.

    External forces parallel to the plane:
    Due to weight of body = (19.6)sin(20deg). It is negative because it is going down.
    Due to Y = Y*sin(45-20 deg) = Y*sin(25deg). It is positve because it is going up.
    So, resultant of forces parallel to plane = [Y*sin(25deg) -19.6sin(20deg)] --------***

    External forces normal or perpendicular to the plane:
    Due to weight of body = (19.6)cos(20deg).
    Due to Y = Y*cos(25deg).
    So, resultant of forces normal to the plane = [Y*cos(25deg) +19.6cos(20deg)] --------***

    ----------------------
    a) Now if the body tends to slip up the plane.

    The friction force tends to go down the plane, or it is negative.
    Because the body is still at rest, the sum of all forces parallel to the plane is zero.
    So,
    [Y*sin(25deg) -19.6sin(20deg)] -(0.2)[Y*cos(25deg) +19.6cos(20deg)] = 0
    Y[sin(25deg) -0.2cos(25deg)] -19.6[sin(20deg) +0.2cos(20deg)] = 0
    Y[0.241356704] -19.6[0.529958667] = 0
    Y = 19.6(0.529958667)/(0.241356704) = 43 newtons ---------answer.

    --------------------------
    b) If the body tends to slip down the plane.

    The friction force tends to go up the plane, or it is positive.
    Because the body is still at rest, the sum of all forces parallel to the plane is zero.
    So,
    [Y*sin(25deg) -19.6sin(20deg)] +(0.2)[Y*cos(25deg) +19.6cos(20deg)] = 0
    Y[sin(25deg) +0.2cos(25deg)] -19.6[sin(20deg) -0.2cos(20deg)] = 0
    Y[0.603879819] -19.6[0.154081619] = 0
    Y = 19.6(0.154081619)/(0.603879819) = 5 newtons ---------answer.
    Last edited by ticbol; March 18th 2006 at 07:02 PM. Reason: where are the "= 0"?
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