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Moving Wedges watch

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    A wedge of mass (6kg) and triangular cross-section, with angles of 30°, 60° and 90°, is at rest on a smooth horizontal plane with its inclined face at 30° to the plane. A mass of (3kg) is placed gently onto the rough inclined face of the wedge and allowed to move freely down a line of greatest slope of the face. The coefficient of friction between the wedge and the mass is (mu = 1/(2(root3)).

    By considering the horizontal equation of motion of the wedge and that of the mass perpendicular to the face of the wedge only, show that the acceleration of the wedge is approximately 1 m/s^2.
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    Finally got it!

    No time to write it up properly - do that tonight - but here's a summary.

    Work from the attached figure - but ignore Fig2

    The force appplied against the plane (by the weight of the mass) = mgcos30 = 3g(√3/2) = 3√3.g/2
    Horizontal component of this is (3√3.g/2)cos60 =3√3.g/2.(1/2) = (3√3/4).g, to the right.

    Friction is a force, acting up the plane, by the wedge against the mass.
    i.e. the mass provides a force, numerically equal to Fr, acting down the plane, against the plane.
    Horizontal component of this is Fr.cos30 = (3√3/8).g, to the left.

    accelerating force on the wedge is (3√3/4).g, to the right minus (3√3/8).g, to the left.
    F = (3√3/4).g - (3√3/8).g = (3√3/8).g
    F = Ma
    (3√3/8).g = 6a
    a = (3√3/48).g
    a = 1.06
    a = 1 m/s² (approx)
    =======
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    The mass and wedge are arranged as shown in Fig1 with the forces acting.
    There are two forces acting on the wedge.

    The 1st force, normal to the plane, is the resolved component of the mass perpindicular to the plane and equals mgcos30.(m=3)
    The horizontal component of this force, acting to the right is F1 = mgcos30.cos60, as shown in Fig2.

    The 2nd force comes form the friction.
    Friction is the force, acting up the plane, by the wedge against the mass.
    R is the reaction, opposite and equal to Fr, acting down the plane, by the mass against the wedge. See Fig3.
    The horizontal component of this force, acting to the left is F2 = Rcos30, as shown in Fig4

    The effective force, to the right, on the wedge is F1 - F2 = mgcos30.cos60 - Rcos30.

    F1 = mg.cos30cos60 = 3g√3/2.½ = 3g√3/4
    F1 = 3√3.g/4
    ==========
    NR = mg.cos30 = 3g√3/2 = 3√3.g/2
    Fr = µ.NR = 1/(2√3)*3√3.g/2
    Fr = (3/4)g
    ========
    R = Fr = (3/4)g
    F2 = Rcos30 = (3/4)g.√3/2
    F2 = 3√3.g/8
    ==========

    Accelerating force, F = F1 - F2
    F = 3√3.g/4 - 3√3.g/8
    F = 3√3.g/8
    F = Ma (M=6)
    3√3.g/8 = 6a
    a = 3√3.g/48
    a = 1.06
    a = 1 m/s² (approx)
    ========
    Attached Images
     
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    Fermat are you really as old as you claim to be in your profile? :rolleyes:
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    (Original post by Vhisko)
    Fermat are you really as old as you claim to be in your profile? :rolleyes:
    yep
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    (Original post by Fermat)
    yep
    At what age did you teach yourself Mechanics?
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    (Original post by Fermat)
    yep
    truly are a mechanics king.
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    (Original post by Widowmaker)
    truly are a mechanics king.
    Widowmaker your no less :rolleyes:

    I've seen many times you solving retrospec Mechanics problems

    As for me ... I'm in the process of starting the module.
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    (Original post by Vhisko)
    At what age did you teach yourself Mechanics?
    Well, mechanics learned at high school, university, (usual ages) and at odd times thereafter.
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    (Original post by Vhisko)
    Widowmaker your no less :rolleyes:

    I've seen many times you solving retrospec Mechanics problems
    ....
    ditto
 
 
 
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