solución primer punto parcial dinámica.pdf

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  • 7/28/2019 Solucin primer punto parcial dinmica.pdf

    1/2

    COSMOS: Complete Online Solutions Manual Organization System

    Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

    Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

    CChhaapptteerr1155,,SSoolluuttiioonn223388..

    Geometry. ( ) ( )120 mm 180 mmA = +r j k

    Method 1

    Let the rigid bodyDCB be a rotating frame of reference.

    Its angular velocity is ( ) ( )1 2 1.2 rad/s 1.5 rad/s . = + = i k i k

    Its angular acceleration is ( )21 2 1 2 1.8 rad/s . = = =i k j j

    Motion of the coinciding pointA in the frame.

    ( ) ( )

    ( ) ( ) ( )

    1.2 1.5 120 180

    144 216 180 180 mm/s 216 mm/s 144 mm/s

    A A = = +

    = + = +

    v r i k j k

    k j i i j k

    ( ) ( )2 2

    0 1.8 0 1.2 0 1.5

    0 120 180 180 216 144

    324 324 442.8 259.2 442.8 mm/s 259.2 mm/s

    A A A = +

    = +

    +

    = + =

    a r v

    i j k i j k

    i i j k j k

    Motion of point A relative to the frame.

    ( )/ /60 mm/s , 0A F A Fu= = =v j j a

    Velocity of point A. /A A A F= +v v v

    180 216 144 60A

    = + +v i j k j

    ( ) ( ) ( )180.0 mm/s 156.0 mm/s 144.0 mm/sA = +v i j k Coriolis acceleration. /2 A F v

    ( )( ) ( ) ( )2 2/2 2 1.2 1.5 60 180 mm/s 144 mm/sA F = = +v i k j i k

    Acceleration of point A. / /2A A A F A F= + + a a a v

    442.8 259.2 180 144A

    = + +a j k i k

    ( ) ( ) ( )2 2 2180 mm/s 443 mm/s 115.2 mm/sA = a i j k

  • 7/28/2019 Solucin primer punto parcial dinmica.pdf

    2/2

    COSMOS: Complete Online Solutions Manual Organization System

    Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

    Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

    Method 2

    Let frame Dxyz, which at instant shown coincides with DXYZ, rotate with an angular velocity

    11.2 rad/s.= =i i Then the motion relative to the frame consists of the rotation of body DCB

    about the z axis with angular velocity ( )2 1.5 rad/s = k kplus the sliding motion ( )60 mmu= =u i j of the rodAB relative to the bodyDCB.

    Motion of the coinciding point in the frame.

    ( ) ( ) ( )

    ( ) ( ) ( )2 21.2 120 180 216 mm/s 144 mm/s

    1.2 216 144 172.8 mm/s 259.2 mm/s

    A A

    A A

    = = + = +

    = = + =

    v r i j k j k

    a v i j k j k

    Motion of point A relative to the frame.

    ( ) ( ) ( ) ( )

    ( ) ( )

    ( ) ( ) ( ) ( ) ( )

    ( ) ( )

    / 2

    / 2 2 2 2

    2 2

    1.5 120 180 60 180 mm/s 60 mm/s

    2

    0 1.5 180 0 2 1.5 60

    270 180 180 mm/s 270 mm/s

    A F A

    A F A A

    u

    u u

    = + = + + = +

    = + + +

    = + + +

    = + =

    v k r j k j k j i j

    a k r k k r j k j

    k i k j

    j i i j

    &

    Velocity of point A./A A A F= +v v v

    216 144 180 60A

    = + +v j k i j

    ( ) ( ) ( )180.0 mm/s 156.0 mm/s 144.0 mm/sA = +v i j k Coriolis acceleration. /2 F v

    ( )( ) ( ) ( )2

    /2 2 1.2 180 60 144 mm/s

    A F = + =v i i j k

    Acceleration of point A./ /

    2A A A F A F= + + a a a v

    172.8 259.2 180 270 144A

    = + +a j k i j k

    ( ) ( ) ( )2 2 2180.0 mm/s 443 m m/s 115.2 mm/sA = a i j k