Explanation Given information: The weight ( W ) of the gear A is 2 lb. The length ( L ) of the axle AD is 20 inch. The radius ( r ) of gear A is 4 inch. The angular velocity ( ω 1 ) of shaft DE is 4 rad/s. The angular velocity of axle AD is ( ω 2 ) . Calculation: Let point D be the common point of rotation ω 1 and ω 2 and consider principal axes x , y , and z with origin at point O . Sketch the axes of the system as shown in Figure (1). Refer Figure (1), Write the vector equation of angular velocity ( ω 1 ) of shaft DE using kinematics: ω 1 = ω 1 sin β i + ω 1 cos β j Write the vector equation of angular velocity ( ω 2 ) of axle AD using kinematics: ω 2 = ω 2 j Find the vector equation of angular velocity ( ω ) of system using kinematics: ω = ω 1 + ω 2 Substitute ω 1 sin β i + ω 1 cos β j for ω 1 and ω 2 j for ω 2 . ω = ω 1 sin β i + ω 1 cos β j + ω 2 j = ω 1 sin β i + ( ω 1 cos β + ω 2 ) j (1) Write the magnitude of x -component of angular velocity ( ω x ) from equation (1). ω x = ω 1 sin β Write the magnitude of y -component of angular velocity ( ω y ) from equation (1). ω y = ω 1 cos β + ω 2 Write the equation of length ( r G / D ) of axle AD corresponding to point D . r G / D = L sin β i Find the equation of velocity ( v G ) of the system at mass center G . v G = ω 1 × r G / D Substitute L sin β i for r G / D . v G = ω 1 j × L sin β i = − ω 1 L sin β k Find the equation of acceleration ( a G ) of the system at mass center G . a G = ω 1 × v G Substitute − ω 1 L sin β for v G a G = ω 1 × ( − ω 1 L sin β ) = − ω 1 2 L sin β = ω 1 2 L sin β ( ← ) The velocity at C is zero. Find the magnitude of y -component of angular velocity ( ω y ) in terms of ω 1 using the equation: v C = v G + ω × r C / G Here, r C / G the distance between mass center and point C . Substitute 0 for v C , − ω 1 L sin β k for v G , ( ω x i + ω y j ) for ω , and − r i for r C / G . 0 = − ω 1 L sin β k + ( ω x i + ω y j ) × ( − r i ) 0 = − ω 1 L sin β k + 0 + ω y r k ω y r k = ω 1 L sin β k Equate the k -vector coefficients. ω y r = ω 1 L sin β ω y = ω 1 L r sin β Write the equation of angular momentum of system about G ( H G ) . H G = I ¯ x ω x i + I ¯ y ω y j Write the angular velocity of reference frame ( Ω ) . Ω = ω 1 Substitute ω 1 sin β i + ω 1 cos β j for ω 1 . Ω = ω 1 sin β i + ω 1 cos β j The rate of change of angular momentum ( H ˙ G ) G x y z about the reference frame is 0. Write the equation of rate of change of angular momentum ( H ˙ G ) of the system. H ˙ G = ( H ˙ G ) G x y z + Ω × H G Substitute 0 for ( H ˙ G ) G x y z , ω 1 sin β i + ω 1 cos β j for Ω , and I ¯ x ω x i + I ¯ y ω y j for H O . H ˙ G = 0 + ( ω 1 sin β i + ω 1 cos β j ) × ( I ¯ x ω x i + I ¯ y ω y j ) = 0 + I ¯ y ω y ω 1 sin β k − I ¯ x ω x ω 1 cos β k = ( I ¯ y ω y sin β − I ¯ x ω x cos β ) ω 1 k (2) Find the mass moment of inertia ( I ¯ x ) of the system about x -axis using the equation: I ¯ x = 1 4 m r 2 Here, m is the mass of the gear A

Vector Mechanics for Engineers: Statics and Dynamics

11th Edition

ISBN: 9780073398242

Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self

Publisher: McGraw-Hill Education

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Chapter 18.2, Problem 18.89P

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The force F exerted by gear B on gear A.

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Chapter 18.2, Problem 18.88P

Chapter 18.2, Problem 18.90P

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Vector Mechanics for Engineers: Statics and Dynamics

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The force F exerted by gear B on gear A .

Solution Summary: The author explains the force F exerted by gear B on gear A. The angular velocity of shaft DE is 4 rad/s.

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The force F exerted by gear B on gear A.

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Chapter 18 Solutions