Explanation Given information: The weight per unit length of wire is w . Calculation: Sketch the brass wire as shown in Figure 1. Express the mass of each component as shown below: m = W g = 1 g w L Here, w is a weight per unit length. Find the mass of component 1 as follows: m 1 = W g ( L ) (1) Substitute ( π × 3 2 a ) for L in Equation (1). m 1 = w g ( π × 3 2 a ) = 3 2 π w g a Find the mass of component 2 as follows: m 2 = w g ( L ) (2) Substitute ( 3 a ) for L in Equation (2). m 2 = w g ( 3 a ) = 3 w g a Find the mass of component 3 as follows: m 3 = w g ( L ) (3) Substitute ( π × a ) for L in Equation (3). m 3 = w g ( ( π × a ) ) = π w g a Find the centroid section ( x 1 ) as shown below: ( x 1 ) = − 2 π ( 3 2 a ) Find the centroid section ( y 1 ) as shown below: ( y 1 ) = 2 a Find the centroid section ( z 1 ) as shown below: ( z 1 ) = 1 2 a Find the total section ( m x ¯ y ¯ ) 1 as shown below: ( m x ¯ y ¯ ) 1 = − 2 π ( 3 2 a ) ( 2 a ) ( 1 2 a ) Similarly calculate the remaining values as shown in Table 1. m x ¯ y ¯ z ¯ m x ¯ y ¯ m y ¯ z ¯ m x ¯ z ¯ 1 3 2 π w g a − 2 π ( 3 2 a ) 2 a 1 2 a − 9 w g a 3 3 2 π w g a 3 − 9 4 w g a 3 2 3 w g a 0 1 2 a 2 a 0 3 w g a 3 0 3 π w g a 2 π a a a − 2 w g a 3 π w g a 3 2 w g a 3 ∑ − 11 w g a 3 w g ( π 2 + 3 ) a 3 − 1 4 w g a 3 Find the mass products of inertia I x y of the machine component. I x y = ∑ ( I x ′ y ′ + m x ¯ y ¯ ) = ( ( I x ′ y ′ ) 1 + ( I x ′ y ′ ) 2 + ( I x ′ y ′ ) 3 + ( m x ¯ y ¯ ) ) (4) Substitute 0 for ( ( I x ′ y ′ ) 1 + ( I x ′ y ′ ) 2 + ( I x ′ y ′ ) 3 ) and − 11 w g a 3 for ∑ ( m x ¯ y ¯ ) in Equation (4)

Vector Mechanics for Engineers: Statics and Dynamics

12th Edition

ISBN: 9781259638091

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

Publisher: McGraw-Hill Education

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Chapter 9.6, Problem 9.160P

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Find the mass products of inertia Ixy,Iyz, and Izx of the wire.

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Chapter 9.6, Problem 9.159P

Chapter 9.6, Problem 9.161P

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