Explanation Given information: The diameter ( d ) of aluminum wire is 1.5 mm The density ( ρ AL ) of steel is 2 , 800 kg / m 3 . Calculation: Sketch the aluminum wire as shown in Figure 1. Find the mass of the component 1 and 4 as shown below: m 1 = m 4 = ρ AL V = ρ AL ( A L ) (1) Here, ρ AL is density of aluminum, V is the volume of component 1 and 4, L is length of aluminum wire, and A is the cross sectional area of aluminum wire. Find the area of aluminum wire ( A ) as shown below: A = ( π 4 d 2 ) L (2) Substitute 1.5 mm for d and 250 mm for L in Equation (2). A = ( π 4 ( 1.5 mm × ( 1 m 10 3 mm ) ) 2 ) ( 250 mm × ( 1 m 10 3 mm ) ) = π 4 × ( 5.625 × 10 − 7 ) = 4.417 × 10 − 7 m 2 Substitute 2 , 800 kg / m 3 for ( ρ AL ) and 4.417 × 10 − 7 m 2 for A in Equation (1). m 1 = 2 , 800 × 4.417 × 10 − 7 = 1.23700 × 10 − 3 kg Find the mass of the component 2 and 5 as shown below: m 2 = m 5 = ρ AL V = ρ AL ( A L ) (3) Here, V is the volume of component 2 and 5, L is aluminum wire. Find the area of aluminum wire ( A ) as shown below: A = ( π 4 d 2 ) L (4) Substitute 1.5 mm for d and 180 mm for L in Equation (4). A = ( π 4 ( 1.5 mm × ( 1 m 10 3 mm ) ) 2 ) ( 180 mm × ( 1 m 10 3 mm ) ) = π 4 0.0015 2 × ( 0.18 ) = 3.180 × 10 − 7 m 2 Substitute 2 , 800 kg / m 3 for ( ρ AL ) and 3.180 × 10 − 7 m 2 for A in Equation (3). m 2 = m 5 = 2 , 800 × 3.180 × 10 − 7 = 0.8096 × 10 − 3 kg Find the mass of the component 3 and 6 as shown below: m 3 = m 6 = ρ AL V = ρ AL ( A L ) (5) Here, V is the volume of component 3 and 6, L is aluminum wire. Find the area of aluminum wire ( A ) as shown below: A = ( π 4 d 2 ) L (6) Substitute 1.5 mm for d and 300 mm for L in Equation (6). A = ( π 4 ( 1.5 mm × ( 1 m 10 3 mm ) ) 2 ) ( 300 mm × ( 1 m 10 3 mm ) ) = π 4 0.0015 2 × ( 0.3 ) = 5.3014 × 10 − 7 m 2 Substitute 2 , 800 kg / m 3 for ( ρ AL ) and 5.3014 × 10 − 7 m 2 for A in Equation (5). m 3 = m 6 = 2 , 800 × 5.3014 × 10 − 7 = 1.48440 × 10 − 3 kg Find the centroid of section 1 about x axis ( x 1 ) as shown below: ( x 1 ) = 180 mm = 180 mm ( 1 m 10 3 mm ) = 0.18 m Find the centroid of section 2 about x axis ( x 2 ) as shown below: ( x 2 ) = 180 2 = 90 mm ( 1 m 10 3 mm ) = 0.09 m Find the centroid of section 3 about x axis ( x 3 ) as shown below: x 3 = 0 Find the centroid of section 4 about x axis ( x 4 ) as shown below: ( x 4 ) = 0 Find the centroid of section 5 about x axis ( x 5 ) as shown below: ( x 5 ) = 180 2 = 90 mm ( 1 m 10 3 mm ) = 0.09 m Find the centroid of section 6 about x axis ( x 6 ) as shown below: ( x 6 ) = 180 = 180 mm ( 1 m 10 3 mm ) = 0.18 m Find the centroid of section 1 about y axis ( y 1 ) as shown below: ( y 1 ) = 250 2 = 125 mm ( 1 m 10 3 mm ) = 0.125 m Find the centroid of section 2 about y axis ( y 2 ) as shown below: ( y 2 ) = 250 = 250 mm ( 1 m 10 3 mm ) = 0.25 m Find the centroid of section 3 about y axis ( y 3 ) as shown below: ( y 3 ) = 250 = 250 mm ( 1 m 10 3 mm ) = 0.25 m Find the centroid of section 4 about y axis ( y 4 ) as shown below: ( y 4 ) = 250 2 = 125 mm ( 1 m 10 3 mm ) = 0.125 m Find the centroid of section 5 about y axis ( y 5 ) as shown below: ( y 5 ) = 0 Find the centroid of section 6 about y axis ( y 6 ) as shown below: ( y 6 ) = 0 Find the centroid of section 1 about z axis ( z 1 ) as shown below: ( z 1 ) = 0 Find the centroid of section 2 about z axis ( z 2 ) as shown below: ( z 2 ) = 0 Find the centroid of section 3 about z axis ( z 3 ) as shown below: ( z 3 ) = 300 2 = 150 mm ( 1 m 10 3 mm ) = 0

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 9.6, Problem 9.157P

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

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Chapter 9.6, Problem 9.156P

Chapter 9.6, Problem 9.158P

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Chapter 9 Solutions

Vector Mechanics for Engineers: Statics and Dynamics

Ch. 9.1 - 9.1 through 9.4 Determine by direct integration...Ch. 9.1 - 9.1 through 9.4 Determine by direct integration...Ch. 9.1 - Prob. 9.3P Ch. 9.1 - Prob. 9.4P Ch. 9.1 - 9.5 through 9.8 Determine by direct integration...Ch. 9.1 - 9.5 through 9.8 Determine by direct integration...Ch. 9.1 - Prob. 9.7P Ch. 9.1 - Prob. 9.8P Ch. 9.1 - 9.9 through 9.11 Determine by direct integration...Ch. 9.1 - Prob. 9.10P

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Find the mass products of inertia I x y , I y z , and I z x of the machine component.

Solution Summary: The author calculates the mass of the component 1 and 4 as shown in Figure 1.

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

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