Explanation Given information: The specific weight of the aluminum is 0.100 lb / in 3 . Draw the diagram for the component of the body. Figure-(1) Draw the diagram to show the sections. Figure-(2) Write the expression for the mass of the component 1. m 1 = γ g V 1 ..... (I) Here, the specific weight of the aluminum is γ , the acceleration due to gravity is g and the volume of the component 1 is V 1 . Write the expression for the mass of the component 2. m 2 = γ g V 2 ..... (II) Here, the volume of the component 2 is V 2 . Write the expression for the mass of the component 3. m 3 = γ g V 3 ..... (III) Here, the volume of the component 3 is V 3 . Write the expression for the mass of the component 4. m 4 = γ g V 4 ..... (IV) Here, the volume of the component 4 is V 4 . Write the expression for the volume of the component 1. V 1 = ( l 1 ) ( b 1 ) ( h ) ..... (V) Here, the length of the component 1 is l 1 , the width of the component 1 is b 1 and the height of the component is h . Write the expression for the volume of the component 2. V 2 = ( l 2 ) ( b 2 ) ( h ) ..... (VI) Here, the length of the component 2 is l 2 and the width of the component 2 is b 2 . Write the expression for the volume of the component 3. V 3 = π 2 r 2 ( h ) ..... (VII) Here, the radius of the component 3 is r 3 . Write the expression for the volume of the component 4. V 4 = π r 2 ( h ) ..... (VIII) Here, the radius of the component 4 is r 4 . The centroidal products of inertia, I ¯ x ′ y ′ , I ¯ y ′ z ′ and I ¯ z ′ x ′ of each component are zero due to symmetry. Write the expression for the centre of the gravity. x ¯ 3 = − ( l 1 + 4 r 3 3 π ) in ..... (IX) Write the expression for the products of inertia of the aluminum. I x y = I ¯ x ′ y ′ + m x ¯ y ¯ ..... (X) Write the expression for the products of inertia of the aluminum. I y z = I ¯ y ′ z ′ + m y ¯ z ¯ ..... (XI) Write the expression for the products of inertia of the aluminum. I z x = I ¯ z ′ x ′ + m z ¯ x ¯ ..... (XII) Calculation: Substitute 7.8 in for l 1 , 1.6 in for b 1 and 0.8 in for h in Equation (V). V 1 = ( 7.8 in ) ( 1.6 in ) ( 0.8 in ) = 9.984 in 3 Substitute 2.4 in for l 2 , 2 in for b 2 and 0.8 in for h in Equation (VI). V 2 = ( 2.4 in ) ( 2 in ) ( 0.8 in ) = 3.84 in 3 Substitute 0.8 in for h and 0.8 in for r 3 in Equation (VII). V 3 = π 2 ( 0.8 in ) 2 ( 0.8 in ) = π 2 ( 0.64 in 2 ) ( 0.8 in ) = 0.8042 in 3 Substitute 0.6 in for h and 0.55 in for r 4 in Equation (VIII). V 4 = π ( 0.55 in ) 2 ( 0.6 in ) = π ( 0.3025 in 2 ) ( 0.6 in ) = 0.5701 in 3 Substitute 9.984 in 3 for V 1 , 0.1 lb / in 3 for γ and 32.2 ft / s 2 g in Equation (I). m 1 = 0.1 lb / in 3 ( 9.984 in 3 ) 32.2 ft / s 2 = ( 0.9984 lb ) 32.2 ft / s 2 = 0.03100 lb ⋅ s 2 / ft Substitute 3.84 in 3 for V 2 , 0.1 lb / in 3 for γ and 32.2 ft / s 2 g in Equation (II). m 2 = 0.1 lb / in 3 ( 3.84 in 3 ) 32.2 ft / s 2 = ( 0.384 lb ) 32.2 ft / s 2 = 0.01192 lb ⋅ s 2 / ft Substitute 0.8042 in 3 for V 3 , 0.1 lb / in 3 for γ and 32.2 ft / s 2 g in Equation (III). m 3 = 0.1 lb / in 3 ( 0.8042 in 3 ) 32.2 ft / s 2 = ( 0

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Moment of Inertia

Inertia can be termed as a resistance offered by a body to change its state. A rigid body initially at rest will apply resistance when acted by an external force that tends to change its state of rest or velocity, or a body initially at motion will provi…

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Chapter B, Problem B.42P

To determine

The mass product of the inertia Ixy.

The mass product of the inertia Iyz.

The mass product of the inertia Izx.

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Chapter B, Problem B.41P

Chapter B, Problem B.43P

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