Explanation Given: The radius of the circular plate ( r ) is 0.75 m. The diameter of the plate ( d ) is 1.5 m. The height of the tank ( h t ) is 5 m. The height of center of the plate from bottom of the tank ( h ′ ) is 2.5 m. The density of the vegetable oil ( ρ v o ) is 932 kg / m 3 . Calculation: Draw the cross section of the plate establish the x and y axis as shown in Figure (1). The differential element is the shaded area shown in Figure (1). Establish the equation for the circumference of the circular plate. x 2 + y 2 = ( d 2 ) 2 x 2 + y 2 = r 2 x 2 + y 2 = ( 0.75 ) 2 x 2 = ( 0.75 ) 2 − y 2 x = ( 0.75 ) 2 − y 2 (I) Calculate the area of the differential element ( dA ). d A = ( x + x ) d y = 2 x d y Substitute Equation (I) for x in the above equation. d A = 2 ( ( 0.75 ) 2 − y 2 ) d y (II) Calculate the depth of centroid for the circular plate from the top of the tank ( h ¯ ) . h ¯ = h t − h ′ h ¯ = 5 − 2.5 = 2.5 m Calculate the height of the differential element from top of the tank ( h ). h = h ¯ − y h = 2.5 − y Here, y is the location of differential element from the x axis. Calculate the pressure acting on the differential element ( p ). p = ρ v o g h p = 932 kg/m 3 × 9.81 m / s 2 × ( 2.5 m − y ) = 22 , 857.3 − 9 , 142.92 y Calculate the resultant force on the plate ( F R ) by integrating the relation for area. F R = ∫ − 0.75 m 0.75 m p d A Substitute p = ( 22 , 857.3 − 9 , 142.92 y ) and d A = 2 ( ( 0.75 ) 2 − y 2 ) d y in the above equation. F R = ∫ − 0.75 m 0.75 m ( 22 , 857.3 − 9 , 142.92 y ) × 2 ( ( 0.75 ) 2 − y 2 ) d y = 18 , 285.84 ∫ − 0.75 0.75 ( 2.5 − y ) ( ( 0.75 ) 2 − y 2 ) d y Multiply and divide by 8. F R = 18 , 285.84 ∫ − 0.75 0.75 ( 5 − 2 y ) ( − 16 y 2 + 9 ) 8 d y = − 2 , 285.73 ∫ − 0.75 0.75 ( 2 y − 5 ) ( − 16 y 2 + 9 ) d y = − 2 , 285.73 [ ∫ − 0.75 0.75 2 y ( − 16 y 2 + 9 ) d y − 5 ∫ − 0.75 0.75 ( − 16 y 2 + 9 ) d y ] (III) Solve first part of Equation (I); P ( I ) = ∫ y ( − 16 y 2 + 9 ) d y (IV) Use substitution method. u = − 16 y 2 + 9 d u = − 32 y d y y = ( − 1 32 d u d y ) Substitute ( − 1 32 d u d y ) in Equation (II). P ( I ) = − 1 32 ∫ ( u ) d y = − 2 32 ( u 3 2 3 ) = − 1 48 u 3 2 Use back substitution. Substitute ( − 16 y 2 + 9 ) for u . P ( I ) = − 1 48 ( − 16 y 2 + 9 ) 3 2 (V) Solve second part of Equation (I); P ( I I ) = ∫ ( − 16 y 2 + 9 ) d y (VI) Use substitution method. y = 3 sin u 4 u = sin − 1 4 y 3 d y = 3 cos u 4 d u d y d u = 3 cos u 4 Substitute ( 3 sin u 4 ) for y in Equation (III). P ( I I ) = ∫ ( − 16 ( 3 sin u 4 ) 2 + 9 ) ( 3 cos u 4 ) d u = ∫ − 9 sin 2 u + 9 ( 3 cos u 4 ) d u = ∫ 9 cos 2 u 4 d u = 9 4 ∫ cos 2 u d u (VII) Solve Equation (IV). P ( I I ) = 9 4 ∫ cos 2 u d u = 9 4 ( 1 2 ∫ ( cos 0 u ) d u + cos u sin u 2 ) = 9 8 u + 9 cos u sin u 8 Use back substitution. Substitute u = ( sin − 1 4 y 3 ) in the above equation. P ( I I ) = 9 8 ( sin − 1 4 y 3 ) + 9 cos ( sin − 1 4 y 3 ) sin ( sin − 1 4 y 3 ) 8 = 9 8 ( sin − 1 4 y 3 ) + 9 8 ( 4 y 3 ) 1 − 16 y 2 9 = 9 8 ( sin − 1 4 y 3 ) + 3 y 2 1 − 16 y 2 9 (VIII) Substitute Equation (V) and Equation (VIII) in Equation (III). F R = − 2 , 285.73 [ ∫ − 0.75 0.75 2 y ( − 16 y 2 + 9 ) d y − 5 ∫ − 0.75 0.75 ( − 16 y 2 + 9 ) d y ] = − 2 , 285.73 [ 2 ( − 1 48 ( − 16 y 2 + 9 ) 3 2 ) − 5 ( 9 8 ( sin − 1 4 y 3 ) + 3 y 2 1 − 16 y 2 9 ) ] = − 2 , 285.73 [ − 1 24 ( − 16 y 2 + 9 ) 3 2 − 45 8 ( sin − 1 4 y 3 ) − 15 y 2 1 − 16 y 2 9 ] = 2 , 285.73 [ 1 24 ( − 16 y 2 + 9 ) 3 2 + 45 8 ( sin − 1 4 y 3 ) + 15 y 2 1 − 16 y 2 9 ] − 0

2nd Edition

ISBN: 9780134649290

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 2, Problem 76P

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The resultant force acting on the circular plate and the location of the resultant force from the bottom of the tank.

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Chapter 2, Problem 75P

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Fluid Mechanics (2nd Edition)

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