Explanation Given: The height of the tank ( h ′ ) is 0.75. The density of water ( ρ w ) is 1 , 000 kg / m 3 Calculation: Calculate the location of differential element from the top of the tank ( h ). h = h ′ − y h = ( 0.75 − y ) Here, the height of the differential element from the bottom is y . Establish the equation for the elliptical tank. y 2 0.75 2 + x 2 = 1 x 2 = 1 − y 2 0.75 2 x = 1 − y 2 0.75 2 Calculate the area of the differential element ( dA ). d A = ( x + x ) d y Substitute Equation (1) for x in the above equation. d A = ( 1 − y 2 0.75 2 + 1 − y 2 0.75 2 ) d y d A = 2 1 − y 2 0.75 2 d y Calculate the pressure acting on the elliptical plate ( p ). p = ρ w g h p = 1 , 000 kg m 3 × 9.81 m/s 2 × ( 0.75 − y ) p = 9 , 810 ( 0.75 − y ) Calculate the resultant force acting on the elliptical plate ( F R ) . d F = p d A Substitute p = 9 , 810 ( 0.75 − y ) and d A = 2 1 − y 2 0.75 2 d y in the above equation. d F = 9 , 810 ( 0.75 − y ) ( 2 1 − y 2 0.75 2 d y ) = 19 , 620 ( 0.75 − y ) ( 1 − y 2 0.75 2 d y ) Integrate both sides of the equation. ∫ d F = ∫ 19 , 620 ( 0.75 − y ) ( 1 − y 2 0.75 2 d y ) F R = 19 , 620 { 0.75 ∫ 1 − y 2 0.75 2 d y − ∫ y 1 − y 2 0.75 2 d y } = 19 , 620 { ∫ 0.75 2 − y 2 d y − 1 0.75 ∫ y 0.75 2 − y 2 d y } Apply the boundary limit between –0.75 m to 0.75 m. F R = 19 , 620 { ∫ − 0.75 0.75 0.75 2 − y 2 d y − 1 0.75 ∫ − 0.75 0.75 y 0.75 2 − y 2 d y } Solve the integral. P ( I I ) = ∫ y 0.75 2 − y 2 d y (I) Multiply and divide by 8. P ( I I ) = 1 4 ∫ y 9 − 16 y 2 d y Use substitution method. u = − 16 y 2 + 9 d u = − 32 y d y y = ( − 1 32 d u d y ) Substitute y = ( − 1 32 d u d y ) in equation (I). P ( I 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 I ) = 1 4 ( − 1 48 ( − 16 y 2 + 9 ) 3 2 ) (II) Solve the integral. P ( I ) = ∫ 0.75 2 − y 2 d y (III) Multiply and divide by 8. P ( I ) = 1 4 ∫ 9 − 16 y 2 d y 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 y = ( 3 sin u 4 ) and d y = 3 cos u 4 d u in equation (III) P ( 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 (IV) Solve Equation (IV). P ( 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 ( sin − 1 4 y 3 ) for u in the above equation. P ( I ) = 1 4 { 9 8 ( sin − 1 4 y 3 ) + 9 cos ( sin − 1 4 y 3 ) sin ( sin − 1 4 y 3 ) 8 } = 1 4 { 9 8 ( sin − 1 4 y 3 ) + 9 8 ( 4 y 3 ) 1 − 16 y 2 9 } = 1 4 { 9 8 ( sin − 1 4 y 3 ) + 3 y 2 1 − 16 y 2 9 } (V) Substitute the respective values in the above equation. F R = 19 , 620 { 1 4 { 9 8 ( sin − 1 4 y 3 ) + 3 y 2 1 − 16 y 2 9 } − 0.75 0.75 − 1 0.75 { 1 4 ( − 1 48 ( − 16 y 2 + 9 ) 3 2 ) } − 0.75 0.75 } = 19 , 620 ( 1 4 ( 1.767 − ( − 1.767 ) ) + 1 144 ( 0 ) ) = ( 17 , 334.27 N ) × 1 kN 1 , 000 N = 17

2nd Edition

ISBN: 9780134649290

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 2, Problem 81P

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The resultant force and its location acting on the elliptical plate.

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Chapter 2, Problem 80P

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Chapter 2 Solutions

Fluid Mechanics (2nd Edition)

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