Explanation Express the equation for the circumference of the elliptical plate. x 2 + y 2 ( 0.75 ) 2 = 1 x 2 = 1 − y 2 ( 0.75 ) 2 x = 1 − y 2 ( 0.75 ) 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 ( 1 − y 2 ( 0.75 ) 2 ) d y (II) Calculate the pressure acting on the differential element ( p ). p = ρ w g ( − y ) p = 1 , 000 × 9.81 ( − y ) p = − 9 , 810 y Here, the height at which the differential element is located is y . Calculate the resultant force acting on the differential element ( F R ) . d F = p d A Substitute p = − 9 , 810 y and d A = 2 ( 1 − y 2 ( 0.75 ) 2 ) d y in the above equation. d F = ( − 9 , 810 y ) 2 ( 1 − y 2 ( 0.75 ) 2 ) d y = − 26 , 160 ( y ( 0.75 ) 2 − y 2 ) d y Integrate both sides of the above equation and apply the value of y varies between (–0.75 m) to 0 m. ∫ d F = ∫ − 26 , 160 ( y ( 0.75 ) 2 − y 2 ) d y F R = − 26 , 160 ∫ − 0.75 0 y ( ( 0.75 ) 2 − y 2 ) d y (III) Multiply and divide Equation (III) by 8. F R = − 26 , 160 ∫ − 0.75 0 y ( ( 0.75 ) 2 − y 2 ) d y = − 26 , 160 4 ∫ − 0.75 0 y ( 9 − 16 y 2 ) d y = − 6 , 540 ∫ − 0.75 0 y ( 9 − 16 y 2 ) d y (IV) Solve the integral. P ( I ) = ∫ y ( − 16 y 2 + 9 ) 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 the above equation. 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) Substitute Equation (V) in Equation (IV). F R = − 6 , 540 ( − 1 48 ( − 16 y 2 + 9 ) 3 2 ) − 0.75 0 = − 6 , 540 ( − 1 48 ( 27 ) ) = ( 3 , 678.75 N ) × 1 kN 1 , 000 N = 3.68 kN Thus, the resultant force acting on the plate is 3.68 kN _ . Calculate the location where the resultants force acts ( y p ) . y p = − 19 , 620 ∫ y ( y ) ( 1 − y 2 ( 0.75 ) 2 ) d y F R y p = − 19 , 620 ∫ y ( y ) ( 1 − y 2 ( 0.75 ) 2 ) d y 3 , 678.75 = 5.3333 0.75 ∫ y 2 ( 0.75 ) 2 − y 2 d y = 7.1111 ∫ y 2 ( 0.75 ) 2 − y 2 d y Multiply and divide the equation with 8. y p = 7.1111 4 ∫ y 2 9 − 16 y 2 d y = 1.7778 ∫ y 2 9 − 16 y 2 d y The value of y varies between (-0.75 m) to 0 m. y p = 1.7778 ∫ − 0.75 0 y 2 9 − 16 y 2 d y (VI) Integrate the integral. ∫ y 2 9 − 16 y 2 d y Substitute y 2 = ( − 1 16 ( − 16 y 2 + 9 ) + 9 16 ) in the above equation. I = ∫ y 2 9 − 16 y 2 d y = − 1 16 ∫ ( − 16 y 2 + 9 ) 3 2 d y + 9 16 ∫ − 16 y 2 + 9 d y (VII) Solve integral. P ( I I ) = ∫ ( − 16 y 2 + 9 ) 3 2 d y Take 3 sin u 4 for y . y = 3 sin u 4 u = sin − 1 ( 4 y 3 ) d y d u = 3 cos u 4 Substitute y = 3 sin u 4 and d y = 3 cos u 4 d u in the above equation

2nd Edition

ISBN: 9780134649290

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 2, Problem 83P

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The resultant force and the location where the resultant force acting on the plate.

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Chapter 2, Problem 82P

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