Explanation Given: The width of the channel ( b ) is 2 m. Mass of the gate ( m ) is 6 Mg . Calculation: Draw the free body diagram of the sluice gate as shown in Figure (1). Calculate the intensity of load at point A ( w 1 ) . w 1 = ρ w g h A b w A = 1 , 000 kg / m 3 × 9.81 m / s 2 × 3 m × 2 m = 58.86 × 10 3 N / m Calculate the intensity of load at point B ( w 2 ) . w B = ρ w g h C b w B = 1 , 000 kg / m 3 × 9.81 m/s 2 × ( 3 + 3 sin 60 ° ) m × 2 m = 109.83 × 10 3 N / m Calculate the horizontal force acting due to load w A , ( F h ) 1 . ( F h ) 1 = w A h A B ( F h ) 1 = ( 58.86 × 10 3 ) N/m × ( 3 sin 60 ° ) m = 52.92 × 10 3 N Calculate the horizontal force acting due to load w B , ( F h ) 2 . ( F h ) 2 = 1 2 ( w B − w A ) h A B ( F h ) 2 = 1 2 ( 109.83 − 58.86 ) × 10 3 × ( 3 sin 60 ° ) = 66.22 × 10 3 N Calculate the resultant horizontal force ( F h ) . F h = ( F h ) 1 + ( F h ) 2 F h = 52.92 × 10 3 N + 66.22 × 10 3 N = 219.14 × 10 3 N = 219.14 kN Calculate the point of action of y ¯ 1 and y ¯ 2 . y ¯ 1 = 1 2 h A B = 1 2 ( 3 sin 60 ° ) = 1.3 m y ¯ 2 = 2 3 h A B = 2 3 ( 3 sin 60 ° ) = 1.732 m Calculate volume of the section ( V 1 ) . V 1 = h A × L A D × b V 1 = ( 3 × 1

2nd Edition

ISBN: 9780134649290

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 2, Problem 113P

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The magnitude of the resultant force acting on the gate and the smallest torque required to open the gate.

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Chapter 2, Problem 112P

Chapter 2, Problem 114P

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