Explanation Given: The height of water above the cylinder ( h ) is 0 ft. The radius of the cylinder ( r ) is 0.35 ft. The length of point E from F ( L E F ) is 2 r . The length of the cylinder ( l ) is 1 ft. The unit weight of steel ( γ s ) is 490 lb / ft 3 Calculation: Calculate height of the point O from top of the water ( h O ) . h O = h + r h O = 0 ft + 0.35 ft = 0.35 ft Draw the cross section of the tank and mark the dimensions as shown in Figure (1). Calculate volume of the portion JGFEJ ( V 1 ) . V 1 = h O × L E F × l = h O × 2 r × l V 1 = 0.35 ft × 2 × 0.35 ft × 1 ft = 0.245 ft 3 Calculate the volume of the upper semicircle ( V 2 ) . V 2 = π 2 r 2 l V 2 = π 2 ( 0.35 ft ) 2 × 1 ft = 0.1924 ft 3 Calculate the volume of the water above the cylinder ( V ′ ) . V ′ = V 1 − V 2 V ′ = 0.245 ft 3 − 0.1924 ft 3 = 0.0526 ft 3 Calculate the length of the side HG ( L H G ) . L H G = d − L A B 2 = 2 r − L A B 2 L H G = 2 × 0.35 ft − 0.5 ft 2 = 0.1 ft Calculate the volume of the strip HGFCH ( V 3 ) . V 3 = L H G × h O × l V 3 = 0.1 ft × 0.35 ft × 1 ft = 0.035 ft 3 Calculate the length of side OC ( L O C ) . L O C = L A B 2 L O C = 0.5 ft 2 = 0.25 ft Calculate the angle BOC of triangle OBCO ( θ ) . cos θ = L O C r cos θ = 0.25 ft 0.35 ft θ = cos − 1 ( 0.714 ) = 44.42 ° Calculate volume of the sector OBFO ( V 4 ) . V 4 = θ 360 ° π ( r ) 2 l V 4 = 44

2nd Edition

ISBN: 9780134649290

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 2, Problem 115P

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The resultant force the bottom of the tank exerts on the cylinder

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Chapter 2, Problem 114P

Chapter 2, Problem 116P

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