HW_SET_3.soln - 126 Solutions Manual Fluid Mechanics, Fifth...

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126 Solutions Manual Fluid Mechanics, Fifth Edition Solution: The horizontal component is  HC G v e r t F h A (9790)(6)(2 6) (a) Ans. 705000 N Fig. P2.85 The vertical component is the weight of the fluid above the quarter-circle panel:   2 (9790)(2 7 6) (9790)( /4)(2) (6) 822360 184537 (b) Ans. 638000 N V F W(2 by 7 rectangle) W(quarter-circle) 2.86 The quarter circle gate BC in Fig. P2.86 is hinged at C. Find the horizontal force P required to hold the gate stationary. The width b into the paper is 3 m. Neglect the weight of the gate. Solution: The horizontal component of water force is Fig. P2.86 3 G F h A (9790 N/m )(1 m)[(2 m)(3 m)] 58,740 N This force acts 2/3 of the way down or 1.333 m down from the surface (0.667 m up from C). The vertical force is the weight of the quarter-circle of water above gate BC: 32 V water F (Vol) (9790 N/m )[( /4)(2 m) (3 m)] 92,270 N  F V acts down at (4R/3 ) 0.849 m to the left of C. Sum moments clockwise about point C:
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Chapter 2 Pressure Distribution in a Fluid 127 C M 0 (2 m)P (58740 N)(0.667 m) – (92270 N)(0.849 m) 2P 117480  Solve for 58,700 N . PA n s 58 7 kN 2.87 The bottle of champagne (SG 0.96) in Fig. P2.87 is under pressure as shown by the mercury manometer reading. Compute the net force on the 2-in-radius hemispherical end cap at the bottom of the bottle. Solution: First, from the manometer, com- pute the gage pressure at section AA in the Fig. P2.87 champagne 6 inches above the bottom: AA atmosphere 24 p (0.96 62.4) ft (13.56 62.4) ft p 0 (gage), 12 12     2 AA or: P 272 lbf/ft (gage) Then the force on the bottom end cap is vertical only (due to symmetry) and equals the force at section AA plus the weight of the champagne below AA: V AA AA 6-in cylinder 2-in hemisphere 3 FF p( A r e a ) W W  22 (272) (4/12) (0.96 62.4) (2/12) (6/12) (0.96 62.4)(2 /3)(2/12) 4 23.74 2.61 0.58 . Ans   25.8 lbf 2.88 Circular-arc Tainter gate ABC pivots about point O. For the position shown, determine (a) the hydrostatic force on the gate (per meter of width into the paper); and (b) its line of action. Does the force pass through point O? Solution: The horizontal hydrostatic force is based on vertical projection: Fig. P2.88
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HC G v e r t F h A (9790)(3)(6 1) 176220 N at 4 m below C  The vertical force is upward and equal to the weight of the missing water in the segment ABC shown shaded below. Reference to a good handbook will give you the geometric properties of a circular segment, and you may compute that the segment area is 3.261 m 2 and its centroid is 5.5196 m from point O, or 0.3235 m from vertical line AC, as shown in the figure. The vertical (upward) hydrostatic force on gate ABC is thus VA B C F A (unit width) (9790)(3.2611) 31926 N at 0.4804 m from B The net force is thus per meter of width, acting upward to the right at an angle of 10.27 and passing through a point 1.0 m below and 0.4804 m to the right of point B. This force passes, as expected, right through point O .
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HW_SET_3.soln - 126 Solutions Manual Fluid Mechanics, Fifth...

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