HW_SET_3.soln

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

This preview shows pages 1–4. Sign up to view the full content.

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:

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
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 .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 10/12/2011 for the course CIVIL 2000 taught by Professor Dong during the Spring '11 term at Miami University.

### Page1 / 13

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online