fluid_mechanics
66 Chapter 2 ■ Fluid Statics SOLUTION The pressure distribution acting on the inside surface of the plate is shown in Fig. E2.8b. The pressure at a given point on the plate is due to the air pressure, p s , at the oil surface, and the pressure due to the oil, which varies linearly with depth as is shown in the figure. The resultant force on the plate 1having an area A2 is due to the components, F 1 and F 2 , where F 1 and F 2 are due to the rectangular and triangular portions of the pressure distribution, respectively. Thus, and F 1 1 p s gh 1 2 A 350 10 3 Nm 2 10.90219.81 10 3 Nm 3 212 m2410.36 m 2 2 24.4 10 3 N F 2 g a h 2 h 1 b A 2 10.902 19.81 10 3 Nm 3 2 a 0.6 m 2 b 10.36 m2 2 0.954 10 3 N The magnitude of the resultant force, F R , is therefore (Ans) The vertical location of can be obtained by summing moments around an axis through point O so that or F R F 1 F 2 25.4 10 3 N 25.4 kN F R F R y O F 1 10.3 m2 F 2 10.2 m2 y O 124.4 103 N210.3 m2 10.954 10 3 N210.2 m2 25.4 10 3 N 0.296 m (Ans) Thus, the force acts at a distance of 0.296 m above the bottom of the plate along the vertical axis of symmetry. COMMENT Note that the air pressure used in the calculation of the force was gage pressure. Atmospheric pressure does not affect the resultant force 1magnitude or location2, since it acts on both sides of the plate, thereby canceling its effect. 2.10 Hydrostatic Force on a Curved Surface V2.5 Pop bottle The equations developed in Section 2.8 for the magnitude and location of the resultant force acting on a submerged surface only apply to plane surfaces. However, many surfaces of interest 1such as those associated with dams, pipes, and tanks2 are nonplanar. The domed bottom of the beverage bottle shown in the figure in the margin shows a typical curved surface example. Although the resultant fluid force can be determined by integration, as was done for the plane surfaces, this is generally a rather tedious process and no simple, general formulas can be developed. As an alternative approach we will consider the equilibrium of the fluid volume enclosed by the curved surface of interest and the horizontal and vertical projections of this surface. For example, consider a curved portion of the swimming pool shown in Fig. 2.23a. We wish to find the resultant fluid force acting on section BC (which has a unit length perpendicular to the plane of the paper) shown in Fig. 2.23b. We first isolate a volume of fluid that is bounded by the surface of interest, in this instance section BC, the horizontal plane surface AB, and the vertical plane surface AC. The free-body diagram for this volume is shown in Fig. 2.23c. The magnitude and location of forces F 1 and F 2 can be determined from the relationships for planar surfaces. The weight, w, is simply the specific weight of the fluid times the enclosed volume and acts through the center of gravity 1CG2 of the mass of fluid contained within the volume. The forces F H and F V represent the components of the force that the tank exerts on the fluid. In order for this force system to be in equilibrium, the horizontal component F H must be equal in magnitude and collinear with F 2 , and the vertical component F V equal in magnitude and collinear with the resultant of the vertical forces F 1 and w. This follows since the three forces acting on the fluid mass 1F 2 , the resultant of F 1 and w, and the resultant force that the tank exerts on the mass2 must form a concurrent force system. That is, from the principles of statics, it is known that when a body is held in equilibrium by three nonparallel forces, they must be concurrent 1their lines of action intersect at a common point2, and coplanar. Thus, F H F 2 F V F 1 w and the magnitude of the resultant is obtained from the equation F R 21F H 2 2 1F V 2 2
2.10 Hydrostatic Force on a Curved Surface 67 A B F 1 A CG F 2 O B F H (a) F √(F H ) 2 + (F V ) 2 R = O B C C C (b) (c) (d) F V F I G U R E 2.23 force on a curved surface. Hydrostatic The resultant F R passes through the point O, which can be located by summing moments about an appropriate axis. The resultant force of the fluid acting on the curved surface BC is equal and opposite in direction to that obtained from the free-body diagram of Fig. 2.23c. The desired fluid force is shown in Fig. 2.23d. E XAMPLE 2.9 GIVEN A 6-ft-diameter drainage conduit of the type shown in Fig. E2.9a is half full of water at rest, as shown in Fig. E2.9b. Hydrostatic Pressure Force on a Curved Surface FIND Determine the magnitude and line of action of the resultant force that the water exerts on a 1-ft length of the curved section BC of the conduit wall. 3 ft A B F 1 1 ft A CG B F H 1.27 ft A F R = 523 lb 32.5° O 1 ft C C F V (a) (b) (c) (d) F I G U R E E2.9 (Photograph courtesy of CONTECH Construction Products, Inc.) SOLUTION We first isolate a volume of fluid bounded by the curved section BC, the horizontal surface AB, and the vertical surface AC, as shown in Fig. E2.9c. The volume has a length of 1 ft. The forces acting on the volume are the horizontal force, F 1 , which acts on the vertical surface AC, the weight, w, of the fluid contained within the volume, and the horizontal and vertical components of the force of the conduit wall on the fluid, F H and F V , respectively. The magnitude of F 1 is found from the equation F 1 gh c A 162.4 lbft 3 2 1 3 2 ft2 13 ft 2 2 281 lb and this force acts 1 ft above C as shown. The weight w gV , where V is the fluid volume, is w g V 162.4 lbft 3 2 19p4 ft 2 2 11 ft2 441 lb and acts through the center of gravity of the mass of fluid, which according to Fig. 2.18 is located 1.27 ft to the right of AC as shown. Therefore, to satisfy equilibrium F H F 1 281 lb F V w 441 lb and the magnitude of the resultant force is F R 21F H 2 2 1F V 2 2 21281 lb2 2 1441 lb2 2 523 lb (Ans) The force the water exerts on the conduit wall is equal, but opposite in direction, to the forces F H and F V shown in Fig. E2.9c. Thus, the resultant force on the conduit wall is shown in Fig. E2.9d. This force acts through the point O at the angle shown.
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2.10 Hydrostatic Force on a Curved Surface 67<br />
A<br />
B<br />
F 1<br />
<br />
A<br />
CG<br />
F 2<br />
O<br />
B<br />
F H<br />
(a)<br />
F √(F H ) 2 + (F V ) 2<br />
R =<br />
O<br />
B<br />
C<br />
C<br />
C<br />
(b) (c) (d)<br />
F V<br />
F I G U R E 2.23<br />
force on a curved surface.<br />
Hydrostatic<br />
The resultant F R passes through the point O, which can be located by summing moments about an<br />
appropriate axis. The resultant force of the <strong>fluid</strong> acting on the curved surface BC is equal and opposite<br />
in direction to that obtained from the free-body diagram of Fig. 2.23c. The desired <strong>fluid</strong><br />
force is shown in Fig. 2.23d.<br />
E XAMPLE 2.9<br />
GIVEN A 6-ft-diameter drainage conduit of the type shown in<br />
Fig. E2.9a is half full of water at rest, as shown in Fig. E2.9b.<br />
Hydrostatic Pressure Force on a Curved Surface<br />
FIND Determine the magnitude and line of action of the resultant<br />
force that the water exerts on a 1-ft length of the curved section<br />
BC of the conduit wall.<br />
3 ft<br />
A<br />
B<br />
F 1<br />
1 ft<br />
A<br />
CG<br />
<br />
B<br />
F H<br />
1.27 ft<br />
A<br />
F R = 523 lb<br />
32.5°<br />
O<br />
1 ft<br />
C<br />
C<br />
F V<br />
(a)<br />
(b)<br />
(c)<br />
(d)<br />
F I G U R E E2.9<br />
(Photograph courtesy of CONTECH Construction Products, Inc.)<br />
SOLUTION<br />
We first isolate a volume of <strong>fluid</strong> bounded by the curved section<br />
BC, the horizontal surface AB, and the vertical surface AC, as<br />
shown in Fig. E2.9c. The volume has a length of 1 ft. The forces<br />
acting on the volume are the horizontal force, F 1 , which acts on<br />
the vertical surface AC, the weight, w, of the <strong>fluid</strong> contained<br />
within the volume, and the horizontal and vertical components of<br />
the force of the conduit wall on the <strong>fluid</strong>, F H and F V , respectively.<br />
The magnitude of F 1 is found from the equation<br />
F 1 gh c A 162.4 lbft 3 2 1 3 2 ft2 13 ft 2 2 281 lb<br />
and this force acts 1 ft above C as shown. The weight w gV ,<br />
where V is the <strong>fluid</strong> volume, is<br />
w g V 162.4 lbft 3 2 19p4 ft 2 2 11 ft2 441 lb<br />
and acts through the center of gravity of the mass of <strong>fluid</strong>, which<br />
according to Fig. 2.18 is located 1.27 ft to the right of AC as<br />
shown. Therefore, to satisfy equilibrium<br />
F H F 1 281 lb F V w 441 lb<br />
and the magnitude of the resultant force is<br />
F R 21F H 2 2 1F V 2 2<br />
21281 lb2 2 1441 lb2 2 523 lb<br />
(Ans)<br />
The force the water exerts on the conduit wall is equal, but opposite<br />
in direction, to the forces F H and F V shown in Fig. E2.9c.<br />
Thus, the resultant force on the conduit wall is shown in<br />
Fig. E2.9d. This force acts through the point O at the angle shown.