HYDROSTATIC FORCE
ON A CURVED SURFACEThe cylinder in Fig. 3a rests on an inclined plane and is immersed in liquid up to its top,
as shown. Find the hydrostatic force on a 1-ft (30.48-cm) length of cylinder in terms of w
and the radius R; locate the pressure center.
Calculation Procedure:
- Evaluate the horizontal and vertical component of the force dP
on an elemental surface having a central angle d0
Refer to Fig. 3b. Adopt this sign convention: A horizontal force is positive if directed to
the right; a vertical force is positive if directed upward. The first three steps constitute
method 1.
Evaluating dP yields dPH = wR^2 (sin O - sin O cos 0) dO\ dPv = wR^2 (- cos 0 + cos^2 0)
dO. - Integrate these equations to obtain the resultant forces PH
and Pv; then find P
Here, PH = wR^2 [(- cos 0 + Y 2 cos^2 0)]^/6 = w^^2 [- (- 0.866 - 1) = /2(0.7S - I)] =
1.742wU^2 , to right; Pv=wR^2 (- sin 0 + V 2 B + % sin 20^^6 = w£^2 (0.5) + 1.833 + 0.217) =
2.55Ow^^2 , upward; P = wR^2 (l.742^2 + 2.550^2 )^05 = 3.QSIwR^2. - Determine the value of O at the pressure center
Since each elemental force dP passes through the center of the cylinder, the resultant
force P also passes through the center. Thus, tan (180° - (^0) PC) = PnIPy = 1.741/2.550;
0pc=145°41'.
- Evaluate P 11 and Pv
Apply these principles: Pn = force on an imaginary surface obtained by projecting the
wetted surface on a vertical plane; Pv = ± weight of real or imaginary liquid lying be-
tween the wetted surface and the liquid surface. Use the plus sign if the real liquid lies be-
low the wetted surface and the minus sign if it lies above this surface.
Then Pn = force, to right, on AC + force, to left, on EC = force, to right, on AE; AE =
l£66R;pm = 0.933 wR; Pn= 0.933 wR(lM6R) = 1.741 wR^2 ; Pv = weight of imaginary
FIGURE 3.