PLANE SURFACE 6.) A vertical triangular surface has a horizontal 11.) The rectangular gate in Fig.

iangular surface has a horizontal 11.) The rectangular gate in Fig. B is hinged at B and rests on a
1.) Water stands on one side of the vertical gates shown base of 4 ft and an altitude of 9 ft, the vertex smooth floor at A, the horizontal component of the reaction at A
in the figure A. Find by integration the total pressure on being below the base. If the center of pressure is therefore being zero. The gate is 5 ft wide perpendicular to the
each gate and the location of the center pressure. 6 in. below the center of gravity, how far is the paper. Determine the vertical component of the reaction at A and
base below the liquid surface? the horizontal and vertical components of the reaction at B for
a. P=∂hcgA the following sets of conditions:
= 62.4 lb/ft^3 (1/2(9)) (4x9)
= 10108.8 lb

b. P=∂hcgA
= 62.4 lb/ft^3 (1/3(9)) (1/2(4x9))
= 3369.6 lb

c. P=∂hcgA
= 62.4 lb/ft^3 (3/5(9)) (4/3(2x9))
= 8087.04 lb

2.) Water stands on one side on the vertical gates 7.) A vertical, trapezoidal gate in the face of a dike is
shown in figure A., the water surface being 10 ft. above subjected to sea- water pressure (w = 64.0) on one
the top of the gates. Find by integration the total side. The upper edge is in the water surface and is 5
pressure on each gate and the location of the center of ft long. Two edges are vertical and measure 6 ft and
pressure. 9 ft each. Determine the total pressure on the gate
and the location of the center of pressure.
a. P=∂hcgA
= 62.4 lb/ft^3 (1/2(9)+10) (4x9)
= 32572.8 lb

b. P=∂hcgA
= 62.4 lb/ft^3 (1/3(9)+10) (1/2(4x9))
= 14601.6 lb

c. P=∂hcgA e= 5.13 ft from water surface

= 62.4 lb/ft^3 (3/5(9)+10) (4/3(2x9)) e= 2.17 ft from long side
= 23063.04 lb Ay= 24252.8 lb By= 9276.76 lb Bx= 19967.97 lb

3.) A gate 2 ft. square lies in a plane making an angle 8.) A rectangular gate of height h, with
of 30° with the vertical. Its upper edge is horizontal upper and lower edges horizontal is
and 3 ft. below the surface of the liquid (sp. gr 3.0). inclined at any angle. (< 90°) with the
Find by integration the total pressure on the gate vertical. Liquid stands on one side of the
and the location of the center of pressure. gate, the upper edge of the gate being in
the liquid surface. Show (and remember)
that under these conditions the distance
from the upper edge of the gate to the
center of pressure is two-thirds of h.

e = 0.075 ft

4.) A vertical circular gate 3 ft. in diameter is 9.) A vertical rectangular gate 4 ft wide and 6
subjected to pressure of molasses (sg 1.5) on one ft high, hinged at the top, has water on one
side. The free surface of the molasses is 8ft above side. What force applied at the bottom of the
the top of the gate. Determine the total pressure gate, at an angle of 45° with the vertical, is
and the Location of the center of pressure. required to open the gate when the water
surface is (a) at the top of the gate; (b) 3 ft
above the top of the gate; (c) 3 ft below the
top of the gate?

5.) A circular gate 5 ft in diameter is inclined at an angle of 10.) On one side, water stands level with 12.) The flashboard gate shown in Fig. C consists of a
45°. Sea water stands on one side of the gate to a height of the top of a vertical rectangular gate 4 ft plane face bd resting in a groove at d and supported by
30 ft above the center of the gate. Determine the total wide and 6 ft high, hinged at the bottom; the strut ce which is pinned at the ends. Neglecting the
pressure on the gate and the location of the center of on the other side water stands 3 ft below weight of the gate, determine the greatest depth, h,
pressure. the top. What horizontal force applied at which the water can have without causing the gate to
the top of the gate is required to open it? collapse.
15.) The gate in Fig. D is hinged at A and rests on a smooth floor at B. The gate is 10 ft DAMS
square. Oil stands on the left side of the gate to a height of 5 ft. above A. Above the oil EXAMPLE PROBLEM:
surface is gas under a gage pressure of -1 lb per sq in. Determine the amount of the
vertical force F applied at B that would be required to open the gate.

CURVED SURFACE 1.) A masonry dam of trapezoidal cross

1-2.) The curved surface represented by AB in Fig 4 is the section, with one face vertical, has a A.) THERE IS NO HYDROSTATIC UPLIFT
surface of the quadrant of a circular cylinder 10 ft long. thickness of 2 ft at the top and 10 ft at the
Determine the horizontal & vertical components of the total bottom. It is 22 ft high and has a horizontal
hydrostatic force on the surface locate the horizontal and base. The vertical face is subjected to water
vertical components of total pressure in problem 1. pressure, the water standing 15 ft above the
base. lb The weight of the masonry is 15016
a.) r=10ft z=0 L=10ft b.) r=8ft z=5ft L=10ft per cu. ft. Where will the resultant pressure
Px=∂hcgA Py = ∂V Px=∂hcgA Py1 = ∂V assuming:
𝜋(10)2
Px = 62.4(5)(10x10) Py = 62.4 ( 4 (10)) Px = 62.4(9)(8x10) Py1 = 62.4 (5x8x10)
Px = 31200 lb Py = 15600𝝅 lb Px= 44928 lb Py1 = 24960 lb

B.) THERE IS HYDROSTATIC UPLIFT

3.) The corner plate of the hull of a ship (AB in Fig. B) is curved on 2.) A masonry Dam of trapezoidal Cross section, with one face vertical, has a thickness
the arc of a circle with a 5-ft radius. With submersion in sea water of 2 ft at the top and 10 ft at the bottom, His 22 feet high and has a horizontal base.
as shown, compute for a 1-ft length perpendicular to the sketch The inclined face is subjected to water pressure, the water standing to a depth of Is ft
the amount and location of the horizontal and vertical dare the base. The weight of the masonry is 150 lb per cu.ft. If there is no hydrostatic
components of total pressure on AB. Determine graphically the uplift, where will the resultant pressure intersect the bare?
amount and location of the result- ant pressure.

4.) In the crest gate on a dam shown in Fig. C, surface AB forms

the area of a circle of 10-ft radius subtending 45° at the hinge.
The gate is 10 ft long. With water surface at B compute the
amount and location of the horizontal and vertical
components of total pressure on AB. Determine graphically
the amount and direction of their resultant.
GENERAL PROBLEMS (b) directly, by equation 5. Locate the center of pressure.
1.) Water stands 40 ft above the top of a vertical
gate which is 6 ft square and weighs 3000 lb.
What vertical lift will be required to open the
gate if the coefficient of friction between gate
and guides is 0.3?

2.) A plane parabolic gate with axis vertical 7.) The rectangular gate in Fig. B is hinged at A
and vertex down is sub- merged in oil (sp and rests against a smooth vertical wall at B. The
gr 0.80) to a depth of 9 ft. The width of the gate is 5 ft wide, perpendicular to the paper.
gate at the oil surface is 4 ft. Determine by Determine the horizontal and vertical
integration the total pressure on the gate components of the reactions at A and B when:
and the location of the center of pressure. a.) x=6 ft, y=8 ft, d1=12 ft, d2=0

𝟖 𝟖
d= √𝟖𝟐 + 𝟔𝟐 ; d=10 ̅=
𝒚 =
𝒔𝒊𝒏𝜽 𝟖/𝟏𝟎
= 10
P=∂hcgA sin𝜃 = 8/10 =53.13°
=62.4 (8)(10x5) w= 4/sin𝜃 = 4/(8/10) = 5ft
P= 24960 lb
𝟓(𝟏𝟎)𝟑
𝑰𝒄 ̅̅̅̅
𝟏𝟐
̅ + ̅̅̅̅ = 10 + ̅̅̅̅̅̅̅̅̅̅̅̅̅
Ycp= 𝒚 = 10.833 ft
𝒚𝑨 𝟏𝟎(𝟓𝒙𝟏𝟎)
Ycp = 10.833 – 5 = 5.833 from B and 4.167 from A

ΣFy= 0+ (upward)
Ay - Pcos𝜃 = 0
Ay – 24960cos(53.13)=0
Ay = 14976.036 lb

4.) A vertical triangular gate has a horizontal base 8 ft b.) x=6 ft, y=8 ft, d1=12 ft, d2=8ft y1= 10ft , y2= 5ft
long and 6 ft below the water surface. Its vertex is 2 P1=∂hcgA = 62.4(8)(5x10) = 24960 lb
ft above the water surface. What normal force must P2=∂hcgA = 62.4(4)(5x10) = 12480 lb
be applied at the vertex to open the gate? sin𝜃 = 8/10 =53.13°
w= 4/sin𝜃 = 4/(8/10) = 5ft

𝟓(𝟏𝟎)𝟑
𝑰𝒄 ̅̅̅̅
𝟏𝟐
Ycp1= ̅
𝒚 + ̅̅̅̅ = 10 + ̅̅̅̅̅̅̅̅̅̅̅̅̅ = 10.833 ft
𝒚𝑨 𝟏𝟎(𝟓𝒙𝟏𝟎)
= 5.833 ft from B
= 4.167 ft from A
𝟓(𝟏𝟎)𝟑
𝑰𝒄 ̅̅̅̅
𝟏𝟐
̅ + ̅̅̅̅ = 10 + ̅̅̅̅̅̅̅̅̅̅̅̅ = 6.667 ft from B
Ycp2= 𝒚 𝒚𝑨 𝟓(𝟓𝒙𝟏𝟎)
= 3.333 ft from A

ΣMa = 0+(counter clockwise) ΣFy= 0+ (upward)

-P1(4.167)+P2(3.333)+Bx(8)=0 Ay-P1cos(53.13)+P2cos(53.13)=0
Bx= 7801.56 Ay= 7488.018 lb

ΣFx= 0+ (to the right)

P1sin(53.13)-7801.56-P2sin(53.13)+Ax=0
Ax= -2182.427 lb
Ax= 2182.427 lb
5.) Find the horizontal and vertical components
of the hydrostatic pressure per foot of length on
the plane face of a dam inclined at an angle of
40° with the vertical if water stands 50 ft deep
above the base?

6.) Te plane surface AB in Fig. A is rectangular in shape,

6 ft wide, and 10 ft long perpendicular to the paper.
With oil on one side as shown, determine the total
pressure on the gate, (a) by computing the horizontal
and vertical components and their resultant

Cos 30 = 5/w ; w= 5.774 cos y/6 ; y= 3√𝟑

Cos 30 = x/6 ; x=3