LECTURE 4
3.5 HYDROSTATIC FORCES ON SUBMERGED SURFACE
Now that we have determined the manner in which the pressure varies in a static fluid, we
can examine how these pressure produce forces on surfaces submerged in a liquid.
In order to determine completely the force acting on a submerged surface, we
must specify:
1. The magnitude of the force.
2. The direction of the force.
3. The line of action of the resultant force.
3.5.1 Hydrostatic Force on Plane Submerged surfaces
z
Ambience of application
O
θ

dA
h


dF
FR
y
Liquid
density ρ
r'
dy
dA
x'
y
x
y'
y
Point of application

FR

The pressure force acting on an element of the upper surface, d A , is given by


d F  p d A
(3.10)
The resultant force acting on the surface is found by summing the contribution of the
infinitesimal forces over the entire area. Thus
1


FR    pdA
(3.11)
A
The basic pressure-height relation for a static fluid can be written as
dp
 g
dh
where h is measured positive downward from the liquid free surface. Then, if the pressure
at the free surface (h = 0) is po, the pressure-height relation is obtain by an expression for
the pressure, p, at any depth, h.

p
po
h
dp    g dh
o
and hence
h
p  po    g dh
o
If the position vector, from an arbitrary origin of coordinates to the point of application of

the resultant force, is designated as r ' , then






r '  F R   r  d F   r  p d A
(3.12)
A




Referring to Figure above, it can be seen that r '  ix ' jy' , r  ix  jy , and d A  dA k

Since the resultant force, F R , acts against the surface (in a direction opposite to that of



d A ), then F R   FR k . Substituting into Eq. 3.12 gives



ix ' jy'   FR k   ix  jy  d F  A ix  jy  p dA k
Therefore,
jx' FR  iy ' FR    jxp  iyp  dA
A
This is a vector equation, so the components must be equal. Thus
y' FR   y p dA
(3.13a)
x' FR   x p dA
(3.13b)
A
and
A
Example 3.4
The inclined surface shown, hinged along A, is 5 m wide. Determine the resultant force,

F R , of the water on the incline surface.
2

D=2m
h
A
L=4m
30o
GIVEN:
Rectangular gate, hinged along A. w = 5 m
FIND:

Resultant force, F R , of the water on the gate.
SOLUTION:

In order to completely determine F R , we must specify: (a) the magnitude, (b) the
direction, and (c) the line of action, of the resultant force.

Basic equations:
dp
 g
dh
F R    p dA
A




r '  FR   r  p d A
A
Consider the gate, hinged at A, lying in the xy plane, with coordinates as shown.



 

 d A  w dy k 



FR   p d A   p w dy k
A
A
z

dF
y
3
We now need ρ as a function of y to perform the integration. From the basic pressureheight relation,
dp
 g
dh
so
dp   g dh

and
p
po
h
dp    g dh
o
By assuming ρ = constant,
This gives
p  po  g dh
p  ph . We need p  p y 
From the diagram
h  D  y sin 30 o
where D = 2 m
Since we are interested in the force of the water on the gate, then

p  g D  y sin 30 o

Thus


L



FR   p d A   g D  y sin 30 o w dy k
A
0
L

 


y2
L2
  gw Dy 
sin 30 o  k   gw DL  sin 30 o  k
2
2

0


2

kg
m
16 m 1  N.sec 2 
  999 3  9.81 2  5m2m  4m 
 
k
2
2  kg.m
m
sec



Force acts in negative
FR  588 k kN

z dirction  F R

To find the line of action of the resultant force, F R , we recognize that the line of action
of the resultant force must be such that the moment of the resultant force about an axis
through point A equals the moment of distributed force about the same axis. That is,






r '  F R   r  d F   r  p d A
A
Then on substituting, we obtain


ix ' jy'  FR k  A ix  jy p dA k
and
jx' FR  iy ' FR  j  xp dA  i  yp dA
A
A
Then
4
y' FR   y p dA
A
1
y' 
FR

A
y p dA 
gw  D
1
FR

L
0
ypw dy 
gw
FR
 yD  y sin 30 dy
L
0
0
L

y3
gw  DL2 L3
0

y

sin
30

 sin 30 o 



FR  2
3
FR  2
3
0

2
3
 2m 16m
kg
m
5m
64m 1  N.sec
 999 3  9.81 2 

 

5
2
3
2  kg.m
m
sec 5.88 10 N 
y '  2.22m
2
Also
In calculating the moment of the distributed force (right side), recall from your earlier
courses in statics, that the centroid of the area element must be used for “x”. Since the
area element is of constant width, x = w/2, and
p dA
x' FR   x p dA
A
1
x' 
FR
x
w
w
A 2 p dA  2 FR
w
A p dA  2  2.5 m
w/2
F

r '  ix ' jy'





r '  2.5i  2.22 j m  Line of action of F R is along negative z axis through r 



This problem illustrates the procedure utilized in determining the resultant force, F R ,
equivalent to a distributed force on a plane submerged surface.
Example 3.5
The rectangular gate, AB, is 5 ft wide (w = 5 ft) and 10 ft long (L = 10 ft). The gate is
hinged along B. Neglecting the weight of the gate; calculate the force per unit width
exerted against the stop along A.

patm
D=5’
h
10’
S.O.S
L
A
patm
30o
B
5
GIVEN:
Rectangular gate, AB, hinged along B. width, w = 5 ft, and length, L = 10 ft. Neglected
weight of the gate
FIND:
Force per unit width against stop along A.
SOLUTION:

Basic equation:

dp
 g  
dh
FR  F R   p d A
A

Moment, M  M  Fd
Where d is the moment arm (counterclockwise moment assumed positive).
For equilibrium,
M  0
If we consider the gate hinged along B as a free body, lying in the xy plane, with
coordinate as shown, then:
z

FA
A
L

dF
B
L-y
y

(a). F A , the force of the stop on the gate, acts along the line y = 0 as shown.

(b). dF is an element of the distributed force due to the water acting as shown.
Note: If in calculating the distributed force due to the water, we do not need to include the
force due to atmospheric pressure acting on the upper surface of the gate.
Moment about x axis through B:

(a). of F A is FA • L
(b). of total distributed force is   L  y dF   L  y p dA
A
Since
 M  0 , then
FA  L   L  y dF  0
6
FA  L   L  y p dA  0
A
FA 
1
L  y p dA
L A
The element of are, dA, of the gate is dA = w dy, where w is the gate width.
 FA 
1 L
L  y  pw dy
L 0
We need p as a function of y in order to perform the integration. From the basic pressureheight relation,
dp
 g  
dh
dp   dh

and
p
po
h
dp    dh
o
By assuming γ = constant,
p  po  h
this gives
p  ph. We need p  p y 
From the diagram,
h  D  y sin 30 o
where D = 5 ft

p  p o   D  y sin 30 o

When we wrote the moment equation (  M  0 ), we did not include the moment of the
force due to atmospheric pressure acting on the top of the gate. Consequently, in figuring
the moment due to the water, we should not include the effect of the atmospheric pressure
at the free surface. Hence the pressure due to the water alone is

p   D  y sin 30 o

Thus
FA 
or


1 L
w L


L  y  D  y sin 30 o dy
L

y
pw
dy



0
0
L
L
FA  L
  DL  Ly sin 30 o  Dy  y 2 sin 30 o dy
w L 0


 
L

L
D
y3
  DLy  y 2 sin 30 o  y 2 
sin 30 o 
L
2
2
3
0



L3
DL2 L3
2
o
DL

sin
30

 sin 30 o 

L
2
2
3


FA   DL2 1 3
 
 L sin 30 o 
w L 2
16

With D = 5 ft, L = 10 ft, and γ = 62.4 lbf/ft3,
7
FA
1 62.4 lbf  5ft 100ft 2 1 1000ft 3 1 
lbf


 
   2080

3
w 10ft
2
16
1
2
ft
ft

Where FA is the force of the top on the gate. The force on the stop acts in the opposite
direction. Therefore,



FA
on stop   FA k  2080 k lbf
w
w
ft
this problem illustrate s the direct use of the distribute d moment wit hout evaluating 


 the resultant force and line of applicatio n separately .

3.5.2 Hydrostatic Force on Curve Submerged surfaces

The pressure force acting on the element of area, d A , is given by


d F  p d A
(3.10)
The resultant force is again given by


FR    pdA
(3.11)
A
z

z = zo
dAx

dA
dA y
y
dAz
x
We can write


F R  iFRx  jFRy  k FRz
8

where FRx ,
FRy , and FRz , are components of F R in the positive x, y, and z directions,
respectively.
To evaluate the components of the force in a given direction, we take the dot product of
the force with the unit vector in the given direction. Then,



FRx   dFx  FR  i   d F  i   p d A  i
A
FRx   p dA cos x    p dAx
A
A
Likewise,
FRy   p dA cos y    p dAy
A
A
FRz   p dA cos z    p dAz
A
A

θx is the angle between d A and i
where:

θy is the angle between d A and j

θz is the angle between d A and k
dAx  dA cos  x is the projection of the area element dA on a plane perpendicular to the x
axis
dAy  dA cos  y is the projection of the area element dA on a plane perpendicular to the y
axis
dAz  dA cos  z is the projection of the area element dA on a plane perpendicular to the z
axis
In general the component of the resultant force in the l direction is given by
FRl   p dAl
Al
In considering the vertical component, FRz , of resultant force we note that the pressure
exerted by the liquid is given by
zo
p   g dz
zs
Where zs is the vertical coordinate of the surface.
zo
dFz   pdAz    gdz dA
 zz

9
The vertical component of the resultant force is obtained by integrating over the entire
surface.
Fz   
Az

zo
zz
g dz dAz
In working with cylindrical surface, that is, surfaces with a constant radius of curvature,
than dA  wR d , where R is the radius, and w the width, of the cylindrical surface. In
these cases it is often easier to use θ as the variable of integration. Then
2
FRl    p dA cos θ    p cos θ wR d
1
A

where θ is the angle between d A and the unit vector in the l direction.
To find the line of action of each of the components of the resultant force on a curved
surface, we would write


rx'  i FRx   r  dFx i


ry'  j FR y   r  dFy j




rz'  k FRz   r  dFz k



where rx' , ry' , and rz' are vectors to the lines of action of the components of the resultant
force in the x, y, and z direction, respectively.
To find the components of the force:


FRy   dFy   p d A  j   p dA cos y   p dA cos    p dAy


FRz   dFz   p d A  k   p dA cos z   p dA sin    p dAz

k
θx
z
z
θy



β
β
dA


j
dA
dA cos
 dA cos y
 dAy
y
dA sin 
 dA cos  z
FRz
FRy
z'
y'
y
 dAz
10

The z component of F R has magnitude
 p dA
z
and acts in the negative z direction. By
designing the vertical component of the resultant force as FR v , then FR v   FRz . To find
the lines of action of the components:



z ' k  FRy j   z k  dFy j   z k  p dAy j





y' j  FRZ k  y' j   FRV k   yj  dFz k   yj   p dAz k
From these two equations we obtain
z' 
1
FR y

y' 
1
FRZ

Ay
z p dAy
AZ
y p dAz
y' 
or
1
FRV

AZ
y p dAz
Example 3.6
The gate shown has a constant width, w, of 5 m. The equation of the surface is x = y2/a,
where a = 4 m. The depth of water to the right of the gate is 4 m. Find the components
FRx and FRy of the resultant force due to the water and the line of action each.
y

x
0
y2
a
D=4m
water
x
GIVEN:
Gate with constant width, w = 5 m.
Equation of surface in xy plane is x = y2/a, where a = 4 m.
Water stands at depth, D. of 4 m to the right of the gate.
FIND:
FRx , FRy and line of action of each.
SOLUTION:
11


dp
 g
dh
d F  p d A
Basic equation



Moment of a force, M  r  F
y


D
FRx    p d A  i    p dAx    pwdy
A
AX

FRy    p d A  j    p dAy  
A
FRy
o
Ay
D2 / a
o
FR X
p w dx
y'
x'
FRx and FRx are 


assumed positive
x
In order to perform the necessary integrations, we need expressions for p(y) and p(x)
along the surface of the gate.
dp
 g ,
dh
dp  g dh

and
p
po
h
dp   g dh
o
If we assume ρ = constant, then
p  po  gh
Since atmospheric pressure acts on both the top of the gate and the free surface of the
liquid, there is no net contribution of the atmospheric pressure force. Thus, in determining
the force due to the liquid, we take p = ρgh.
We now need an expression for h = h(y) and h = h(x) along the surface of the gate.
Along the surface the gate, h = D-y. Since the equation of the gate surface is x = y2/a, then
along gate y  a x 1 / 2 and thus h can also be written as h  D  a x1 / 2 . Substituting the
appropriate equations for h info the expressions for FRx and FRy , gives
FRx   pw dy   ghw dy   g  h dy   gw D  y dy
D
D
D
D
o
0
0
0
D

 2 D2 
y2 
gwD 3
  gw Dy    gw D 

2 0
2 
2


4 m3  N  sec 2  392 kN
999 kg 9.81 m
FRx  


5
m

m3
sec 2
2
kg  m
3
The minus sign indicates

FRx actually acts to the left .
12
FRy  
D2 / a
o
p w dx  
D2 / a
0
 gw
D2 / a
0
ghw dx  gw
D2 / a
0
D  y  dx  gw0
D2 / a
D2 / a
2


 gw Dx 
ax3/ 2 
3

0
D 
h dx

a x1/ 2 dx
 D3 2
D 3  gwD 3
 gw

a 3/ 2  
3
3a
a 
 a
4 m3  1  N  sec 2  261 kN
999 kg 9.81 m


5
m

m3
sec 2
3
4m kg  m
3
FRy  
To find the line of action of FRx .



y' j  FRx i   yj  dFx i  yj   p dAx i
Therefore,
y ' FRx  
Ax
x' 
1
FRy


x' 

D2 / a
o
xp w dx 
gw  D
FRy
and
y p dAx
1
FRy

D2 / a
0
y'  

Ax
xghw dx 
D2 / a
2
5/ 2 
 2 x  5 a x 
0
2
1
FRx

y p dAz x
gw
FRy

D2 / a
0


x D  a x1 / 2 dx
gw  D 5
FRy
2
D 5  gwD 5
a 5/ 2  
 2
5
a  10 FRy a 2
 2a
gwD5  3a 
10a 2  gwD3 
3D 2
3
1
2
  4 m 2 
 1.2 m
10a 10
4m
This problem illustrate s the calculatio n of resultant force components on 


a curved submerged surface .

ANY QUESTIONS?
13