FLUID MECHANICS EXERCISES

SOLUTION:

EXAMPLE 7:

EXAMPLE 8:

Assume that the concrete dam shown weighs 23.6 kN/m3 and rests on a solid foundation. Determine the minimum

coefficient of friction between the dam and the foundation required to keep the dam from sliding at the water depth

shown. Neglect a possible uplift along the base. Base your analysis on a unit length of the dam.

SOLUTION:

The angular portion of the dam decreases by 4 m (from 6 m to 2 m) as the

dam rises 5 m. Thus, the angle at the base of the dam = tan-1(5 m/4 m) =

0.8961 radians = 51.34o. This is also the angle, θ, that the water makes

with the sloping surface of the dam, which is used in our formula for the

resultant force. See the sketch at the left for the forces acting on the dam.

The slanted length of the dam, below the surface of the water, is (4 m)/sin

θ = (4 m)/sin(0.8961) = 5.122 m. The area of a unit width of the dam is

(5.122 m)(1 m)2 = 5.122 m2. The value of yc = (5.122/2 m) = 2.5561 m.

With these data we can find the resultant force of the water on a unit

length of the dam.

( )( ) ( ) kNmmm

kNAyF cR 4.1008931.0sin122.55561.2

80.9sin 2

3===

For force equilibrium in the x direction, the x component of the resultant force, FRsin θ = (100.4 kN)sin(0.8931)

= 78.40 kN, must equal the friction force, Ff. Force equilibrium in the y direction requires the weight of the dam,

which is the product of the specific weight times the volume of the unit length of the dam. The cross sectional

area has two parts: (1) a rectangle with an area of (2 m)(5 m) = 10 m2 and (2) a triangle with an area of (4 m)(5

m)/2 = 10 m2. The volume of a unit length of the dam is thus (10 m2 + 10 m2)(1 m) = 20 m3. Thus the weight of

concrete in the unit length of he dam is (23.6 kN/m3)(20 m3) = 472 kN. With this weight we have the following

force balance in the y direction.

( ) kNkNkNFN R 5344728931.0cos4.100cos =+=+= W

The friction coefficient, ɳ, is simply the friction force divided by the normal force.

kN

kN

N

F f

534

40.78== ɳ = 0.146

EXAMPLE 9:

SOLUTION:

EXAMPLE 10:

A swimming pool has a width of 2 m and a side profile as

shown. Determine the resultant force the pressure of the water

exerts on walls AB and CD, and on the bottom BC. Also, find

line of actions for resultant forces.

SOLUTION:

• FAB = γhcA = 9810 (4/2) (4x2) = 156960 N

yR,AB = yC + 𝐼𝑥𝑐

𝑦𝑐𝐴 = 2 +

1

12𝑥2𝑥43

2𝑥(4𝑥2) = 2.667 m

• FBC = γhcA = 9810 (1+4) (6x2) = 588600 N

yR,BC = yC + 𝐼𝑥𝑐

𝑦𝑐𝐴 =

ℎ𝑐

𝑠𝑖𝑛θ +

𝐼𝑥𝑐 𝑠𝑖𝑛θ

ℎ𝑐𝐴 =

5

(1

3) +

1

12𝑥2𝑥63(

1

3)

5𝑥(6𝑥2) = 15.2 m

• FCD = γhcA = 9810 (3) (8x2) = 470880 N

yR,CD = yC + 𝐼𝑥𝑐

𝑦𝑐𝐴 = 4 +

1

12𝑥2𝑥83

4𝑥(8𝑥2) = 5.334 m

sinθ = 1/3