Question

The water in a 25-m-deep reservoir is kept inside by a 140-m-wide wall whose cross section is an equilateral triangle as shown in the figure Water 25 m 60° 60° Determine the total force (hydrostatic+ atmospheric) acting on the inner surface of the wall and its line of action. Take the value of g as 9.81 m/s2, the atmospheric pressure as 100,000 N/m2, and the density of water to be 1000 kg/m3 throughout. x 108 N The total force (hydrostatic + atmospheric) acting on the inner surface of the wall is The distance of the pressure center from the free surface of water along the wall surface is m.

Answer 1

Answer:  (a) 9.00 Mega Newtons or 9.00 * 10^6 N

               (b)  17.1 m

Explanation:  The length of wall under the surface can be given by

                                            $b=25m/sin(60)\\=28.867$

The average pressure on the surface of the wall is the pressure at the centeroid of the equilateral triangular block which can be then be calculated by multiplying it with the Plate Area which will provide us with the Resultant force.

$F(resultant) = Pavg ( A) = (Patm + \rho g h c)*A \\= [100000 N/m^2 + (1000 kg/m^3 * 9.81 m/s^2 * 25m/2)]* (140*25m/sin60)\\= 8.997*10^8 N \\= 9.0*10^8 N$

Noting from the Bernoulli  equation that

$Po/\rho g sin60 = 100000/1000 * 9.81* sin(60) = 11.77 m$

From the second image attached the distance of the pressure center from the free surface of the water along the surface of the wall is given by:

$Yp = s+\frac{b}{2} +\frac{b^2}{s+\frac{b}{2}+Po/\rho g sin60}= 0+\frac{28.87}{2} +\frac{28.87^2}{0+\frac{28.87}{2}+100000 /1000 *9.81 sin60} = 17.1 m$

Substituting the values gives us the the distance of the surface to be equal to = 17.1 m

Answer 2

The total force acting on the inner surface of the wall is 8.99*10^8N, the distance of the pressure center for the free surface of the water along the surface is y(p) = 17.09m

Given data

• The depth of the reservoir d = 25m
• The atmospheric pressure P$_a_t_m$ = 100kPa
• Width of the wall (w) = 140m
• Density of water = 1000 kg/m^3

Assuming the atmospheric pressure is acting on both sides of the gate which cancels each other.

From the figure attached;

$b=\frac{h}{sin60^0}=\frac{25}{sin60^0} =28.87m$

The vertical distance of the centroid from the free surface of the liquid

$h_c=\frac{h}{2} =\frac{25}{2}=12.5m$

The Hydrostatic Force Acting on The Wall

$F_R=P_a_v_gA=(P_a_t_m+Pgh_c)A$

Where area of the surface is

$A=w*b=140*28.87\\A = 4041.8m^2$

substituting the known value in the above equation

$F_R=[(100*10^3)+(1000*9.81*12.49)]*4041.8\\F_R=899231202.9\\F_R=8.99*10^8N$

The total force acting on the inner surface of the wall

$F_R=8.99*10^8N$

The line of action of the total force is

$y_p=s+\frac{b}{2} + \frac{b^2}{12(s+\frac{b}{2}+ \frac{p_o}{pgsinO}}\\y_p=0+\frac{28.87}{2}+\frac{28.87^2}{12[0+\frac{28.87}{2} +\frac{100*10^3}{1000*9.81*sin60^0}} \\y_p=14.44+2.65\\y_p=17.09m$

The line of action of the center of pressure acting along the wall surface

$y_p=17.09m$

The Horizontal Component of Hydrostatic Force

$F_H=F_Rsin60^0\\F_H=(8.99*10^8)*sin60^0\\F_H=778757065.59N\\F_H=7.787*10^8N$

The horizontal component of hydrostatic force $F_H=7.787*10^8N$

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