SOLUTION:

Given,

Width,`=2m`

Gauge Pressure,`P=-117.72 KPa`

Thus pressure head,`h=P/(rho*g)=-117.72/(1.2**9.81)`

`h=-10 m`

Thus the free surface of the tank will be at a height of 10m further below the top of the given oil surface.

This free surface is known as imaginary free surface. 

Figure below shows the equivalent free surface of water.

Now, 

The total horizontal component of force acting on the curved surface ABCDEF   is equal to the force on the projected area of the curved surface on the vertical plane

I.e,

`F_x=rho*g*A*bar (h)`

Where,

`bar (h)=`distance of the c.g of the area from free water surface

`bar (h)=(3+2+2**1.6)/2=4.10 m`

and projected area,`A=(3+2+2**1.6)**10=82 m^2`

Thus,

`F_x=1.2**9.81**82**4.10`

or,`F_x=3957.7 KN->`

and this acts at a distance of,

`h^**=(I_G sin^2theta)/(A*bar (h))`

`I_G=`moment of inertia

`I_G=(10**8.2^3)/12=459.47 m^4`

Thus`I_G=459.47/(82**4.10)+4.10`

`=5.47 m` from the imaginary surface.

Vertical force

 In portion C'D = 0 KN

In portion DD',

`F_(y1)=rho*g*10*Area`

`=9.81**1.2**3**2**10`

`=706.32 KN`(downward)

In portion D'E =0 KN

In portion EF,

`F_(y2)=rho*g*h*Area`

`=9.81**1.2**10**pi**1.6^2/2`

`=473.38 KN`(upward)

Thus, the net vertical force is,

`F_y=473.38-706.32+0`

`=232.94 KN`(Downward)

SOLUTION:

Given,

Width,`=2m`

Gauge Pressure,`P=-117.72 KPa`

Thus pressure head,`h=P/(rho*g)=-117.72/(1.2**9.81)`

`h=-10 m`

Thus the free surface of the tank will be at a height of 10m further below the top of the given oil surface.

This free surface is known as imaginary free surface. 

Figure below shows the equivalent free surface of water.

Now, 

The total horizontal component of force acting on the curved surface ABCDEF   is equal to the force on the projected area of the curved surface on the vertical plane

I.e,

`F_x=rho*g*A*bar (h)`

Where,

`bar (h)=`distance of the c.g of the area from free water surface

`bar (h)=(3+2+2**1.6)/2=4.10 m`

and projected area,`A=(3+2+2**1.6)**10=82 m^2`

Thus,

`F_x=1.2**9.81**82**4.10`

or,`F_x=3957.7 KN->`

and this acts at a distance of,

`h^**=(I_G sin^2theta)/(A*bar (h))`

`I_G=`moment of inertia

`I_G=(10**8.2^3)/12=459.47 m^4`

Thus`I_G=459.47/(82**4.10)+4.10`

`=5.47 m` from the imaginary surface.

Vertical force

 In portion C'D = 0 KN

In portion DD',

`F_(y1)=rho*g*10*Area`

`=9.81**1.2**3**2**10`

`=706.32 KN`(downward)

In portion D'E =0 KN

In portion EF,

`F_(y2)=rho*g*h*Area`

`=9.81**1.2**10**pi**1.6^2/2`

`=473.38 KN`(upward)

Thus, the net vertical force is,

`F_y=473.38-706.32+0`

`=232.94 KN`(Downward)

SOLUTION:

Given,

Radius of the cylinder,`R=1 m`

Length,`l=2m`

Gauge Pressure,`P=0.2 (kg_f)/(cm)^2`

`=0.2**9.81**10^4`

`P=1.962**10^4`N/`m^2`

And pressure head,`h=P/(rho*g)=(1.962**10^4)/(9810)`

`h=2 m`

Thus the free surface of the tank will be at a height of 2m from the top of the tank.

This free surface is known as imaginary free surface. 

Figure below shows the equivalent free surface of water.

Now, 

The total horizontal component of force acting on the curved surface ABC of the cylinder is equal to the force on the projected area of the curved surface on the vertical plane

I.e,

`F_x=rho*g*A*bar (h)`

Where,

`bar (h)=`distance of the c.g of the area from free water surface

`bar (h)=(2+1.5/2)`

Thus,

`F_x=9810**(1.5**2)**(2+1.5/2)`

`F_x=80932.5N`

Also, the vertical component of the force is,

`F_y=`weight of water enclosed by the curved surface upto the free surface  (real or imaginary )

`F_y=` weight of water in portion `CODEABC`

`=`Weight of water in CODFBC - Weight of water in AEFB

But,

Weight of water in CODEFBC = Weight of water on [COB + BFDOB]

`=rho *g ((pir^2)/2+BO**OD)*Wi dth `

`=9810(pi/4*1^2+1**2.5)**2`

`=64458.5 N`

Again, the weight of water in AEFB = Weight of water in  [AEFG + AGBH - AHB] 

`=rho g*2(2BH+0.5BH-AHB)`

From figure,

`sintheta=(AH)/(AO)=0.5/1=0.5`

And, `BH=BO-HO=1-AO*cos30`

`BH=1-1cos30=0.134`

And the area of `triangle ABH`=Area of [ABO - AHO]

`=pir^2*30/360-(AH**HO)/2`

`=(pir^2)/12-(0.5**0.866)/2`

`=0.0453`

Thus using all these values,we get,

Weight of water in AEFB `=9810**2 (2**0.134+0.134**0.5-0.0453`

`=5684 N`

Hence,

`F_y=64458.5-5684=58774.5 N`

SOLUTION:

Given,

Radius of the cylinder,`R=1 m`

Length,`l=2m`

Gauge Pressure,`P=0.2 (kg_f)/(cm)^2`

`=0.2**9.81**10^4`

`P=1.962**10^4`N/`m^2`

And pressure head,`h=P/(rho*g)=(1.962**10^4)/(9810)`

`h=2 m`

Thus the free surface of the tank will be at a height of 2m from the top of the tank.

This free surface is known as imaginary free surface. 

Figure below shows the equivalent free surface of water.

Now, 

The total horizontal component of force acting on the curved surface ABC of the cylinder is equal to the force on the projected area of the curved surface on the vertical plane

I.e,

`F_x=rho*g*A*bar (h)`

Where,

`bar (h)=`distance of the c.g of the area from free water surface

`bar (h)=(2+1.5/2)`

Thus,

`F_x=9810**(1.5**2)**(2+1.5/2)`

`F_x=80932.5N`

Also, the vertical component of the force is,

`F_y=`weight of water enclosed by the curved surface upto the free surface  (real or imaginary )

`F_y=` weight of water in portion `CODEABC`

`=`Weight of water in CODFBC - Weight of water in AEFB

But,

Weight of water in CODEFBC = Weight of water on [COB + BFDOB]

`=rho *g ((pir^2)/2+BO**OD)*Wi dth `

`=9810(pi/4*1^2+1**2.5)**2`

`=64458.5 N`

Again, the weight of water in AEFB = Weight of water in  [AEFG + AGBH - AHB] 

`=rho g*2(2BH+0.5BH-AHB)`

From figure,

`sintheta=(AH)/(AO)=0.5/1=0.5`

And, `BH=BO-HO=1-AO*cos30`

`BH=1-1cos30=0.134`

And the area of `triangle ABH`=Area of [ABO - AHO]

`=pir^2*30/360-(AH**HO)/2`

`=(pir^2)/12-(0.5**0.866)/2`

`=0.0453`

Thus using all these values,we get,

Weight of water in AEFB `=9810**2 (2**0.134+0.134**0.5-0.0453`

`=5684 N`

Hence,

`F_y=64458.5-5684=58774.5 N`

SOLUTION:

Given,

Diagonal of aperture,PR=QS= 2.4 m

Area of square aperture, A= area of `triangle`PQR + area of `triangle`PSR

`A=2**1/2**2.4**2.4/2`

`A=2.88 m^2`

Now,  the thrust exerted on the plate by the liquid is,

`F=rho*g*A*bar (h)`

`=1.2**1000**9.81**2.88**1.8`

`F=61026.05 N`

Again,

And the point of application of this force is,

`h^** =I_G/(A*bar (h))+bar (h)`

Where,

`I_G=`moment of inertia of aperture PQRS about the diagonal PR

=M.I of `triangle`PQR+M.I of `triangle`PSR about PR

`=(2.4**(1.2)^3)/12+(2.4**1.2^3)/12`

`=0.6912 m^4`

Thus,

`h^**=0.6912/(2.88**1.8)+1.8`

`h^**=1.933 m`

SOLUTION:

Given,

Diagonal of aperture,PR=QS= 2.4 m

Area of square aperture, A= area of `triangle`PQR + area of `triangle`PSR

`A=2**1/2**2.4**2.4/2`

`A=2.88 m^2`

Now,  the thrust exerted on the plate by the liquid is,

`F=rho*g*A*bar (h)`

`=1.2**1000**9.81**2.88**1.8`

`F=61026.05 N`

Again,

And the point of application of this force is,

`h^** =I_G/(A*bar (h))+bar (h)`

Where,

`I_G=`moment of inertia of aperture PQRS about the diagonal PR

=M.I of `triangle`PQR+M.I of `triangle`PSR about PR

`=(2.4**(1.2)^3)/12+(2.4**1.2^3)/12`

`=0.6912 m^4`

Thus,

`h^**=0.6912/(2.88**1.8)+1.8`

`h^**=1.933 m`

SOLUTION:

We know, the pressure force on the gate is,

`F=rho*g*A*bar (h)`... (i)

Now,

Area of gate,`=6**2=12m^2`

Now,

Length of gate submerged in water = AD

From figure,

`AC=AD*sin60`

or,`AD=(AC)/(sin60)`

`AD=2/(root ()(3))*h`

Now, the area of the gate immersed in water is,

`A=AD**WIDTH`

`A=2/(root ()(3))*h*2`

`A=4/(root ()(3))*h`

Also,depth of C.G. of the immersed area is,

`bar (h)=h/2`

Thus,the  pressure force on the gate is,

`F=rho*g*A*bar (h)`

`=9.81**1000**4/(root ()(3))*h*h/2`

`F=19620/(root ()(3))*h^2`

Also,

`h^**=(I_G sin^2theta)/(A*bar (h))`

Where,`I_G=`moment of inertia

`I_G=(b*(AD)^3)/12`

`=2/12*(2/(root ()(3))*h)^3`

`I_G=(4h^3)/(9root ()(3))`

Thus,

`h^**=((4h^3)/(9root ()(3))*(root ()(3)/2)^2)/((4h)/(root ()(3)*h/2)+h/2`

or,`h^**=(2h)/3`

Also, the distance of centre of pressure from the hinge along the length of the gate` = AE`

We know,

`h-h^**=AE*sin60`

or,`AE=(h-h^**)/(sin60)`

`=(h-(2h)/3)/(root ()(3)/2)`

`=(2h)/(3root ()(3))`

Now taking the momentum about the hinge,

`60000**6-F**AE=0`

or,`60000**6=19620/(root ()(3))*h^2*(2h)/(3root ()(3))`

or,`h=4.36 m`

Note: This type of problem can be solved easily by plotting pressure diagram as explained in next problem.

SOLUTION:

We know, the pressure force on the gate is,

`F=rho*g*A*bar (h)`... (i)

Now,

Area of gate,`=6**2=12m^2`

Now,

Length of gate submerged in water = AD

From figure,

`AC=AD*sin60`

or,`AD=(AC)/(sin60)`

`AD=2/(root ()(3))*h`

Now, the area of the gate immersed in water is,

`A=AD**WIDTH`

`A=2/(root ()(3))*h*2`

`A=4/(root ()(3))*h`

Also,depth of C.G. of the immersed area is,

`bar (h)=h/2`

Thus,the  pressure force on the gate is,

`F=rho*g*A*bar (h)`

`=9.81**1000**4/(root ()(3))*h*h/2`

`F=19620/(root ()(3))*h^2`

Also,

`h^**=(I_G sin^2theta)/(A*bar (h))`

Where,`I_G=`moment of inertia

`I_G=(b*(AD)^3)/12`

`=2/12*(2/(root ()(3))*h)^3`

`I_G=(4h^3)/(9root ()(3))`

Thus,

`h^**=((4h^3)/(9root ()(3))*(root ()(3)/2)^2)/((4h)/(root ()(3)*h/2)+h/2`

or,`h^**=(2h)/3`

Also, the distance of centre of pressure from the hinge along the length of the gate` = AE`

We know,

`h-h^**=AE*sin60`

or,`AE=(h-h^**)/(sin60)`

`=(h-(2h)/3)/(root ()(3)/2)`

`=(2h)/(3root ()(3))`

Now taking the momentum about the hinge,

`60000**6-F**AE=0`

or,`60000**6=19620/(root ()(3))*h^2*(2h)/(3root ()(3))`

or,`h=4.36 m`

Note: This type of problem can be solved easily by plotting pressure diagram as explained in next problem.

SOLUTION:

Given,

Size of the gate = 1.2 m ** 5 m

Area of the gate, `A=1.2**5=6 m^2`

Now,

The depth of C.G of the gate from free water surface is,

`bar (h)=5-(BG*cos45)`

`=5-(AB)/2cos45`

`=5-(1.2/2*1/root ()(2))`

`bar (h)=4.576 m`

Now, Total pressure force acting on the gate is,

`P=rho*g*A*bar (h)`

`=(1000**9.81**6**4.576)/1000`

`=269.3 KN`

Let, this force acts at a depth  of `h^**`.

`h^**=(I_G sin^2theta)/(A*bar (h)+bar (h)`

`I_G=`moment of inertia

`I_G=(b*d^3)/12=(5**1.3^3)/12=0.72 m^4`

Thus,

`h^**=(0.72sin^2theta)/(6**4.576)+4.576`

`=4.589 m`

Now, taking moments about hinge A,we get,

`P**AC-F**1.2=0`

or,`269.3**AC=F**1.2`

From figure,

`AC=AB-BC`

`=AB-(OB-OC)`

`=AB-(5/(sin45)-(h^**)/(sin45))`

`=1.2-(5/(sin45)-4.589/(sin45))`

`=0.618`

Thus,

`F=(269.3**0.618)/1.2`

`F=138.69 KN`

SOLUTION:

Given,

Size of the gate = 1.2 m ** 5 m

Area of the gate, `A=1.2**5=6 m^2`

Now,

The depth of C.G of the gate from free water surface is,

`bar (h)=5-(BG*cos45)`

`=5-(AB)/2cos45`

`=5-(1.2/2*1/root ()(2))`

`bar (h)=4.576 m`

Now, Total pressure force acting on the gate is,

`P=rho*g*A*bar (h)`

`=(1000**9.81**6**4.576)/1000`

`=269.3 KN`

Let, this force acts at a depth  of `h^**`.

`h^**=(I_G sin^2theta)/(A*bar (h)+bar (h)`

`I_G=`moment of inertia

`I_G=(b*d^3)/12=(5**1.3^3)/12=0.72 m^4`

Thus,

`h^**=(0.72sin^2theta)/(6**4.576)+4.576`

`=4.589 m`

Now, taking moments about hinge A,we get,

`P**AC-F**1.2=0`

or,`269.3**AC=F**1.2`

From figure,

`AC=AB-BC`

`=AB-(OB-OC)`

`=AB-(5/(sin45)-(h^**)/(sin45))`

`=1.2-(5/(sin45)-4.589/(sin45))`

`=0.618`

Thus,

`F=(269.3**0.618)/1.2`

`F=138.69 KN`

Solution:

Here the force acting on the left sides of the cylinder are, 

horizontal component of force `F_(x1)`= force on the projected vertical area AOC

` =rhog*A*bar (h)`

`=9810**(4**2)**2`

`=156960 N`

and this acts at a distance of,

`(h_1)^**=(I_G sin^2theta)/(A*bar (h))`

`=(2**4^3)/(12**4**2**2)+2`

`=2.67 m` from free surface (point A)

Similarly,Vertical force is,

`F_(y1)=`weight of water enclosed by the ABCA

`F_(y1)=rhog*Area**L`

`=9810**2**(pi*2^2)/2`

`=123276.1 N`

Which acts at a distance,

`=(4r)/(3pi)=0.8488m`

Similarly the forces acting on the right side of the cylinder are,

horizontal component of force `F_(x2)`= force on the projected vertical area AOC

` =rhog*A*bar (h)`

`=9810**(2**2)**2/2`

`=39240 N`

and this acts at a distance of,

`(h_2)^**=(I_G sin^2theta)/(A*bar (h))`

`=(2**2^3)/(12**2**2**1)+1`

`=1.33 m` from free surface (point A)

Similarly,Vertical force is,

`F_(y2)=`weight of water enclosed by the COD

`F_(y2)=rhog*Area**L`

`=9810**2**(pi*2^2**90/360)`

`=61638.05 N`

Which acts at a distance,

`=(4r)/(3pi)=0.8488m`

Now, Resultant force in X-direction is,

`F_x=15690-39240` (force acts towards each other. So one is taken as positive and another is taken negative)

`=117720 N`

Similarly Resultant force in Y-direction is,

`F_y=123276.1+61638.05` (both are positive since acting upward )

`=184914.15 KN` (upward)

Thus, Resultant force, 

`F=root ()(F_x^2+F_y^2)`

`=219206 N`

And, `theta=tan^-1 (F_y/F_x)=57^0` with the X-axis.

Taking moment about C,

`F_x*y=F_(x1)*(4-2.67)-F_(x2)*(2-1.33)`

or,`y=1.55 m` from bottom.

Similarly,

`F_y*x=F_(y1)**0.8848-F_(y2)**0.8848`

`x=0.282 m` from AOC.

we know, the resulting force in upward direction is,

`F_y=184914.15 N`

Thus the weight of the cylinder should not be less than the upward force `F_y`. Hence the least weight of the cylinder should be at least equal to `184914.15 N`

Solution:

Here the force acting on the left sides of the cylinder are, 

horizontal component of force `F_(x1)`= force on the projected vertical area AOC

` =rhog*A*bar (h)`

`=9810**(4**2)**2`

`=156960 N`

and this acts at a distance of,

`(h_1)^**=(I_G sin^2theta)/(A*bar (h))`

`=(2**4^3)/(12**4**2**2)+2`

`=2.67 m` from free surface (point A)

Similarly,Vertical force is,

`F_(y1)=`weight of water enclosed by the ABCA

`F_(y1)=rhog*Area**L`

`=9810**2**(pi*2^2)/2`

`=123276.1 N`

Which acts at a distance,

`=(4r)/(3pi)=0.8488m`

Similarly the forces acting on the right side of the cylinder are,

horizontal component of force `F_(x2)`= force on the projected vertical area AOC

` =rhog*A*bar (h)`

`=9810**(2**2)**2/2`

`=39240 N`

and this acts at a distance of,

`(h_2)^**=(I_G sin^2theta)/(A*bar (h))`

`=(2**2^3)/(12**2**2**1)+1`

`=1.33 m` from free surface (point A)

Similarly,Vertical force is,

`F_(y2)=`weight of water enclosed by the COD

`F_(y2)=rhog*Area**L`

`=9810**2**(pi*2^2**90/360)`

`=61638.05 N`

Which acts at a distance,

`=(4r)/(3pi)=0.8488m`

Now, Resultant force in X-direction is,

`F_x=15690-39240` (force acts towards each other. So one is taken as positive and another is taken negative)

`=117720 N`

Similarly Resultant force in Y-direction is,

`F_y=123276.1+61638.05` (both are positive since acting upward )

`=184914.15 KN` (upward)

Thus, Resultant force, 

`F=root ()(F_x^2+F_y^2)`

`=219206 N`

And, `theta=tan^-1 (F_y/F_x)=57^0` with the X-axis.

Taking moment about C,

`F_x*y=F_(x1)*(4-2.67)-F_(x2)*(2-1.33)`

or,`y=1.55 m` from bottom.

Similarly,

`F_y*x=F_(y1)**0.8848-F_(y2)**0.8848`

`x=0.282 m` from AOC.

we know, the resulting force in upward direction is,

`F_y=184914.15 N`

Thus the weight of the cylinder should not be less than the upward force `F_y`. Hence the least weight of the cylinder should be at least equal to `184914.15 N`

SOLUTION:

The gate would tip about the hinge point O when the line of action of the resultant pressure force lies from O to B anywhere on the gate; the limiting condition being the situation when the resultant force passes through the hinged point O. Hence for the given condition point O becomes the centre of pressure.

The resultant also passes through the centroid of the pressure diagram, and the centroid lies at a distance `1/3** (AB) ` from the bottom point A.

Thus,

`x=â??*(AB)`

or, `AB=3*x`... (i)

Also,

`h=AB*sin45`

or,`h=3*x*sin45`

or,`h=2.4 m`

Now, the hydrostatic pressure force P is,

P = area of pressure diagram ** width of gate.

Assuming unit width of the gate,

`P=(1/2 *AB*wh*h)*1`

`=180.11 KN`

SOLUTION:

The gate would tip about the hinge point O when the line of action of the resultant pressure force lies from O to B anywhere on the gate; the limiting condition being the situation when the resultant force passes through the hinged point O. Hence for the given condition point O becomes the centre of pressure.

The resultant also passes through the centroid of the pressure diagram, and the centroid lies at a distance `1/3** (AB) ` from the bottom point A.

Thus,

`x=â??*(AB)`

or, `AB=3*x`... (i)

Also,

`h=AB*sin45`

or,`h=3*x*sin45`

or,`h=2.4 m`

Now, the hydrostatic pressure force P is,

P = area of pressure diagram ** width of gate.

Assuming unit width of the gate,

`P=(1/2 *AB*wh*h)*1`

`=180.11 KN`

a) Write the pressure balance equation between points 1 and 2

`P_1+rho_0 g (0.5)-rho_m g(0.1)-rho_a g (0.6)+rho_B g (0.5)=P_2`

b) Find the pressure at point 2

`200** 1000+850**9.81**0.5-13600**9.81**0.1-1**9.81**0.6+1100**9.81**0.5=P_2`

or, `P_2=196217.3` Pa

c) Write the pressure balance equation between points 2 and 3
`196217.3+1100**9.81**0.1-13600**9.81**0.3-1000**9.81**0.3=P_3`

or, `P_3=154328.6` Pa

a) Write the pressure balance equation between points 1 and 2

`P_1+rho_0 g (0.5)-rho_m g(0.1)-rho_a g (0.6)+rho_B g (0.5)=P_2`

b) Find the pressure at point 2

`200** 1000+850**9.81**0.5-13600**9.81**0.1-1**9.81**0.6+1100**9.81**0.5=P_2`

or, `P_2=196217.3` Pa

c) Write the pressure balance equation between points 2 and 3
`196217.3+1100**9.81**0.1-13600**9.81**0.3-1000**9.81**0.3=P_3`

or, `P_3=154328.6` Pa

SOLUTION:

Here, head of oil equivalent to negative pressure of 8250 N/`m^2` is,

`h=p/(rho*g)`

`=8250/(0.7**9810)`

`=1.2 m`

This negative pressure will reduce the oil head above the top edge of the gate from 1.8 m to 1.2 m i.e by 0.6 m.

This free surface is known as imaginary free surface. 

Now, the distance of centroid of plate from the free surface is,

`bar (h)=0.6+(3.6/2*sin45)`

`bar (h)=1.873 m`

And,

Area,`A=3.6**3.6=12.96 m^2`.

The pressure force is,

`P=rho*g*A*bar (h)`

`=700**12.96**9.81**1.873`

`=166690N`

Let, this force acts at a depth  of `h^**` from the imaginary surface.

`h^**=(I_G sin^2theta)/(A*bar (h)+bar (h)`

`I_G=`moment of inertia

`I_G=(b*d^3)/12`

Thus,

`h^**=((3.6**3.6^3)/12*sin^2(45))/(12.96**1.873)+1.873`

`h^**=2.16 m`

Now, taking moments about hinge A,we get,

`F**3.6sin45=P**(2.16-0.6)/(sin45)`

or, `F=144465 N`

This is the vertical force to be applied at the lower end of the gate to open the gate.

SOLUTION:

Here, head of oil equivalent to negative pressure of 8250 N/`m^2` is,

`h=p/(rho*g)`

`=8250/(0.7**9810)`

`=1.2 m`

This negative pressure will reduce the oil head above the top edge of the gate from 1.8 m to 1.2 m i.e by 0.6 m.

This free surface is known as imaginary free surface. 

Now, the distance of centroid of plate from the free surface is,

`bar (h)=0.6+(3.6/2*sin45)`

`bar (h)=1.873 m`

And,

Area,`A=3.6**3.6=12.96 m^2`.

The pressure force is,

`P=rho*g*A*bar (h)`

`=700**12.96**9.81**1.873`

`=166690N`

Let, this force acts at a depth  of `h^**` from the imaginary surface.

`h^**=(I_G sin^2theta)/(A*bar (h)+bar (h)`

`I_G=`moment of inertia

`I_G=(b*d^3)/12`

Thus,

`h^**=((3.6**3.6^3)/12*sin^2(45))/(12.96**1.873)+1.873`

`h^**=2.16 m`

Now, taking moments about hinge A,we get,

`F**3.6sin45=P**(2.16-0.6)/(sin45)`

or, `F=144465 N`

This is the vertical force to be applied at the lower end of the gate to open the gate.

SOLUTION:

Given,

Diameter of circular opening, `d=3 m`

Thus area, `A=(pi*d^2)/4=7.068 m^2`

Depth of centre of gravity, `bar (h) =4 m`

Thus, the force on the disc is,

`F=rho*g*A*bar (h)`

`=1000**9.81**7.068**4`

`F=277368 N`

And the point of application of this force is,

`h^** =I_G/(A*bar (h))+bar (h)`

Where,

`I_G=`moment of inertia of the disc through it's centre of gravity.

For circular disc,

`I_G=(pi*d^4)/64`,

Thus,

`h^**=((pi*d^4)/64)/((pi*d^2)/4*bar (h))+bar (h)`

`=pi* d^2/(16**4)+4`

`h^**=4.14 m`

Thus the force acts at a distance  of 4.14 m from the free surface of water. Now, taking moment of this force (i.e, torque) about a horizontal diameter, we get,

`=F*(h^** - bar (h))`

`=277368**(4.14-4)`

`=38831 Nm`

Hence, a torque of 38831 Nm must be applied on the disc in clockwise direction to maintain the disc in equilibrium position.

SOLUTION:

Given,

Diameter of circular opening, `d=3 m`

Thus area, `A=(pi*d^2)/4=7.068 m^2`

Depth of centre of gravity, `bar (h) =4 m`

Thus, the force on the disc is,

`F=rho*g*A*bar (h)`

`=1000**9.81**7.068**4`

`F=277368 N`

And the point of application of this force is,

`h^** =I_G/(A*bar (h))+bar (h)`

Where,

`I_G=`moment of inertia of the disc through it's centre of gravity.

For circular disc,

`I_G=(pi*d^4)/64`,

Thus,

`h^**=((pi*d^4)/64)/((pi*d^2)/4*bar (h))+bar (h)`

`=pi* d^2/(16**4)+4`

`h^**=4.14 m`

Thus the force acts at a distance  of 4.14 m from the free surface of water. Now, taking moment of this force (i.e, torque) about a horizontal diameter, we get,

`=F*(h^** - bar (h))`

`=277368**(4.14-4)`

`=38831 Nm`

Hence, a torque of 38831 Nm must be applied on the disc in clockwise direction to maintain the disc in equilibrium position.