Gate AB in figure below is a homogeneous mass of 180 kg, 1.2 m wide into paper, resulting in a smooth bottom B. For what depth h will force at point B will be zero?

SOLUTION:

Here,

The total pressure force on the left side of the gate due to glycerine is,

`F_1=rho*g*A*bar (h)`

`=12360**(1**1.2)(2-1sin60+(1sin60)/2`

`=12360**1.2**(2-(sin60)/2)`

`=23241.56 KN`

Also,

`(h_1)^**=(1.2**1^3*sin^2 60)/(12**(1.2**1)(2-(sin60)/2))+(2-(sin60)/2)`

`=1.6069 m`

Distance of this point of application of force from A is,

`=1-(2-1.6069)/(sin60)`

`=0.546 m`

Similarly, Force due to water is,

`F_2=rho*g*A*bar (h)`

`=9.81**(1.2**1)(h-0.5sin60)`

`=11.772 (h-0.....Show More

SOLUTION:

Here,

The total pressure force on the left side of the gate due to glycerine is,

`F_1=rho*g*A*bar (h)`

`=12360**(1**1.2)(2-1sin60+(1sin60)/2`

`=12360**1.2**(2-(sin60)/2)`

`=23241.56 KN`

Also,

`(h_1)^**=(1.2**1^3*sin^2 60)/(12**(1.2**1)(2-(sin60)/2))+(2-(sin60)/2)`

`=1.6069 m`

Distance of this point of application of force from A is,

`=1-(2-1.6069)/(sin60)`

`=0.546 m`

Similarly, Force due to water is,

`F_2=rho*g*A*bar (h)`

`=9.81**(1.2**1)(h-0.5sin60)`

`=11.772 (h-0.5 sin60)`

And,

`(h_2)^**=(1.2**1^3**sin^2 60)/(1.2**1.2**bar (h))+bar (h)`

`=0.0625/(bar (h))+bar (h)`

Also, Distance of this point of application of force from A is,

`1-(h-(h_2)^** )/(sin60)`

`=1-1/(sin60)*(h-(0.0625/(bar (h))+bar (h)))`

Where,

`bar (h)=(h-0.5sin60)`

Thus,

`(h_2)^**=1-1/(sin60)*(h-(0.0625/(h-0.5sin60)+h-0.5sin60))`

`=0.5+0.125/(h-0.5sin60)`

Vertical distance between c.g and A =`0.5cos60`

Now,taking the moment about the hinge A,

`sumM_A=0`

or,`23286.24**0.546+1765.8**0.5cos60 -11772 (h-0.5sin60)*0.5+0.125/(h-0.5sin60)=0`

Solving this,

`h=2.5 m`