A circular opening 3m in diameter in a vertical side of a tank is closed by a disc of 3m diameter which can rotate about a horizontal diameter. Calculate:

• force on disc
• torque required to maintain disc in equilibrium in vertical position ehen head of water above horizontal diameter is 4m

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 gravi....Show More

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.