A liquid of density `rho` is filled upto a geight of 'h' in a container as shown in firgure. Base of the contaimer is a square of side L. Ignoring the atmospheric pressure find

(a) pressure force `F_(1)` on its base.
(b) torque of force `F_(1)` about an axis passing through `C` and perpendicular to plane of paper.
(c ) Pressure force `F_(2)` on the vertical side wall DC.
(d) torque of force `F_(2)` about the same axis mentioned in part (b).
(e) point of application of force `F_(2)`

Answer

A, B, C

Explanation

(a) Base is horizontal. So, we can directy apply
`F_(1)=pA=(rho gh)(L)(L)`
`= rho gh L^(2)`
(b) Point of application of `F_(1)` is at a perpendicular distance `(L)/(2)` from the axis mentioned in the question
`:. tau_(F_(1))=F_(1)xxr_(_|_)=(rho gh L^(2)) ((L)/(2))`
`=(1)/(2) rho ghL^(3)`
(c) and (d) : side wall is vertical. Pressure is non-uniform. so, force and torque both will be obtained by integration.

Pressure at depth x,
`p=rho gx`
Area of small element shown in figure is
`dA =L(dx)`
`:. d F_(2)=(p)(dA)=(rho gx)(L dx)`
Perpendicular distance of this small force `d F_(2)` from the axis mentioned in the question is `(h-x)`. Therefore, small torque of force `d F_(2)` is
`d tau =(d F_(2)) (h-x)`
`=(rho gx)(h-x)(Ldx)`
Now, `F_(2)-int_(x=0)^(x=h) d F_(2)`
`tau_(F_(2))=int_(x=0)^(x=h)(d tau)`
Substituting the value and then integration, we get
`F_(2)=(rho g Lh^(2))/(2)`
and `tau_(F_(2))=(rho gLh^(3))/(6)`
(e) For point of application of `F_(2)`, we can apply
`r_(_|_)=(tau_(F_(2)))/(F_(2))` (from `C)`
`=((rho g LH^(3)//6))/((rho gH^(2)L//2)) =(h)/(3)`
Note that point of application of `F_(2)` is below the centre, as pressure is not uniform. It is increasing with depth.
`F_(1)` and `F_(2)` and their points of application are shown in figure below.
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