Question

A cubical vessel (open from top) of side L is filled with a liquid of density rho. What is the torque of hydrostatic force on a side wall about an axis passing through one of the bottom edges?

Answer 1

Answer:

Torque due to the pressure of liquid is given as

$\tau = \frac{\rho g L^4}{6}$

Explanation:

Lets take a small element of width "dx" in which length is "L"

the position of this element is "x" height above the bottom

So pressure due to liquid at this position is given as

$dP = \rho g (L - x)$

now we can find the torque on this small element with respect to bottom position is given as

$\tau = \int dF . x$

$\tau = \int (\rho g (L - x))(L dx) x$

$\tau =\rho g L \int (L x - x^2) dx$

$\tau = \rho g L(\frac{L^3}{2} - \frac{L^3}{3})$

$\tau = \frac{\rho g L^4}{6}$

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Answer 2

The torque of hydrostatic force on a side wall is (ρgL⁴)/6.

Explanation:

Let us consider small element 'dx' from one side of the cubical vessel.

The 'x' is at a distance L - x from the bottom of the vessel.

The area of the element is:

dA = l.dx

The pressure of the element is:

P = ρgx

The force acting on the element is:

dF = P × dA = ρgx × l.dx

The torque acting on the element is:

dτ = P × dA = ρgx × l.dx (L - x)

dτ = ρgL (x² - lx) dx

⇒ τ = ∫dτ = ρgL $\int^{L}_{0}$ (x² - lx) dx

τ = ρgL $\left[ x^3/3 - Lx^2/2 \right]^L_0$

τ = ρgL [L³/3 - L³/2]

τ = ρgL L³/6

∴τ = (ρgL⁴)/6