Question

A square gate of size 4 mx 4 m is hinged
topmost point. A fluid of density p fills the space
left of it. The force which acting 1 m from lowest
point can held the gate stationary is
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g
256pg​

Answer 1

Answer:

The force which acting 1 m from lowest  point can held the gate stationary is

$F = 2.84 \times 10^5 N/m^2$

Explanation:

Torque due to force of liquid on the gate is given as

$\tau = \vec r \times \vec F$

so we have

force on small element of the gate is given as

$dF = P . dA$

here pressure at y depth below the top is given as

$P = \rho g y$

now the area element is given as

$dA = 4 . dy$

Now we have

$\tau = \int y(\rho g y) (4 dy)$

$\tau = 4 \rho g \int y^2 dy$

$\tau = 4\rho g (\frac{4^3}{3})$

$\tau = (4 \times 1000 \times 10)(\frac{64}{3})$

$\tau = 853333.3 N/m^2$

now this torque is counter balanced by the torque due to applied force

This applied force is at 1 m above the bottom so we have

$\tau = F. (4 - 1)$

so we have

$3 F = 853333.3$

$F = 2.84 \times 10^5 N/m^2$

#Learn

Topic : Torque due to liquid pressure

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