First slide

Pressure in a fluid-mechanical properties of fluids

Question

A smooth gate is kept in equilibrium by applying a horizontal force. What is the value of y so that no horizontal reaction force acts at the pivot?

Difficult

A

$\frac{h}{3}$
B

$\frac{h}{6}$
C

$\frac{2 h}{3}$
D

$z e r o$

Solution

The thrust force due to pressure act at center of gravity'
$∴ F_{t h} = P A = (ρ g \frac{h}{2}) (h w) [h e r e w = w i d t h o f g a t e]$
For translational equilibrium :
$F_{t h} + R_{x} = F I f h o r i z o n t a l r e a c t i o n f o r c e i s a b s e n t, R_{x} = 0 \Rightarrow F_{t h} = F$
For rotational equilibrium about hinge point :
$T o r q u e d u e t o F_{t h} = T o r q u e d u e t o F \Rightarrow \int_{0}^{h} (ρ g (h - y')) (w d y') (y') = (F) (y) \Rightarrow ρ g w \int_{0}^{h} (h y' - y'^{2}) d y' = (F) (y) \Rightarrow ρ g w (\frac{h}{6}) = (F) (y) \Rightarrow (ρ g \frac{h}{2}) (h w) (\frac{h}{3}) = (F) (y) \Rightarrow (F_{t h}) (\frac{h}{3}) = (F) (y) \Rightarrow y = \frac{h}{3}$
The center of liquid thrust force passes through $y = \frac{h}{3}$

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