A triangular gate with a base width of 2 m and a height of 1.5 m lies in a vertical plane. The top vertex of the gate is 1.5 m below the surface of a tank which contains oil of specific gravity 0.8. Considering the density of water and acceleration due to gravity to be $1000 k g / m^{3}$ and $9.81 m / s^{2}$ respectively, the hydrostatic force (in kN) exerted by the oil on the gate is

29.43

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Solution

The correct option is A 29.43

$¯ h = 1.5 + 1.5 \times (\frac{2}{3}) = 2.5 m$

$F_{R}$ = $ρ g ¯ h A$

$= (800) (9.81) (2.5) (\frac{1}{2} \times 2 \times 1.5)$

= 29430 N = 29.43 kN

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The correct option is A 29.43 ¯h=1.5+1.5×(23) =2.5m FR = ρg¯hA =(800)(9.81)(2.5)(12×2×1.5) = 29430 N = 29.43 kN

The correct option is A 29.43

$¯ h = 1.5 + 1.5 \times (\frac{2}{3}) = 2.5 m$

$F_{R}$ = $ρ g ¯ h A$

$= (800) (9.81) (2.5) (\frac{1}{2} \times 2 \times 1.5)$

= 29430 N = 29.43 kN