Find the pressure and force on the curved wall.
Given that the object is filled with liquid.
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see the diagram showing the cross-section (vertical) of the given tank. Width of the tank into the page is 'c'.
Pressure P at a depth h from surface DA = P = ρ g h (in all directions)
h = R(1 - Cos θ)
Infinitesimal area on the curved surface AEB: ΔS = R dθ * c
Since Pressure is a scalar and directionless, we can calculate the average. But we cannot total up the values at different heights to get total pressure.
Average pressure on the curved surface =
= Pressure at E (center) = ρ g R
We can find also by integration.
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The horizontal force on the curved surface will be same as the force on a vertical surface like CD kept jus in front of curved surface AEB.
F_x = ρ g R * (2R c) = 2 ρg R² c.
There is a vertical component of force too, which is vertically upwards. We find this by integration.
Fy = π/2 * ρ g R² c
Net force F on the curved surface AEB: √ [Fₓ² + Fy² ]
F = ρ g R² c √(2² + π²/2² ). This acts through point O (center of circle).
Angle between F and Fₓ : Φ. Tan Φ = Fy / Fx = π/4
We can find Fy by integration as below.
Take area dS1 above OE at θ from OA, and area dS2 below OE at θ from OB. So θ varies from 0 to π/2. Numerically dF2 is more as Pressure is more.
dF1_y = - dF1 Cosθ = - ρ g R² c (1 - Cosθ) cos θ dθ
dF2_y = + dF2 Cosθ = + ρ g R² c (1 + Cos θ) cos θ dθ
dFy = dF1_y + dF2_y = ρ gR² c (1 + Cos 2θ) dθ
On integration from 0 to π/2, we get Fy = π/2 * ρ g R² c (upwards).
Pressure P at a depth h from surface DA = P = ρ g h (in all directions)
h = R(1 - Cos θ)
Infinitesimal area on the curved surface AEB: ΔS = R dθ * c
Since Pressure is a scalar and directionless, we can calculate the average. But we cannot total up the values at different heights to get total pressure.
Average pressure on the curved surface =
= Pressure at E (center) = ρ g R
We can find also by integration.
======
The horizontal force on the curved surface will be same as the force on a vertical surface like CD kept jus in front of curved surface AEB.
F_x = ρ g R * (2R c) = 2 ρg R² c.
There is a vertical component of force too, which is vertically upwards. We find this by integration.
Fy = π/2 * ρ g R² c
Net force F on the curved surface AEB: √ [Fₓ² + Fy² ]
F = ρ g R² c √(2² + π²/2² ). This acts through point O (center of circle).
Angle between F and Fₓ : Φ. Tan Φ = Fy / Fx = π/4
We can find Fy by integration as below.
Take area dS1 above OE at θ from OA, and area dS2 below OE at θ from OB. So θ varies from 0 to π/2. Numerically dF2 is more as Pressure is more.
dF1_y = - dF1 Cosθ = - ρ g R² c (1 - Cosθ) cos θ dθ
dF2_y = + dF2 Cosθ = + ρ g R² c (1 + Cos θ) cos θ dθ
dFy = dF1_y + dF2_y = ρ gR² c (1 + Cos 2θ) dθ
On integration from 0 to π/2, we get Fy = π/2 * ρ g R² c (upwards).
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kvnmurty:
:-) :-) Hope that is right.
The above value can be obtained also by integration. Take two area elements dS1 and dS2 at θ from OA and OB respectively. θ varies from 0 to π/2.
F1_x and F2_x will be as:
dF1_x = dF1 * Sinθ = ρ g R² c (1- cosθ) Sinθ dθ
dF2_x = dF2 * Sinθ = ρ g R² c ( 1 + cos θ) Sin θ dθ
dFx = dF1_x + dF2_x = 2 ρ g R² c Sin θ dθ.
On integration we get Fx = 2 ρ g R² c
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