## Fluid Mechanics - Torque at hinge of closed vessel

The closed vessel contains water with an air pressure of 10 psi at the water surface. One side of the vessel contains a spout closed by a 6-inch diameter circular gate hinged along one side. Horizontal axis of the hinge is located 10 ft below the water surface. Determine the minimum torque that needs to be applied at the hinge to hold the gate shut.

Okay.
Here's what I am thinking, and please let me know if I am doing this correctly.

First, I'm trying to find the resultant Force = F. Here's where I am slightly confused. Do I take this body as a curved surface or not? because the gate is located at a spout which is circular. If not, then what I did was just

A = pi * r^2 = 9*pi in^2 = pi/16 ft^2
hc (vertical distance from fluid surface to centroid of area) = 10 ft
specific weight = 62.4 lb/ft^3

F = sp wt * hc * A = 122.52 lb

Torque (T) = F * radius = F * (3/12)ft = 30.63 lb-ft

Is that right?
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Recognitions:
Homework Help
 Calculate the velocity of efflux from the curved pipe opening using Bernoulli's Theorem. This water-flow will form a force at the circular gate trying to push it outwards, therefore equal and opposite torque is needed to counteract the torque due to water flow. The force due to water flow is given by $dAv^2$ where d is the density of water .The torque will be the moment of this force about the hinge of the circular gate. BJ