Calculating Hydrostatic Forces in a Steel Tank with a Viewing Glass Window

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The discussion focuses on calculating the hydrostatic force exerted on a circular viewing glass window in a steel tank filled with water. The tank dimensions are specified, with water reaching a depth of 3m and the window positioned 0.5m below the surface. The initial calculation of the force at the center of the window yields approximately 20,662.39N. Participants express uncertainty about whether to calculate the hydrostatic force at both the top and bottom of the window and suggest using definite integrals for a more accurate assessment. The conversation emphasizes the need for precise calculations to ensure no water spillage occurs during tank acceleration.
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Homework Statement



A steel tank is 6m long, 2m wide and 3.5m high. A viewing glass window is mounted at one side of the tank as shown. The tank contains water to a depth of 3m. The glass window is 0.2m in radius and its top is 0.5m below the water surface. Calculate the magnitude and location (distance below the water surface) of the horizontal hydrostatic force. If the tank is then accelerated on a level road along its length in the positive y direction, find the maximum acceleration if no water spillage is allowed. Density of water is 1000kg/m^3. The second moment of the area for a circle is I_xc=I_yc=(\PiR^4)/4, where R is the radius of the circle.


Homework Equations


Force=density*gravity*height*area


The Attempt at a Solution



Force at the centre of circular window= 1000*9.8*0.7*[(1.75*6)-0.5*\Pi(2^2)]
=20662.39N


I'm not sure if i need to get the hydrostatic force at the base and top of the circular window in order to get the hydrostatic force. Anyone can help me with this?
 

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I think you need a definite integral with geometric limits at the top and bottom of the window. Consider an elemental strip thickness delta x distance x from the water surface....
 

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