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

[zerogoal]

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=([tex]\Pi[/tex]R^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*[tex]\Pi[/tex](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?

 

[pongo38]

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....