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Firstly, using the principle of hydrostatic pressure, the initial force acting on the door can be calculated using the formula \(F=蟻ghA\), where \(蟻\) is the water density, \(g\) is the gravitational constant, \(h\) is the height of water surface above the centroid of the door and \(A\) the area of the door. \nHere, \(蟻=1000kg/m^3\), \(g=9.81m/s^2\), \(h=1.75m\) and \(A=1*1.5m^2\). Substituting these values into the formula gives \(F=25927.25N\).\nThe line of action is calculated by finding the center of pressure, which is given by the formula \(h_c=\frac{(I_g+Ah^2)}{Ah}\), where in this problem the area moment of inertia (\(I_g\)) of a rectangular surface area about its own centroidal x axis is simply \(bh^3/12\), where \(b\) is the base width of the rectangle and \(h\) the height (of the door in this case). A is the area of the door and hs the distance of the centroid of the door from the water surface. Hence \(h_c = 2.375m\) below the water surface.

Next, the water pressure on the top surface of the door now becomes the sum of the atmospheric pressure and the gauge pressure i.e., 1 atm + 0.3 atm = 1.3 atm = 1.3 * 101325 Pa. If we consider the end of the door (consider it as \(h\)) to be at the 1m beneath the surface of water, then the pressure at that point is \(蟻gh + P_{atm}\) = 1000 * 9.81 * 1 + 1.3 * 101325 = 1.33 * 10^5 Pa. \nSince the pressure varies linearly along the door, the average pressure on the door is the average of the pressure at the top and bottom i.e. (pressure at top + pressure at bottom)/2. Hence the total force exerted is the average pressure times the area of the door. The line of action of the force i.e., center of pressure will still be at the previously calculated \(h_c=2.375m\).

Finally, ratios of the initial resultant force to the new resultant force and the initial line of action to the new line of action for different values of surface pressure can be plotted. \(F/F_0\) and \(y'/y_c\) will be the y and x coordinates respectively.