The question is: for a container with a large cylindrical lowerpart and a long t

Question

The question is: for a container with a large cylindrical lowerpart and a long thin cylindrical neck, the lower part of thecontainer holds 12.5m^3 of water and the surface area of the bottomof the container is 5m^2. The height of the lower part of thecontainer is 2.50 m and the neck contains a column of water 8.50mhigh. The total volume of the column of water in the neck is0.200m^3. What is the magnitude of the force exerted by thewater on the bottom of the container?
In solving this problem, I wanted to find the force of thecolumn of water then the bottom portion of water and add them tofind total force. I used Force = density*gravity*Volume foreach (1000*9.8*0.2 + 1000*9.8*12.5). However, my answer isincorrect (124.46 kN) since the answer is 1 MN. Why does thisnot work? Also, how could using F = (Patm+density*gravity*height)*Area be the solution when there'sonly a skinny column for 8.50 meters of this container?
In solving this problem, I wanted to find the force of thecolumn of water then the bottom portion of water and add them tofind total force. I used Force = density*gravity*Volume foreach (1000*9.8*0.2 + 1000*9.8*12.5). However, my answer isincorrect (124.46 kN) since the answer is 1 MN. Why does thisnot work? Also, how could using F = (Patm+density*gravity*height)*Area be the solution when there'sonly a skinny column for 8.50 meters of this container?

Explanation / Answer

P = g h      pressure due to acolumn of liquid of height h and density P = g h1 + gh2       pressure inbottom cylinder increased by pressure at bottom of top cylinder(Pascal) P = g (2.5 + 8.5) = 1000 * 9.8 * 11 = 1.08 * 10E5 N/ m2
Adding to this the atmospericpressure P = 2.09 * 10E5 N / m2
F = P A = 1.05 *10E6 N     multiplying by area of bottom ofcontainer

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