I want that to be true cost wise but can't wrap my head around it because it seems to fall apart in the extreme hypothetical case where at some point there just isn't enough water to exert 1000lbs. I ask this because the math says that structurally I can double the volume of the tank just by increasing the same materials in the width without any additional support required. And if I theoretically continue to halve until the tank only holds 1 ounce of water the that 1 ounce still exerts 1000 lbs on the long side. Same formula states that since only depth matters in pressure then if I halve the width of the tank (front to back) then the total force on the long side remains the same 1000 lbs. 43 is psi per foot of depth and average psi is at depth 1 foot in this case. If you need more help, please show us your work.I building a tank for water 2ft high, 2ft wide (front to back), 8 feet long (left to right) Common formulas dictate the total force on the long sides is approx 1000lbs (w x h x. Use the relationship \(\displaystyle F = P \times A\) to find the incremental force. column of water above \(\displaystyle y\), and What is the incremental Area (square feet) between \(\displaystyle y\) and \(\displaystyle y+dy\)?įind the pressure \(\displaystyle P(y)\) as the weight of a 1-sq.ft. \(\displaystyle y_2=8 \ln 18\) is the top.Īssuming also that the tank is full, we will find the total force by integrating the incremental Force \(\displaystyle y_1=0\) is the bottom, and IT IS NOT CLEAR WHICH IS THE TOP AND WHICH IS THE BOTTOM, but in the absence of that bit of information I will assume that One of the bounding surfaces is a vertical plane, and the boundaries of that surface are four curves or lines in the xy-plane. Click to expand.The tank is a volume bounded by surfaces.
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