These are different statements because hydrostatic pressure is being applied to the entire surface of the tank (and hence the entire surface is pushing back by Newton's $3$rd law), so the force at the bottom is not the only thing coming into play in the force balance. The point is that the net vertical force on the water must be zero, which is distinct from requiring that the force from the tank bottom be equal to gravity. This might seem surprising, however, because we intuitively expect the tank to only need to feel the force necessary to support the water's weight- so where's the disconnect? This can be calculated in a similar way as AA' was analysed except here the same pressure act on projected area AC.As others have discussed, the computations are computing different things, but only the second approach yields the total force on the bottom of the tank. It is a semicircle of radius r in case of a sphere hence area $\pi r^2/2$Īs pointed out in the comments, there is a vertical downward acting component on surface ABC. It is a rectangle of length L and breadth R-r in case of a cyinder of length L Hence area $L(R-r)$\ This area needs to be visualised from the side view. Secondly, the area you are concerned with can be the cross section (AA') normal to the pressure (refer fig). So I find the average pressure acting on the center of the surface found by The surface AB and BC are identical and have indentical pressures therefore cancelling each other out.īut below A there is hydrostatic force component though the pressure is not constant as one proceeds down. Onwards I would change the perspective of the problem. Hello I would point out here Pressure is defined over an area so you would need to decide between a cylinder or a sphere (instead of circle) to fully define your problem.
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