Computing the Hydrostatic Force



in this video we'll show how to set up an integral that finds the force exerted by the liquid on a vertical plate submerged into the given liquid in our case we have a circle of diameter 3 filled with gasoline halfway so the water level is right here and the first thing we do to compute the force is we create the coordinate system one dimensional pointing down so we get to choose the origin so let's place the origin at the level of the liquid so halfway in the tank so this is the origin this is going to be the XS X and then the boundaries for the liquid are 10 from 0 to 3 halves and now we're going to pick some arbitrary value of x between 0 & 3 halves and we're going to consider a thin strip of height DX and to compute the hydrostatic force we're going first to compute the hydrostatic force exerted on this trip we're going to call it DF and that force is equal to the pressure times the area and pressure is Rho G and the depth the depth here is exactly X because we chose the origin to be at the surface level so the pressure is Rho GX and to find the area we know we're going to be multiplying the width by this height DX and now we have to focus on finding this width so it is pretty clear that as exchanges this W also changes so that's why it's a function of X now let's take this problem out of this context and let's solve it geometrically we have a circle of radius three halves and we have this line segment of total length W so half of it will be half of W and now we're going to bring this measure from the center of the circle to the W which is in this coordinate system just X so the relation between X and W is that x squared plus half of W squared is equal to three half squared meaning W is equal to two times square root of three half squared minus x squared so this is the W as a function of X that we were looking for so that's the expression that goes here in the set up of the integral so what we have here is the force exerted on that one strip the total force will be the integral from zero to three halves and that and the question so the answer to the question is right here so this is the integral that computes the total force exerted on the on one end of the tank

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