Force on the end of a tank Determine the force on a circular e...

Question

Force on the end of a tank Determine the force on a circular end of the tank in Figure 6.78 if the tank is full of gasoline. The density of gasoline is ρ = 737 kg/m³.

Accepted Answer

Welcome back, everyone. In this problem, a flat circular plate of radius R equals 0.50 m is mounted vertically with its topmost point at the free surface of water with a density row of 1000 kg per cubic meters. Find the hydrostatic force on the plate using our acceleration due to gravity G at 9.8 m per second squared. A says it's 2,0075 Newtons, B, 1,375 Newtons, C, 1,225 Newtons, and D 725 Newtons. Now let's try to visualize what's going on here to help us figure out this hydrostatic force. So we're told that we have a flat circular plate. OK, so let's say that this is our plate. And for our plate. OK, it will have a center. Here or we'll have our center here and from the center to any part on the circumference would be or radius R, which we know is 0.5 m. We also know that if we were to look at a depth of Y from the topmost point at the free surface of water, OK, then here, if I just draw a line across horizontally. OK, and let me extend that. Then if we also go from the top and let me just make these lines. Dotted just to distinguish them from our radios. Then from the topmost point to the tip of where the radius meets, whichever depth here in our circle it could be here, it could be here, wherever, then this depth will be why, OK, and no, if we think about it, if we go across as well from our center. Then our remaining distance from our depth Y to the center of the circle would then be equal to R minus Y. OK, so it means here that we can also represent R minus Y on our circle. So that's what we know so far. Now how is this supposed to help us? Well, recall that at a depth. Of Y, OK, and a depth of Y, then our hydrostatic pressure P of Y is going to be equal to row multiplied by G multiplied by Y, OK, where we already know row and G. Now if we think about it here, the plate is a vertical circle of a radius R whose topmost point is at the free surface. So that means the circle center is at a depth, OK, at depth, the circle center YC is going to be at a depth R. OK, where Y. Measuring from our topmost point is going to go from 0 down to 2R, OK? So, from the start, 0 all the way to the bottom of the circle, which would be 2 radii or the circle's diameter. Now, if we also think about it here, we can visualize our horizontal strip at this depth of Y. Because if I were to put that on our diagram here in blue, OK, then at a depth of Y. OK, let me just draw that here. At that depth of Y. Then we're going to have a horizontal strip in blue as X as a function of Y, OK, or we can imagine that we have a strip here. And for the strip, we also know that if we carry it across, OK, it's going to have a width as a function of Y, which we can call doubly of Y, OK? So now for this. For a horizontal strip, OK. OK. At a depth of Y. OK. Then if we look at our diagram, the length of our strip, OK, is going to be this part shaded in blue where, given that it forms our right triangle, we can use the Pythagorean theorem to get the length of our strip X of Y to be equal to the square root of R2 minus Y minus R2, OK. And in this case, since we know that our full width is going to be 2 multiplied by X of Y. Then this tells us that w of Y is going to be equal to 2 multiplied by the square root of R2 minus Y minus R2. Now why does this matter? Well, if we think about it here, remember we're trying to find a hydrostatic force, OK? And given that an infinitesimal strip at a depth Y of thickness Dy would have a force DF of P Y multiplied by W of Y with respect to Y, then our force, OK, is going to be. The integral Between 0 and 2 are of. Or or hydrostatic pressure row G Y multiplied by or with two multiplied by the square root of R2 minus Y minus R squared with respect to Y. So now we have a way to figure out the hydrostatic force. So let's go ahead and do that, OK? Now, let's try and use substitution to simplify our integrate. So solving by substitution. OK. By substitution, let's let you be equal to Y minus R. Now we need to change our bones. What do we know? Well, we know that the center of the circle corresponds to U equals 0. When Y equals 0 at the topmost point of the plate, then U is going to be equal to 0 minus r, which is negative r. When Y equals. Who are at the bottom of the plate, then u is going to be equal to 2 r minus R, which would be equal to R. So that means that our integration limits. Yeah Let me put that here. Thus, this means that our integration limits U is going to go from negative R all the way up to R. OK? These are going to be our new limits. And thus, we can say. That our force F is going to be equal to. The 2 multiplied by row G, take out our constants here multiplied by R multiplied by the integral between negative R and R of the square root of R2 minus U2 with respect to you. Now the good thing about this is that notice that the term, OK. Notice that our term here, the integral of between r and negative r of the square root of R2 minus u squared with respect to U is going to be the area of a semicircle of a radius r, and it equals a half of pi r 2. So in this case. We could rewrite this then as to row G multiplied by a half of pi r cued and when we simplify that, we get row GIR cubed. And now we have this expression for force, we can solve when R is equal to 0.50 m. Given that we have the rest of the information. Thus our force is going to be 1000 multiplied by 9.80, multiplied by pi, multiplied by 0.50 cube, which is going to be equal to 1,225 newtons. Therefore, that is the hydrostatic force. OK, that is going to be acting on the plate. If we look back at our answer choices, that means C is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Force on the end of a tank Determine the force on a circular end of the tank in Figure 6.78 if the tank is full of gasoline. The density of gasoline is ρ = 737 kg/m³.

Key Concepts

Hydrostatic Pressure

Force on a Surface in a Fluid

Density and Its Role in Fluid Statics

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