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Show that the resultant of the hydrostatic forces acting on a sub- merged plane area $A$ is a force $P$ perpendicular to the area and of magnitude $P=\gamma A \bar{y} \sin \theta=\bar{p} A,$ where $\gamma$ is the specific weight of the liquid and $\bar{p}$ is the pressure at the centroid $C$ of the area. Show that $P$ is applied at a point $C_{P},$ called the center of pressure, whose coordinates are $x_{p}=I_{y} / A \bar{y}$ and $y_{p}=I_{x} / A \bar{y},$ where $I_{y}=\int x y d A$ (see Sec. 9.3 ). Show also that the difference of ordinates $y_{p}-\bar{y}$ is equal to $\vec{k}_{x} / y$ and thus depends upon the depth at which the area is submerged.

Physics 101 Mechanics

Chapter 9

Distributed Forces: Moments of Inertia

Section 2

Parallel-Axis Theorem and Composite Areas

Moment, Impulse, and Collisions

Cornell University

Hope College

University of Sheffield

University of Winnipeg

Lectures

04:30

In classical mechanics, impulse is the integral of a force, F, over the time interval, t, for which it acts. In the case of a constant force, the resulting change in momentum is equal to the force itself, and the impulse is the change in momentum divided by the time during which the force acts. Impulse applied to an object produces an equivalent force to that of the object's mass multiplied by its velocity. In an inertial reference frame, an object that has no net force on it will continue at a constant velocity forever. In classical mechanics, the change in an object's motion, due to a force applied, is called its acceleration. The SI unit of measure for impulse is the newton second.

03:30

In physics, impulse is the integral of a force, F, over the time interval, t, for which it acts. Given a force, F, applied for a time, t, the resulting change in momentum, p, is equal to the impulse, I. Impulse applied to a mass, m, is also equal to the change in the object's kinetic energy, T, as a result of the force acting on it.

08:32

Show that the system of hy…

05:19

Let $R$ be a lamina of uni…

So in this problem, where ask to show that the results of the hydrostatic forces acting on a submerged area a plane area A is a force perpendicular to the area of the magnitude that we were given before. And now we have to find it. Um, we need to find it as p bar A where again Gamma is the specific gravity and P bar is the pressure at the century of the area. Um and then we want to show that p is applied at the centre at the point C p called the senator pressure. So we're basically looking for a senator of pressure in this kind of on this surface that's underwater here at on some plane, slow plane. So again, I have Why prime here is a vertical down and we have from before just that, you know, P is all of those things. Okay, so here's how the pressure various, with the depth or the distance down the slope. And the force is again given as it was before. And we can integrate that and we get what we had before. But now if we if we define P bar as P evaluated at Why bar? So we could just say P p evaluator, Why? Bar is P bar, and we can see that this just this all stuff comes, becomes Pete bar, and then we have people are a Now, in this case, we have no moments acting, so we can use just the definition of the moment that was acting in the previous problem to find the, um the distance where the center of pressure is. So from with respect to where the central it is. So we have the moment about about the ex prime access or the access through that the X axis through the central is the force times the distance from the center, the Y position of the centre prep pressure minus y bar. And then we had this from before in the previous problem. So now we can just start and we have We know this right here and so we can plug in and we can see that the specific gravity cancels out a specific weight, cancels out, um, the sign the angle cancels out. And what we're left with is, um, why bar a times a center of pressure? The Y center, the y coordinate of the center of pressure times. Um, the area moment about the X access through the central and then plus the area times Why bar squared? Well, this is then banking. See back He was in the parallel. Access TM is the area moment about the X axis that is at the water level. So again, we're kind of going backwards here because we had this before, and then we had to subtract this and get this. So this just becomes what was given that center of pressure in the Y direction is the area moment about the X axis at the surface of the water divided by the, um why position of the central I know the distance down the slope of the central Times, the area of that, um, surface. We can do the exact same thing for the the other moment. So we have the other moment here. The force that would create that moment is the net force times the difference between the center of pressure in the X direction and the X coordinate of the centrally again plugging things in from before we and you see castling out. We get that and doing some rearranging we see that this exposition of the Senate of the senator pressure times the why, um the distance from the down the slope to the central eight times a equals I this area moment, um, and again, I should use at double prime here. Um, times turned the area moment of the, um the cross area moment about the central of between X and y, and then plus a times, uh, y bar expert. Right. And if we look ahead in the book, this is now just across moment about the X Y access that was defined at the at the surface of the water. And so then x the position of the exposition of the senator. Pressure is I X y all over. Why bar times a so we can figure out the the, uh, area moment properties about the water level at the surface of the and at the surface of the lake, or whatever it is, and then figure out how where the senator of pressure is. And again it will probably it will be down on a little bit below the centrally and again, a little shifted away in into the plane. And, um, just because the pressure varies linearly as we get deeper

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