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The panel shown forms the end of a trough that is filled with water to the line $A A^{\prime}$. Referring to Sec. $9.1 \mathrm{A}$, determine the depth of the point of application of the resultant of the hydrostatic forces acting on the panel (the center of pressure).

Physics 101 Mechanics

Chapter 9

Distributed Forces: Moments of Inertia

Section 2

Parallel-Axis Theorem and Composite Areas

Moment, Impulse, and Collisions

Cornell University

Rutgers, The State University of New Jersey

Simon Fraser University

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.

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The panel shown forms the …

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An open tank is filled wit…

So now we need to find the center of pressure of a trough that is 1/4 circle in a side wall here that's filled to a level of just to the top of the court quarter circle. So the radius of the quarter circle is our and so I have set up a Y axis here at the top of the water level on positive down. So we need to determine the, um, the center, uh, the central right. And we can actually just look that up. And that is for our over three pie For a we'll freeze for 1/4 circle or for 1/2 circle is the same I And so now we have that and we know can look up the area moment about this axis here. And we looked that up. When we get pie are to the fourth over 16. Now we can get the area and that's pile before r squared. And so now we just need to do some algebra here. We need to take this so here and divided by these things. So all of, um, this part here, right? And lots of things cancel out. We get three R's down here. So we get just a are in the numerator and we get pi squared up with top But we get pied out here so we get a pie And then this four cancels with this 16 and we get that the whole overall we get that this is and then So this is four times four. So three pie are over 16. Is that where the center of pressure is? So this is a little over a little over nine, so this is a little over half way down, so somewhere around here, which seems reasonable

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