(III) A beaker of liquid accelerates from rest, on a horizonta...

Question

(III) A beaker of liquid accelerates from rest, on a horizontal surface, with acceleration α to the right. How does the pressure vary with depth below the surface?

Accepted Answer

Welcome back. Everyone in this problem. A rectangular tank filled with oil is accelerating uniformly from rest along a straight path on a flat surface with an acceleration A to the lift, the surface of the oil forms an angle phi equals the arc tangent of A divided by G with the horizontal, determine the variation in pressure with depth below the oil surface. For our answer choices A says it's P not plus row A where P is the pressure at the surface of the oil row is the density A is the acceleration to the lift and H is the depth below the oil surface. B says it's PN row multiplied by the square root of G squared plus A squared multiplied by H where G is the acceleration due to gravity. C says it's P nat row. GH and D says it's PN row multiplied by G A multiplied by H. Now, if we're going to figure out the variation in pressure, OK. It would probably help to try to visualize what's going on here. So we're saying that we have a rectangular tank. OK. Let's say that this is our tank and the surface of oil the surface of the oil forms an angle of five degrees. OK with the horizontal. So let's say that's five. OK. And we also know that it has an acceleration to the lift, it's been accelerated to the lift. And from this, we want to figure out what the variation in pressure would be. So how are we going to determine the variation in pressure? Well, assuming that the tank, as the problem says, it is filled with oil, it's accelerating uniformly from rest along a straight path on a flat surface. And that the acceleration is to the left. Then that pressure, that pressure variation is going to be equal to the initial pressure or the pressure at the surface of the oil plus. OK. The pressure at a depth edge below the oil surface which we can call P prime. Now what do we know about our pressure? Well, again, recall that pressure is generally equal to the density multiplied by the acceleration due to gravity multiplied by the height. So now that tells us then that our pressure P is going to be equal to peanut plus or density which remains the same multiplied by G in this case or effective acceleration given that we have the acceleration a to the left acting on our frame multiplied by H prime. OK. With where H prime is going to be the depth measured perpendicular to the tilted surface of the oil. So in order to figure out what our, our variation in pressure is we'll need to find the effective acceleration. OK. As, as as G prime. And we'll also need to find the depth measured perpendicular to the tilted surface of the oil prime. Let's start with our effective acceleration. What do we know here? Well, in the accelerated frame of the tank? OK. Auo force a in the direction opposite to the acceleration resulting in a pseudo acceleration. So the effective acceleration is going to be the result and force of the acceleration A and the acceleration due to gravity G. In other words, on our diagram, it's going to be G prime which is shown here. OK. And we know that G prime is at an angle of five degrees with our vertical, which is where G is. So how can we figure out what G prime is going to be? Well, if we think about it here. OK. Notice that or acceleration to the left, if we were to bring it down, it forms a right angle with G. And thus we can use that to figure out what G prime is going to be. OK? Let me just to bring it down a bit and make it look a bit neater. But we can use this to figure out what G prime is going to be because here at this point using what we know about right triangles solving for G prime. Yeah, then G prime squared is going to be equal to G squared plus A squared, which means that G prime is going to be equal to the square root of G squared plus A squared. So now we have an expression for G prime. Next, we need to find one for H prime. Since it's the depth measured perpendicular to the tilted surface of the oil. Again, let's try to put that on our diagram. I'm going to use a different diagram. OK. So let me come down here, let me just make a new one just, just so it looks a bit clearer. But again, we have our oil, OK? And we know it makes an Anglo phi with the horizontal but no in red. OK. We're interested with the distance that's perpendicular to the tilted surface. So if I were to just put this here, a parallel line below, OK, then we're interested in this distance prime. Now how can we figure out what age prime is? Well, so far we know that there is also going to be a distance vertically going downwards a distance from our tilted surface. And when we think about it, since prime is perpendicular to the surface, a red angle is formed between them. OK. And the angle that H prime forms with the vertical is going to be phi. In other words, this shaded part of our diagram here shaded in red, OK is going to be, let me put an arrow there. It's going to be an Phi. OK. Now since we have our angle, Phi here, if we can figure out what age prime is or sorry, if we can substitute Phi here, what we know about Phi use what we know about Phi that should be able to help us to solve for H prime. Now, when we look at it here, since Phi is right here and H prime is adjacent to Phi, then solving for prime, that means H prime is going to be equal to H multiplied by the cosine of Phi. OK. Because H is our hypotenuse in this situation, it's opposite to the right angle. Now, we don't know what the cosine of Phi is, but we also know, OK, before we were told that Phi equals the inverse tangent of a divided by G. So that means the tangent of Phi equals a divided by G. And we can find Phi from what we know about the acceleration and the acceleration due to gravity because it's going to be the same angle. Let's go back to our previous diagram. How could we find Phi here? Well, no, OK. Notice that for Phi Phi is equal to sorry, the cosine of Phi from our previous diagram is going to be equal to the adjacent side to Phi G divided by the hypotenuse G prime. So in other words, in other words, we also know that the cosine of Phi is equal to G divided by G prime. So with that idea that must mean then that H prime is going to be equal to H multiplied by the cosine of Phi, which we can replace as G divided by G prime, which we found to be G squared plus A squared. And thus, we could use this expression for H prime. So now we have an expression for H prime, we already found one for G prime. Thus, we can solve for our variation in pressure because no substitute in what we know. OK. Because now we have expressions purely in terms of GAH and our density then P is going to be equal to P, not plus our density multiplied by G prime, which is G squared plus A squared multiplied by O by H prime, which we have now found to be GH divided by the square root of G squared plus A squared. Now we can cancel our squared terms here. OK. And then that means we're going to get P to be equal to PN plus row multiplied by G multiplied by H. Thus this expression, this equation is going to give us the is going to give us the variation in pressure with depth below the oil surface. If we go back to our answer choices, that means C is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

(III) A beaker of liquid accelerates from rest, on a horizontal surface, with acceleration α to the right. How does the pressure vary with depth below the surface?

Key Concepts

Hydrostatic Pressure

Acceleration and Fluid Dynamics

Pascal's Principle

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