 So what we're going to do now is we're going to solve a problem for a fluid container undergoing uniform linear acceleration. And if you recall from the last segment what we did is we came up with an equation that enabled us to determine the angle at which the free surface would be at for the condition of uniform linear acceleration. So we're going to apply that now and the problem that we're going to look at is one whereby we have a container with fluid in it that happens to be on an incline and there are wheels underneath the container so we originally start with some fluid that looks like this but then it's undergoing gravity the effects of gravity and so the container starts to slide down a hill or roll down a hill and it is at some acceleration a what we're told is that the fluid has weight w and we're also told that the angle of this incline plane that the container is going down is at angle phi and the wheels we are told are frictionless the gravity vector is the earth's gravitational vector so that's minus g times k and the last thing second last our coordinate system we have x in that direction and z in that direction which would be the initial condition of the container would be in that frame and what we're told to solve for is this angle here theta which is the angle that the fluid when it is moving and accelerating down this incline plane in this container makes with respect to the horizontal and so what we're looking for we want to find theta and so that is the problem that we have and what we're going to do we're going to apply the relationship that we derived in the earlier segment which I'll flip back to it was this relationship here that enabled us to determine the angle for a container undergoing uniform linear acceleration so let's take a look at applying that now the first thing we're going to do we're going to resolve the acceleration components for the container going down the incline so let's work that out and in doing this we have a lot of trigonometry here we have our incline it's at an angle phi we have the gravity vector acting in the vertical direction and if that is phi then we know that this here is 90 minus phi that means that this angle here on this side is 90 minus phi and this angle here is phi I'm going to draw a right angle here so what we're going to do is decompose that what I'm going to do I'm going to get rid of that little phi in there because it's going to get in the way of my acceleration okay so we have that and that means that this angle here is phi so that's applying the geometry to figure out what is going on now what we're trying to do we want to decompose the acceleration we want to know the acceleration that the container is undergoing as it goes down the surface and for that we know the gravity vector we know phi and so we can determine that directly from a trig relationship that is g sine phi and so that is equal to the acceleration that the cart is going down this incline surface we can go ahead and what we can do is we can look at the forces so we know some of forces is equal to mass times acceleration and if we express this in the plane of the incline surface itself we get some of forces is equal to mass times the acceleration the mass is weight divided by the gravitational constant and the acceleration here is just g sine phi the g's cancel out and so what we are left with is the sum of forces is the weight times sine phi so that gives us one piece of information that is not really exactly useful in terms of determining theta so the next thing that we're going to do we're going to break down the acceleration vector in terms of our x and z coordinates and we're going to refer it back to what we did in the earlier segment where we're comparing to the free surface where there would be no acceleration of our container so let's go through and do that and then we'll apply the equation to determine the angle so acceleration with respect to fluid at rest and so we just saw that the acceleration was g sine phi and if we want to decompose that into the x and the z what we find is that we have ax and we will have an az now ax and the angle of this by the way is phi so we can write out that ax is equal to it will be the adjacent over the hypotenuse so that is cos phi multiplied by the vector here itself which was g sine phi and az is going to be it's going in the negative direction remember our coordinate system let's go back what did we say z was up and x was to the right so z is up and x is in that direction so x is positive that makes sense az is going to be a negative and that's opposite over the hypotenuse so that is sine phi multiplied by the hypotenuse which is g sine phi okay so that gives us the acceleration components and if you recall back let me go back a couple here this relationship here for the free surface with uniform acceleration we're now going to use that relationship so recall for uniform linear acceleration we had that the free surface could be defined in this way ax of the x component of acceleration divided by the gravity vector and scalar form and az so broken down into the two different components so let's take those and we take ax az and g and we'll go through and see what we get so plugging in the values we have theta is equal to inverse tan and in the numerator we have cos phi multiplied by g sine phi and then in the denominator we have g minus sine phi times g sine phi and what I'm going to do I'm going to rearrange these a little bit simplify so we can pull the g out and then we have a sine phi cos phi and that is divided by again I'll pull the g out and then we have one minus sine squared phi okay so if you remember back to your math courses we had trig relationships this is equal to cos squared phi so I can rewrite it with that substitution and here in the G's cancel out we have a cos and a cos so what we end up with is theta is inverse tan sine phi over cos phi which is the tan of phi and consequently what we end up with is theta is equal to phi which is kind of an odd result but what that tells us is that the angle of the surface we go back to our original problem statement if theta is equal to phi that means that the angle of the fluid does not actually go up like we've drawn it it actually just remains like this so it remains parallel to the surface under uniform linear acceleration so that's what we get from applying the equation for the free surface to an accelerating cart that is going down an incline so that's an example of applying the uniform linear acceleration equation to determine the shape of a free surface