Dot product and cross product of two vectors (Fluid Dynamics w/ Olivier Cleynen)

In this problem we study two concepts that are important for mechanics: the dot product and the cross product. What we’re looking at is a boat from above that is moving at a velocity V in this direction towards the top here. And it’s a sailboat and the wind is pushing the boat; and the wind is coming from the bottom left and pushing the boat with a force F over here. And we know the length of F, the angle theta, the speed V, and we try to answer two questions. One is: what is the power that is contributed by the wind? The wind is helping the boat along, which power is the wind helping [with]? and the second is: what is the moment exerted by the force due to the wind (twisting), the moment around the center of gravity which is here: how much twisting, twisting force if you want, this force is exerting around the center of gravity? You may be thinking this has in fact little to do with fluid mechanics and I would agree with you: this is more solid mechanics.

It’s a boat, its wind for sure, but it’s not a lot of fluid mechanics. Not a lot of fluid flow in there. But the fact is this exercise helps us very much to practice with two very important concepts: the dot product and the cross product. I’ve added [them] in the appendix at the end of the lecture notes where you can check out definitions for those two things. We’ll practice a little bit with those two concepts right here. Okay so let me move this down here so we can keep it in sight, and let’s take a look first at the power, the power that’s requi… that’s contributed by the wind on the boat, yeah? The work created by an object pushing another one is written W and the power as work would be then W dot, you know, the rate of work per second, if you want, the rate of work in time.

So 01:56 - the power W dot is the dot product of the force and the velocity. The dot product of two vectors is a number —it’s a scalar, a scalar value— and so it’s okay to have two vectors on the right side of the equation and on the left side just have a number over here. The unit of work is would be joules and the unit of work per second would be joules per second or watt, yeah? Joules per second may sound intuitive but the unit of “watt” is a bit confusing because it’s also written W — so you have W dot that’s written in W which may be a bit confusing. What is the dot product of F and V? Well let me draw a diagram perhaps down here below, of V first –this is the velocity at which the boat is moving, this is V here– and then let’s take a look at F. F is to the side like so, yes? At an angle theta right here. So this is F here.

What is the dot product of F and V? Well it is the amount by which F is 03:13 - contributing to V. The force exerted by the wind it has basically two components. One is sideways, and the part of the force that’s sideways is just pushing the boat sideways, it’s not helping the boat because the boat is moving straight along. And the component that’s along with the velocity of the boat is the one that’s actually contributing energy to the boat, yes? So what we want to do is to take the length of F, I will take the vector F and project it onto V and this would be here, this will be “F in the vertical direction” which in this case, if I look at my coordinates, is Fy over here. This would be the vector Fy like so. So we project F onto V, and the component of F which is along V multiplied by V, as lengths, that would be then the power as work provided by F on the boat, like so.

04:08 - So what we’re going to do here is to take a length F, yes? And project this length to take that length here and since we know that the angle theta here is on this side, to get that value here I need to take the sine of theta like so. So this is the component of F that’s along V, and I have to multiply this now by the length of V like so. So now we went from a dot product of two vectors, a number, to just lengths, numbers, over there. And so now it’s just a matter of adding the values, and we have F is equal to 13 kilonewtonsn so 13 times 10 to the power 3 here, times sine of the angle which is 30 degrees, like so, multiplied by the length on V, and V happens to be 1.5 meters per second like so. So you type this thing into your calculator, and I did this for you before, and you get 9 point 75 times 10 to the power 3 yes? And now comes the time to write the unit, and we had here power as work so that’s joules per second, or better said, watts, like so.

05:22 - And in engineering like always we have watts, kilowatts megawatts, and so and so forth, so I could just rewrite this as 9.75 here kilowatts like so. So 9.75, about 10 kilowatts would be about 13 or 14 horsepower, so not a very high power projected by the wind but this is just a sailboat so it makes sense to have this over here. All right so this is the power contributed by one force over one velocity, The dot product of two vectors, again check out the appendix if you need some more help on this. Let’s take a look now at the moment. The moment that is exerted by the force F around the center of gravity or more formally said about the center of gravity. The moment exerted by F about the center of gravity.

All right, let me first remake this diagram here, so that we can 06:25 - see better the situation. We have the center of gravity here like so and we have the force, the force is applying at a point that’s slightly shifted compared to the center of gravity, and this is F here. I want to represent what we call the arm in mechanics, which is the distance away from the center of gravity at which the force is applied. And this arm I’m going to represent it with a vector, and we call it this vector R, the radius vector or the arm vector if you want. And I’m going to say that the moment is a vector. And this vector is the cross product of two other vectors.

It is the cross product of 07:07 - R and F, in this order. F put at the end of R creates a torsion, an amount of twisting, which we call the moment, which is called M like so. The cross product of two vectors is a vector. It’s not a length, it’s a vector, and so we will have a direction as well as a length. We’re going to take a look at the length first, and then we’ll talk about the direction.

The length of M we’re going to write it 07:36 - like so, yeah? The length of M turns out to be the component of F which is perpendicular to R multiplied by R So I’m gonna say it is R multiplied by the length of F that is perpendicular to R. I’m going to write it like so? F perpendicular, why? Because this force here twists around the center of gravity so exerts a moment in this direction like so, this is M here, that is proportional to the component of it that’s perpendicular to R, like so. This force here F perpendicular and F contribute the same amount of twisting as would for example forces that long but it with the same perpendicular component over there, yes? R times F here as vectors, is a vector M which we could call, represent here, as a twisting [effort] and the length is R times F perpendicular, like so. And we can rewrite this now as the length of F and we can look at the angles that we have here this would be in this case F cos theta like so. And we can put numbers now in there, the radius turns out to be, [Olivier panics as he realizes he lost the data he needs] The radius turns out to be 2 meters if I remember correctly, the force is 13 kilowatt, erm, kilonewton I’m sorry, so 13 times 10 to the power 3 Newtons and we have over here the cosine of 30 degrees over here.

If you type this into your calculator, like I did before, 09:30 - here, you will get two point 2517 here times 10 to the power 4 and what is the unit of a moment it is a force at the end of a length, yeah? So Newtons at the end of meters: Newton-meters, right? So Newton meters, and again in engineering we like Newton-meters, kilonewton-meters, megaNewton-meters and so on so forth, and so I can rewrite this as 22.52 kilonewton-meters, and this is the length, the length of the vector M, which is like so. I start now about the direction of M because said M is a vector, and we calculated its length like so, but we want to know in which direction it’s pointing. Well it turns out that we represent moments as vectors, as arrows, yes? And we position the arrow so that as seen from the arrow then the moment will turn clockwise, like so. In this case M is turning in this direction here clockwise down into the paper, yeah, and so I put an arrow down towards the paper and this is the direction of M here.

So M is down 10:56 - into the paper through the table that’s sitting here and below my finger and we could represent it as a vector like so, like a cross like so, because that’s the tail end of an arrow that you see entering the paper like so. This would be the vector M here. And how do we write this more formally now? You could say M has actually three coordinates. M here as coordinates in X Y & Z. And let’s take a look at the coordinates that we have here. we have X to the left, Y to the top, and then Z into the paper like so. so the whole length of M which we calculated here at this point here, this whole length here is into the paper. It is in the Z direction.

And so M has 0 in X, 0 in Y, and then in 11:49 - the positive Z direction it has 22 point 52 as kilonewton-meter, yeah? So it’s a three dimensional vector problem, if you want, which sounds a bit overdone for such a simple mechanical problem yes? but it’s quite important that you are familiar with these notions before we move on further mechanics of fluids because at some point we’ll need both dot products and cross products of vectors. If you’re still insecure about these things there is a really good book I recommend to practice those things and it’s called Higher Engineering Mathematics by John Bird right here like so. And this is mathematics, but it’s mathematics for engineers, and the tone of it is very modest. And it’s quite easy to go through, it is mostly written as “Engineering Mathematics for People Who Have Seen It Before But It Was a Very Long Time Ago”. It’s certainly very helpful for me when I’m a bit rusted over these things.

So this is how you 12:53 - calculate the dot product and the cross product of two vectors. .