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
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
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.
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.
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
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
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
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,
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
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
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
calculate the dot product and the cross
product of two vectors.
