Good morning. Bobby, please remind me, what is our current equation for work,
the one from AP Physics 1? ♪ Flipping Physics ♪
Work equals F d cosine theta, where F is the force doing the work,
d is the displacement of the object, and θ is the angle between those two vectors.
Remember we use the magnitudes of force and displacement in the work equation.
(And Bo, what are the units for work?) Newtons times meters or joules.
Great. Thanks. Now, it is very important you realize this equation is for the work
done by a constant force. We will discuss work done by a non-constant force at a later date,
but not today. So, you can look forward to that. Yea! Yea? Yea.
Now that we are doing calculus-based physics, we have the r-position vector which identifies the
position of an object in 3 dimensions using unit vectors. Therefore we can define the work done
by a constant force as work equals force times change in r-position of the object times the
cosine of the angle between those two vectors. So, we are just replacing d with delta r?
Yes. Okay?
Also, now that we are doing calculus-based physics, work done by a constant force is defined
using the dot product which is also called the scalar product because the dot product is a
scalar. In other words, work done by a constant force equals the dot product of the force doing
the work and the displacement of the object. Uh … Can we review the dot product?
Yeah, I do not remember the scalar product at all. The scalar product of vector A dot vector B
equals A times B times cosine theta. Do you not remember that?
Billy and Uh… … Therefore, work done by a constant force
equals the dot product of force and displacement which equals force times displacement times cosine
theta. Just like we defined before. Billy and Right.
But what is the dot product? Thank you Bobby. Billy,
the dot product is the magnitude of one vector times the magnitude of the projection of the
second vector onto the first vector. What?
Right. Let’s look at two example vectors A and B. The dot product of A and B equals the magnitude
of A times B cosine theta. Where B cosine theta is the length of the projection of B onto A or,
in this case, the value of B in the x-direction. In other words, the scalar
product of A and B equals A times B cosine theta. Okay. So, A dot B equals the length of A times the
length of the parallel side of the right triangle or B cosine theta. I guess that makes sense.
Also, please remember that the dot product is commutative.
… What does commutative mean again? The order of the two vectors does
not matter. That is what commutative means. Oh right, A dot B equals B dot A. Commutative.
The order is irrelevant. Thanks. … That means we could look at our
vector diagram example as B dot A or the magnitude of vector B times the projection
of A onto B or A cosine theta. … That kind of makes sense. I mean I see how
A B cosine theta equals B A cosine theta. Sure. Right. Thank you. … Now let’s do a simple unit
vector example to help us recall how to solve the dot product using unit vectors. Vector A equals 4i
plus 3j and vector B equals 2i plus 1j. Bobby, please solve for the dot product of A and B.
Uh … I think … Is it like 4 times 2 plus 3 times 1?
Yeah, that sounds familiar. You multiply all the i values together and all the j values together
and then add those together. But, … Actually, I do not know why we do it that way, but it works
out to be 11. Yeah.
Okay. Let’s understand why this works. First off, just like if this were two standard
binomial expressions like the quantity 4 plus 3 times the quantity 2 plus 1,
we use the FOIL method to get 4 times 2 plus 4 times 1 plus 3 times 2 plus 3 times 1. We also
use the FOIL method on the dot product to get 4i dot 2i plus 4i dot 1j plus 3j dot 2i plus 3j dot
1j. Bobby, see if you can continue from there. …What does FOIL stand for again?
First, Outer Inner, Last. Oh right. Multiply the first
terms of the binomial expressions together, … then the outer terms of the binomial
expressions, … then the inner terms, … then the last terms, … and add that together. Math is fun!
Yeah. Yep. Bobby?
Uh. Okay. 4i dot 2i equals 4 times 2 times the cosine of the angle between those two directions.
Both i vectors are in the same positive x-direction, so the angle is zero.
The cosine of zero equals 1, so that is just 4 times 2. … 4i dot 1j equals 4 times 1 times
cosine of … well i and j are in the positive x and positive y directions respectively,
so the angle between those two is 90 degrees. … And the cosine of 90 degrees equals zero. Right,
so that term equals zero. … And the angle is the same for the next term, that is 3
times 2 times cosine of 90 and therefore equals zero. … The last term 3j dot 1j equals 3 times 1
times … again the cosine of zero degrees because j and j are in the same positive y-direction. … So,
that is why the dot product of A and B works out to be all the i terms multiplied together plus
all the j terms multiplied together. The other terms all work out to be zero
because they all have a cosine of 90 degrees. Correct Bobby. And going back to the projection
description of the scalar product. When the angle between the two vectors is 90 degrees,
the projection of one vector onto the other vector equals zero, so the dot product equals zero. There
is no projection of one vector onto the other one when the two vectors are normal to one another.
So, the dot product of two vectors at 90 degrees to one another equals zero. … Okay, let’s do an
example finding the work done by a constant force using unit vectors. … Let’s say we have an object
which goes through a displacement equal to 3.0 i plus 1.0 j meters while experiencing a force
equal to 4.0 i plus 2.0 k newtons. Bo, please find the work done on the object by the force.
Sure. Work equals the dot product of force and displacement
or the dot product of those two vectors. So, … 3 times 4 plus … 1
times 2 or … 14 joules. … … Uh, what? … Oh, that’s a k unit vector not a j unit
vector. Whoops. So, it’s 3 times 4 plus … 1 times zero plus zero times 2 and that equals 12 joules.
Correct Bo. That is something to be really careful to notice. You have to multiply all the i unit
vector coefficients together, or 3 times 4. … And then multiply all the j unit vector coefficients
together, or 1 times zero. Zero because the r position vector does not have a component in the
j direction. … And then multiply all the k unit vector coefficients together, or zero times 2.
Zero, because the force vector does not have a component in the k direction. … This really is
something my students often overlook, so please make sure to read the unit vectors carefully.
(Mr.p?) Yes Bobby?
Bo showed a lot less work than I did. Can we just show what Bo did?
Oh, yeah. Thanks Bobby. I had you do it that way in order to help us remember
why the shorter version Bo just did works. It’s fine to just show the work Bo did.
(Great. Thanks.) Alright, so far,
everything we have done has just been abstract numbers. Let’s look at a more real example.
Let’s say a 6.9 newton force is applied at 59° to the left of the -y axis to a shopping cart. The
shopping cart is displaced 7.0 m to the left. If this force applied keeps the shopping cart moving
at a constant velocity, what is the work done on the shopping cart? Billy, please start by putting
all of our known values into unit vector form. Well, the displacement, delta r, equals 7.0 m to
the left or negative 7.0i meters. … The force applied as given is neither directly in the x
or y-direction, therefore, we need to break it into its components. … The force applied
equals … in the x-direction the force applied is to the left so negative force applied in
the x-direction times unit vector i. … The force applied in the x-direction equals 6.9 times the
sine of 59 degrees because the angle, theta, is with the vertical. … In the y-direction,
the force applied is down, so minus the force applied in the y direction times unit vector j.
… The force applied in the y-direction equals 6.9 times the cosine of 59 degrees, cosine because,
again, theta is with the vertical. … Do you want me to find those numbers?
No, that’s okay Billy. Bobby can determine the work done by the force applied from here. Bobby?
Yeah. … Uh … The work done by the force applied equals the dot product of the force applied and
the object’s displacement, or the dot product of the two vectors Billy just
determined. … Multiplying the i terms together gives us negative 6.9 sine 59 times negative 7
plus … the j terms or negative 6.9 cosine 59 times … zero. Right, that whole term
equals zero. The i and j directions are normal to one another. And we get
41.401 newton meters or 41 joules with 2 sig figs. Very nice Bobby. You are correct. The work done
by the force applied on the shopping cart is 41 joules. … Billy, please determine the work
done by the force of gravity on the cart. Absolutely. The work done by the force of
gravity equals the scalar product of the force of gravity and the displacement of the cart which
equals … Well, the force of gravity is down. So, … negative mass times acceleration due to
gravity in the j direction … dot … negative 7.0 i. … Oh, that equals zero because i and j are at
90 degrees to one another and the cosine of 90 degrees equals zero. The force of gravity does
zero work on the shopping cart because it is normal to the direction of the displacement
of the cart. That makes sense. And Bo, what about the work
done by the force normal on the cart? Well, … that will also equal zero because the
force normal is up or in the positive j direction and also perpendicular to the displacement of the
cart. The force normal does zero work on the cart. Thanks. Lastly, let’s determine the work done by
the force of friction on the shopping cart. Bobby, please?
Well, … The work done by the force of friction equals the dot product of the force of friction
and the displacement of the cart. … Let’s draw the free body diagram. … Force normal is
up. Force of gravity is down. Force applied is 59 degrees to the left of the negative y-axis.
Force of friction is to the right, opposite the direction the cart is moving. … That means the
work done by the force of friction equals force of friction times unit vector i. … dot … negative 7.0
i. … That’s just negative 7 times the force of friction. So, we need the force of friction. …
Ah, the cart is moving at a constant velocity. So, we sum the forces in the x-direction to get
force of friction minus force applied in the x-direction all equals mass times acceleration
in the x-direction, however, because it is moving at a constant velocity, the acceleration of the
cart in the x-direction equals zero. So, the force of friction equals the force applied in
the x-direction. And that equals … well, force applied times the sine of theta because theta is
with the vertical. … Okay, so the work done by the force of friction equals negative 7
times force applied times sine theta or negative 7 times 6.9 times sine of 59 degrees or
negative 41.401 or … negative 41 joules with 2 sig figs. …
Uh, … why are the magnitudes of the work done by the force applied
and the force of friction equal, … except the work done by the force of friction is negative?
Good question. Let’s talk about what this means. Remember, the shopping cart is moving at a
constant velocity. … What we have shown is that, while the cart is moving at a constant velocity,
all the energy being put into the cart via the force applied,
is being dissipated by the force of friction. … And, notice that when we add up the work done by
every force acting on the cart while it is moving a constant velocity, we get zero. In other words,
the net work on an object equals zero when the object is moving at a constant velocity. … This
is called foreshadowing. … Blatant foreshadowing. I don’t get it.
Do you really think we are supposed to. I don’t know.
Thank you very much for learning with me today, I enjoyed learning with you.
I still want my 2 dollars!
2 dollars!