- Good morning!
Newton's Second Law of Motion is an equation.
- Hey guys.
- Hey Bo.
- Hi Bo.
♫ Flipping physics ♫
- Net force equals mass times acceleration
where both force and acceleration are vectors,
Newton's Second Law of Motion.
The sum of the forces or the net force
equals mass times acceleration
where both force and acceleration are vectors.
That is Newton's Second Law.
Now, let's talk about this symbol.
This symbol is the uppercase Greek letter sigma.
And it means the summation of.
So sigma F means the sum of the forces acting on the object,
or objects you're summing the forces on.
And M and A are the mass and acceleration
of the object or objects you're summing the forces on.
Now, the best way to understand Newton's Second Law
is to do an example problem.
So Bo, could you please read the problem.
Bo, please stop there, we'll get to part B in a little bit.
So we have a book with a mass of 1,627 grams
and we are applying a force of 5.0 newtons on it.
And we're trying to figure out all the forces
that act on the book while we are applying that force on it.
So, Bobby please list all the givens.
- Okay.
The force applied equals 5.0 newtons.
The force of friction equals 3.6 newtons.
The mass of the book equals 1,627 grams.
Oh, and the book starts out at rest,
so the initial velocity equals zero.
- And Billy, we're doing a problem with forces in it,
so what should we do next?
- We need to draw the free body diagram.
- Okay Billy, go ahead.
Tell me what forces to add to the free body diagram.
- Okay.
There is the force of gravity, which is down.
And the force normal, which is normal to
or perpendicular to the table, and a push, so it is up.
There's also the force you applied, which is to the right
and the force of friction is acting to prevent movement
or slow the book down, so it is opposite
the direction the book ends up moving,
so the force of friction is to the left.
- Very nice, Billy.
Part A asks for the magnitudes
of all the forces acting on the book.
And it turns out that two of those forces
are given in the problem statement.
The force applied and the force of friction.
And we have an equation for the force of gravity.
Bo, could you please figure out the magnitude
of the force of gravity acting on the book.
- The force of gravity equals the mass of the object
times the acceleration due to gravity.
So 1,627 times negative 9.81.
- Actually, we need to convert the mass to kilograms first,
because a newton is a kilogram
times a meter per second squared.
- And the acceleration due to gravity
is positive 9.81 meters per second squared.
- Right, so let's convert the mass.
Multiply 1,627 grams by one kilogram over a thousand grams
and we get 1.627 kilograms.
So the force of gravity equals 1.627 times positive 9.81,
which is 15.961
and with two sig figs that is 16 newtons.
- Great, now we have magnitude of the force of gravity.
In order to figure out the magnitude of the force normal
we need to use Newton's Second Law.
Now, because force is a vector, anytime you use
Newton's Second Law and you sum the forces,
you need to identify the direction
in which you are summing the forces.
Let's start by summing the forces in the Y direction.
Bobby, please look at the free body diagram
and tell me what forces are acting on the book
in the Y direction.
- In the Y direction there are two forces,
the force normal and the force of gravity.
- And Bobby, please tell me if those forces
are positive or negative.
- How would I know that?
- Remember, up is positive, down is negative.
- Oh yeah.
So the force normal is up and positive,
and the force of gravity is down and negative.
Thanks, Bo.
- You're welcome.
- So then, the net force in the Y direction is equal to
the positive force normal minus the force of gravity,
which is then equal to
mass times the acceleration in the Y direction.
Now, I know that this particular piece
sometimes can be confusing for students,
so let's take a closer look at it.
The net force in the Y direction
just means the sum of the forces in the Y direction,
so that's the force normal
which is positive because it's up,
minus the force of gravity, negative because it's down.
So that's the net force in the Y direction.
But the net force is always equal to
mass times the acceleration,
so we sum the forces in the Y direction,
that's the acceleration in the Y direction.
So the net force in the Y direction is both equal to
the force normal minus the force of gravity
and it's equal to mass times the acceleration
in the Y direction.
Therefore, the force normal minus the force of gravity
is equal to mass times the acceleration in the Y direction.
- Thanks, I was wondering about that.
But now it makes sense.
- Good.
Now, Billy,
what is the acceleration in the Y direction of the book
while I am applying a force on it?
- Positive 9.81 meters per second squared.
- Actually no Billy, that is the acceleration
due to gravity here on planet earth, or little G.
That's not the acceleration in the Y direction
of the book itself while I'm pushing on it.
So look at it again and tell me what is the acceleration
in the Y direction of the book while I apply a force.
- [Billy] Oh yeah, and the book doesn't move up or down
while you push on it,
so the acceleration in the Y direction of the book is zero.
- Yes.
The book doesn't move up or down,
which means the acceleration in the Y direction of the book
is going to be equal to zero.
Anything times zero is zero, which means
that the force normal minus the force of gravity
is equal to zero and now we can add the force of gravity
to both sides of the equation.
And therefore the magnitude of the force normal
equals the magnitude of the force of gravity,
which is approximately 16 newtons.
So again, Newton's Second Law helped us to figure out
the magnitudes of the forces.
The free body diagram shows us the directions.
That actually completes part A.
Bo, could you please read part B.
- Part B, what is the acceleration of the book?
- To find the acceleration of the book,
we need to use Newton's Second Law again.
Only this time we're going to
sum the forces in the X direction.
Bo, please look at the free body diagram
and tell me which forces are acting in X direction
and whether they are positive or negative.
- The force applied is to the right, so it's positive.
The force of friction is to the left, and so it's negative.
- And we know that the net force in the X direction
is equal to mass times the acceleration in the X direction,
therefore the net force in the X direction
is equal to the force applied minus the force of friction,
and it is also equal to
mass times the acceleration in the X direction.
And we're solving for the acceleration in the X direction,
so we can divide the whole equation by mass.
And we get that the acceleration
in the X direction of the book is equal to
the force applied minus the force of friction,
that quantity divided by the mass of the book.
And now we can substitute in numbers.
And we get 5.0 minus 3.6, all divided by 1.627,
which works out to be
0.86048
or 0.86 meters per second squared.
So Newton's Second Law helped us to figure out
the magnitudes of all the forces
and the acceleration of the object.
- [Bo] Mr. P? - Yes Bo?
- Could you please explain how
the dimensions for acceleration in that equation
work out to be in meters per second squared?
- Actually, you know what, Bobby?
Why don't you explain that one.
- Uh sure, dimensions wise,
the acceleration ends up being newtons divided by kilograms.
We can substitute
kilograms times meters per second squared for newtons.
Oh, we need to flip the guy and multiply,
and the guy in the bottom is actually
kilograms divided by one, so we flip the guy
and multiply by one over kilograms
and kilograms cancel out to leave
just meters per second squared.
- Exactly.
Now please remember, anytime you're working with forces,
to draw the free body diagram or force diagram.
One more thing.
Because all of our forces were constant,
that meant that the net force was constant,
which meant that the acceleration was constant.
The acceleration equaled a number.
Therefore we could use the
uniformly accelerated motion equations with this problem.
We could for example,
because we know the initial velocity of the book is zero,
the acceleration of the book
is 0.86048 meters per second squared,
we could use the UAM equations to figure out
how long it would take the book
to go through a displacement of 1.0 meters.
It's an example.
We're not gonna do that today.
However, we will certainly do things like that
in the future.
Thank you very much for learning with me today,
I enjoyed learning with you.