AP Physics 1 & 2 Newton's 2nd Law of Motion

Hi, I am Dan Fullerton. Welcome back to Educator.com.0000

Let us talk about Newton's Second Law of Motion.0004

Now our objectives are going to be to draw and label a free-body diagram showing all the forces acting on an object and also draw a pseudo-free body diagram showing all components of forces acting on an object.0006

We will explain the relationship between acceleration, net force, and mass of an object, and use Newton's Second Law to solve a variety of problems.0020

Finally, we want to make sure we understand the difference between mass and weight, and the conditions required for equilibrium.0028

Free-body diagrams or FBD's -- these are tools that we use to analyze physical situations.0037

What they do is they show all the forces acting on a single object.0044

The object itself may be drawn as a dot.0049

Some folks like to draw it as a box. It does not matter, either one.0052

Now when you draw a FBD, choose the object of interest and draw it as either a dot or a box.0058

Then you are going to label all the external forces acting on the object and only forces go on that diagram.0064

Finally, sketch a coordinate system showing the direction of the object's motion as one of the positive axis.0070

For example, let us take a look at a circus elephant falling off a tight rope -- sad story -- it is just pretend, do not worry.0077

Neglecting air resistance -- draw a free-body diagram for the falling elephant.0085

I am going to use my amazing physics artistic skills to draw an elephant.0089

There it is in FBD terms and I am going to label all the forces acting on it.0094

The weight of the elephant -- the force of gravity -- which I typically write on FBD as mg -- the force of gravity on an object on a FBD can save yourself a little bit of work if you write weight as mg.0100

Or, we could draw it as a box -- there is our elephant and the one force acting on it is the weight of the elephant pulling it down.0115

How about if we had the falling elephant with air resistance?0127

Well, there is my elephant again.0131

We still have the elephant's weight -- the force of gravity on it -- mg, and we have some force of air resistance.0134

The force of air resistance is going to oppose the force of gravity.0145

The faster something falls, the force of air resistance resists that motion -- pushes up.0149

When eventually the force of air resistance and the object's weight balance out -- they are equal -- the object does not accelerate anymore.0153

It is in equilibrium, like we learned with Newton's First Law. It maintains a constant velocity.0160

We would call that terminal velocity -- when the force of air resistance equals the force of gravity on an object.0166

It maintains a constant velocity; it does not speed up; it does not fall any further.0175

Let's draw a FBD for a glass of soda sitting on a table.0180

Here is our glass of soda.0186

We have the weight pulling it down, but the glass of soda is not moving, it is at rest and it is remaining at rest; it is not accelerating.0189

There must be another force on that glass. What is that force?0197

It is the force of the table on the soda and it must oppose -- absolutely balance that weight.0202

We are going to call this force, the normal force.0209

When we say normal, we are not talking about the opposite of weird, we are talking about the geometric interpretation of normal.0214

It means perpendicular. In this case, the force is perpendicular to the surface of the table.0220

There is our table. If we have our glass sitting on it -- perpendicular to the surface of the table -- the normal force comes out of that surface.0230

Perpendicular to this tablet, a normal force would be coming out this direction.0246

Normal perpendicular -- we will label that fn.0250

Now, in this case it is pretty easy to see that the net force again, must be zero -- the object remains at rest.0256

Example problem: Which diagram here represents a box in equilibrium?0267

Again, the key here is to recognize that equilibrium means that the net force equals 0 -- all the forces are balanced.0272

Well, these are not balanced; it cannot be one.0281

Those are not balanced; it cannot be that one.0284

So, 2 up, 2 down, 5 right, 5 left -- that has to be equilibrium -- Net force = 0.0287

All the forces balance. It will continue in it's current state of motion.0295

All right, example five: Now we have a block sitting on a ramp. Which diagram below best represents the forces acting on the block?0300

If we think about it, let's draw them here first.0310

We have the normal force, which must be perpendicular to the surface -- so there is our normal force this time.0313

The weight -- the force of gravity on all objects -- is down toward the center of the Earth: mg.0320

Or, in this case it is labeled as weight, as fw in these diagrams.0328

Which way does the block want to go?0333

It wants to slide down the incline right?0336

So the force of friction must be up the ramp. Which one of these four choices nears what forces we see on that?0338

Has to be number four here.0347

All right. When forces do not line up with axis, you can draw a pseudo free-body diagram (P-FBD)0353

We are going to break up the forces that do not line up with the axis into components that do.0359

When you do this on an AP exam, your FBD only shows the forces, not components -- then draw a separate P-FBD.0364

If you show components on a FBD, often times they will take off points.0373

You need to have two separate diagrams -- one showing just the forces and then a separate P-FBD where you have broken forces that are not lined up with an axis into their components.0378

Let us draw the FBD for a box sitting on a ramp.0392

We have the applied force -- some force pulling it up the ramp.0397

Of course, we are going to have a normal force perpendicular to our ramp.0401

We have the weight of the box -- straight the force down, the force of gravity.0407

And in this case, if we have a force wanting to pull it up, we can draw this as having a force of friction down that way to balance it -- depending on what all our forces are.0411

If we were to draw a FBD -- let me draw it over here on the right, our FBD.0421

There is one axis -- I am just going to tilt my x and my y.0427

Now when I draw this, I have f pointing up the ramp. I have a normal force that is perpendicular.0433

If we are pulling it up the ramp, we could have a force of friction -- the opposite direction and weight is straight down.0444

That is our FBD. That we do not touch now.0453

We have drawn our diagram showing all the forces.0459

To do the P-FBD, though, we are going to make a separate diagram.0461

We are going to keep our axis y -- x. F already lines up with an axis, so that is fine.0466

Normal force lines are up already. Friction lines up with an axis, but the force of gravity -- the weight of the object -- does not.0475

What are we going to do with that?0480

That one we are going to have to break up into components.0488

If that is weight, we are going to have to do a little bit of Geometry here to see what is going on.0491

If I were to extend the ramp back here -- this is our angle θ -- that is 90 - θ.0497

That angle must be θ again.0506

I am going to break up the weight into an x component -- a component that is parallel with the x axis.0507

Let's call that mg.0515

It is parallel because it is in the same direction as the object's motion, or the direction it wants to move -- mg parallel and we have a component perpendicular to that axis -- mg perpendicular.0518

Now notice mg parallel -- this side is opposite our angle θ, so mg parallel is going to be equal to mg sin θ -- mg perpendicular is the adjacent side -- mg cos θ.0532

My P-FBD, when I go to draw it -- let me label it here -- is going to show the components of the object's weight instead of the weight itself.0548

Now all my forces with line up with the axis. There is my x. There is my y.0551

We have f. We have our normal force -- force of friction.0569

Now this mg parallel and mg cos, or mg cosine θ -- easy enough. Draw it right there, mg cosine θ.0577

Mg parallel right here -- I am just going to shift so it is on the axis; it is already lined up with an axis, but I am going to redraw it over here - mg sine θ.0587

You will see in some cases where teachers prefer to see that all at the same point in force of friction, mg sine θ are right beside each other, almost in the same direction.0597

I prefer to draw mine end-to-end so it is easy to see that they add up.0607

Here is our FBD and here is our P-FBD.0612

You have to show those separately on the AP questions in order to get the full credit for a problem.0617

Let us take the example of a car on a bank turn.0623

Roadways are often angled around steep turns to assist the cars in making the turn.0627

Let us draw the free-body diagram and the pseudo free-body diagram for a car on a banked roadway.0630

I will start off by drawing my curve and we will put our car on it -- pretend we are looking at the car from the back.0638

There are the wheels and there is the license plate.0647

This is at some angle θ. So, the forces acting on it -- we can draw on our FBD.0650

I will draw my x axis and I will draw my y axis -- (y,x).0657

In this case we have a normal force that is coming up out of the ramp -- so our normal force fn -- or capital N if you prefer points that way.0670

We have the weight -- the force of gravity, down.There is our FBD.0680

And we can throw friction in here if we needed to, depending on the situation, but let's keep this one simple for now.0690

If we wanted then to draw the P-FBD for this -- I am first going to redraw this -- y, x, mg -- pointing down -- and just to illustrate this, I am going to draw the normal force up here again.0696

Now as we look at the geometry of the problem, our angle θ is going to be -- Let's draw our components-- there is the x component of the normal force. There is the y.0719

That is going to be our angle θ by geometry, therefore, this side is going to be fN sine θ, the opposite, and this one is going to be fN cos θ.0728

When I do my nice, pretty P-FBD, there is x, there is y, my object -- of course, mg down -- I have this way, fN sine θ and I have fN cos θ.0741

There is my P-FBD. All the forces shown as components.0769

So now, Newton's Second Law of Motion, perhaps the most important formula or relationship in all of physics.0777

We will use this again, and again, and again.0783

The acceleration of the object is in the direction of and directly proportional to the net force applied.0786

It is inversely proportional to the object's mass.0795

If we want to write this in formula form -- acceleration of vector is equal to the net force applied on an object divided by the object's mass -- it's inertial mass.0798

The way that it is more commonly written: F_net equals ma -- Force = mass x acceleration.0811

You can apply this in many, many different ways.0822

As we look at these two laws, there are some interesting observations to take from them.0826

Newton's First Law says an object at rest will remain at rest and an object in motion will remain in motion at constant velocity in a straight line unless acted upon by a net force.0831

Basically what it is saying is, if F_net = 0, then acceleration equals 0.0844

Newton's Second Law says the acceleration of the object is in the direction of and directly proportional to the net force applied and inversely proportional to the object's mass.0851

Notice how if F_net over here is 0, A has to be 0.0863

Newton's First Law is redundant. It is actually a special case of Newton's Second Law where the net force on an object is 0.0869

If you understand Newton's Second Law, through and through, Newton's First Law you do not really need to know.0874

It is already embedded in Newton's Second Law. The First Law is redundant.0883

Let us look at how we apply Newton's Second Law.0890

General strategy -- Draw a FBD -- tremendously helpful tools.0893

For any forces that do not line up with the x or y axis, break those up into components that do and then we are going to draw that P-FBD.0898

Next, write expressions for the net force in the x and y directions.0908

Since the net force equals ma, we can use Newton's Second Law to solve the resulting equations and determine whatever the unknown quantities are that we are after.0912

Let us see how this works.0922

We have a force of 25N East and a force of 25N West acting concurrently on a 5 kg cart.0924

Find the acceleration of the cart.0930

You can probably do this in your head, but it is worth walking through the steps to see how this could applied with a simple situation before we complicate matters.0932

A FBD -- there is our cart.0942

We have a force of 25N to the East and we have a force of 25N to the West.0946

Next, I am going to write my Newton's Second Law equation: F_net = ma.0956

Since this is in the x direction, I am going to specify this and say net force in the x direction is = to mass times acceleration in the x direction.0962

Now, F_net(x) -- all this means is that is says looks at your FBD.0971

Look for all the forces acting in this x direction and write them down.0975

In this case, I have F_net(x) that I am going to replace with 25 to the left, so that is -25.0980

I have 25 to the right, so that is +25.0987

That must be equal to ma(x)-25 + 25 = 0, therefore 0 = ma(x), therefore a(x) must be equal to 0.0990

No acceleration, which you probably knew before we started the problem, but the steps are what is important.0995

Let's take another look here.1013

We have a .15 kg baseball moving 20 m/s stopped by a player in .010 s. What is the average force stopping the ball?1015

The way I would start these sorts of problems -- it is usually a good idea to write down what kind of information you are given.1027

Here I know the mass is equal to .15 kg.1032

It is initially moving at 20 m/s -- 0 = 20 m/s.1037

I am just going to draw in that that is to the right +x.1043

Final velocity is 0. It comes to rest and the time it takes is .01 s.1050

As I look at that -- trying to find the average force stopping the ball -- I know the mass -- it sure would be useful to have acceleration.1058

I do not know acceleration, but I can use my kinematics -- my kinematic equations to find it.1066

Acceleration is change in velocity over time -- δ anything is it's final value minus it's initial divided by time, or 0 - 20 m/s/.01 s is going to be negative 2,000 m/s ².1073

Now that I know acceleration, I can use Newton's Second Law to continue the problem.1096

If the net force equals mass times acceleration, that implies then -- since we know acceleration is -2,000 m/s².1103

I also know that mass is .15 kg, then the net force must be equal to our mass .15 kg x -2,000 m/s².1126

Calculator time -- this implies then that the net force equals -300N.1139

Why the negative? What does the negative mean here?1152

If you think about it, we called to the right the positive direction -- the initial direction the baseball was moving.1157

The negative just implies that this force has to be in the opposite direction in order to stop it.1163

That is the opposite direction of the ball's initial velocity.1168

Moving on -- let us take a look at a block on a surface.1190

Two forces, F1 and F2, are applied to a block on a frictionless, horizontal surface as shown in the diagram.1194

If the magnitude of the block's acceleration is 2 m/s², what is the mass of the block?1200

We know it is accelerating. F1 is bigger than F2, so it must be in that direction at 2 m/s².1207

Well, FBD -- always a great place to start.1218

We have 2N to the right, F2, and we have a lot more -- we have 12N to the left.1220

I am going to write Newton's Second Law, F_net = ma.1230

Since I am interested in just the x direction, the net force in the x direction equals mass times acceleration in the X direction.1234

I am after mass, so net force in the x direction -- 12 to the left, 2 to the right -- let us make this easy and define to the left as positive.1242

That means we have 10N in the positive direction, to the left -- must equal our mass times our acceleration, 2 m/s to the left.1251

So that is positive, therefore mass is going to equal 10/2 or 5 kg.1260

Tremendous. Let us go a little bit further.1272

We have a 25N horizontal force northward and a 35N horizontal force southward acting concurrently -- that means at the same place and at the same time -- on a 15 kg object on a frictionless surface.1277

What is the magnitude of the object's acceleration? Again, let us start with the FBD.1290

There are horizontal forces both North and South, but I am going to draw an overhead view.1296

We have a force of 25N North and we have 35N South.1302

The net force should be pretty easy to see. It is going to be 10N South.1311

The acceleration is going to be the net force divided by the objects mass, which is going to be 10N South, divided by 15 kg, or 0.67 m/s ² South.1320

Let us talk about mass versus weight.1346

Mass is the amount of stuff that something is made up of. It remains constant.1349

Yes, you can change the mass of an object by taking pieces off of it, or adding pieces.1355

For the most part, wherever you go for the same object, it has the same mass -- it is made out of the same stuff.1360

Weight, however, which we are writing as mg is the force of gravity on the object.1366

Weight varies depending on the gravitational field strength, g.1370

Here on the surface of the Earth, g is 9.8 m/s².1375

On the surface of the moon, g is about 1.6 m/s².1380

You would have a different weight if you went to the moon -- 1/6 the weight you would have on Earth.1384

However, regardless of how that works, you have the same mass.1389

Let us look at an example here.1397

An astronaut weighs 1,000N on Earth.1399

What is the weight of the astronaut on Planet X, where the gravitational field strength is 6 m/s ².1402

On Earth, mg, the object's weight, is 1,000N, therefore, we could say that the mass of the object -- what does not change, is going to be 1,000N/g on Earth -- 10, or about 100 kg.1409

If we go over here to Planet X, mg on Planet X must equal the mass, 100 kg -- that does not change, times g on Planet X, 6 m/s² -- 100 x 6 = 600N.1427

An alien on Planet X weighs 400N. What is the mass of the alien?1448

On Planet X, mg(x) for the alien must be 400N, therefore, the mass of the alien on X is 400N/g on x, 6 m/s ², or about 66.7 kg.1454

Take the alien to Earth, it is going to have a different weight.1475

It will not be 400N, but the mass will be the same, 66.7 kg.1478

Coming back to equilibrium. Translational equilibrium occurs when there is no net force on an object, therefore, acceleration is 0.1487

The equilibrant is a name for a single force vector that you add to any unbalanced forces you have on an object in order to bring the object into translational equilibrium.1497

For example, if I have a force that is 25N that direction -- if I want its equilibrant, I need a force that is 25N in that direction so that you add them together -- you get 0.1506

You get no unbalanced forces. You have 0 acceleration.1520

In the diagram here we have a 20N force due North, and a 20N force due East acting concurrently, again at the same place and same time on an object.1530

What additional force is required to bring the object into equilibrium? Or we are looking for the equilibrant.1540

Now the way I do this is, is if I look here we have 20N and 20N.1549

Let us add them together to get the net force.1554

I am just going to slide this vector over so that they are lined up tip to tail so I can add them -- 20N -- and my resultant, the sum of the two vectors is going to be a vector with the length square root of 20 squared plus 20 squared.1556

That is going to be square root of 20 ²⁰ + 20 ² = 28.3N.1572

I could replace that 20N North and 20N East with one vector, 28.3N to the northeast.1583

It's equilibrant, the vector I would have to add to that system to bring it back into equilibrium, must be the exact opposite of that.1589

The equilibrant must be that red vector, which would be 28.3N to the southwest.1596

That is what an equilibrant is.1611

A little bit on translation equilibrium.1615

We have a 3N force and a 4N force, so they are acting concurrently on a point.1618

Which force could not produce equilibrium with those two forces?1621

A 1N force could, because if we have this lined up correctly we could have a 3N, maybe a 4N force and a 1N force that would somehow sum to zero.1626

You would get back to where you started.1638

A 7N force -- if we had 3N this way and 4N this way, an equilibrant that was 7N in the exact opposite direction would bring that into equilibrium.1640

But, we cannot do 9. If we have 3N to the right and 4N to the right, there is no where I can place a 9N force where we end up with no net force when we are all done.1652

No matter what I do, I am going to have a remaining force, an unbalanced force of at least 2N1667

A 9N force cannot combine with the 3N force and a 4N force to give you 0 net force, or to bring you into equilibrium.1673

Determining acceleration.1684

We have a 15 kg wagon that is pulled to the right across a surface by a tension of 100N, an angle of 30 degrees above the horizontal.1687

A frictional force of 20N to the left act simultaneously. What is the acceleration of the wagon?1695

I like to draw a picture first.1703

We will make a little red wagon, because those are the cutest kind really.1704

We have a force of 100N at an angle of 30 degrees, and we know we have a frictional force over here.1718

For my FBD, I am going to have my little red wagon.1720

I have the weight of the wagon, mg. I have some amount of normal force.1728

I also have this applied force that is going to be 100N in an angle of 30 degrees, and a frictional force.1735

There is my FBD, but now I am going to do the P-FBD, where I am going to break up that 100N force that does not line up with the axis into its components: y, x.1748

We will start up with the forces that do line up: mg, force of friction, and normal force.1763

The x component here is just going to be 100N cosine 30 degrees and its y component is going to be 100N sine 30 degrees.1773

If we want the acceleration of the wagon, we are really talking about the acceleration of the wagon in the x direction.1791

I am going to write Newton's Second Law: F_net = ma, and focus on it in the x direction.1796

For F_net, all I do is I go back to my P-FBD diagram and I look for all the forces acting in the x direction.1805

I have the 100N cosine 30 degrees -- that is 86.6N -- and in the opposite direction, I have minus the force of friction, 20N and that must be equal to ma(x) -- 86.6 - 20 = 66.6N = ma(x).1812

Therefore acceleration in the x must be 66.6N over the mass of our cart, 15 kg -- mass of our wagon, which implies that the acceleration must be 66.6/15, or about 4.44 m/s².1836

Let us take a look at one more example problem.1859

We will talk about a suspended mass.1863

In this case we have a traffic light suspended by two cables as shown and we will label them T1 and T2.1866

We have measured the tension and the cables using a spring scale and we found that T1 is 49N and T2 is 85N.1872

Can we find the mass of the traffic light? Of course, the answer is going to be yes.1880

Let us start by drawing our FBD -- x and y.1885

Now, there is our object and of course we have its weight: mg down.1894

We have tension one (T1) and if I do just a little bit of geometry over here -- if that is 30 degrees, that must be 30 degrees and over here, if that is 60 degrees, then that is 60 degrees.1903

So I am going to draw a T1 in that direction at an angle of 30 degrees, and T2 over here --it is a bigger angle at 60 degrees.1914

So my P-FBD, I have to break up T1 and T2 into their components.1926

Take our FBD again -- y, x -- mg of course still points down.1935

Let us start with T2 here. Its x component is going to be T2 cosine 60, so I will have here T2 cosine 60 degrees and its y component will be T2 sine 60 degrees.1945

Now let us deal with T1.1960

Its x component will be to the left and that is going to be T1 cosine 30 degrees, and its y component -- T1 sine 30 degrees.1962

If we are trying to find the mass, I am going to start with Newton's Second Law in the direction that has the mass in it.1977

I am going to write F_net = ma, but I am going to look in the y direction.1986

I am going to replace F_net(y) with all the forces I see over here acting in the y direction.1992

F_net(y) = T1 sine 30 degrees, pointing up + T2 sine 60 degrees and I have mg down - mg.1999

We know that all of that -- since this is just sitting there and the traffic light is not accelerating -- a, it must be 0, so that is all equal to 0.2020

When I do that, I can then say since T1 is 49, then 49 sine 30 degrees + T2 is 85N -- 85 sine 60 degrees must be equal to mg.2030

Or 49 sine 30 = 24.5 + 85 sine 60 = 73.6 must be equal to mg -- 9.8(m).2052

Divide both sides by 9.8 and I come up with a mass of about 10 kg.2069

Newton's Second Law: f = ma and free-body diagrams and pseudo free-body diagrams to help us apply those concepts.2078

Terrific tool that we are going to use all the time here in Physics.2086

Hope you have had a great time here at Educator.com.2090

Make it a great day. We will talk to you soon.2093