AP Physics C: Mechanics Newton's Second & Third Laws of Motion

Hello, everyone, and welcome back to www.educator.com.0000

I am Dan Fullerton and in this lesson we are going to talk about Newton’s s and third laws of motion.0004

Our objective is to understand the relation between the force that acts on an object and the objects acceleration.0009

Perhaps, the most important part of the entire course.0015

Write down and utilize vector equations resulting from the applications of Newton's second law.0018

Applying it in third law and analyzing the forces between two objects.0025

Solve problems in which application of Newton's laws leads to simultaneous linear equations.0029

Let us start off by talking about Newton's s law of motion that is so useful in the course.0036

It says the acceleration of an object is in the direction of indirectly proportional to the net force applied and inversely proportional to the objects mass.0042

It is valid only in inertial reference frames.0052

However, for the purposes of this course we are assuming everything we deal with is an inertial reference frame.0055

The way we write it the net force or if you prefer the sum of all forces on an object is equal the mass × acceleration that = ma.0062

How do we apply it?0072

We start off with free body diagrams like we talk in our last lesson.0074

Then, for any forces that do not line up with the x or y axis we break those in the components that do lie on the x or y axis with those pseudo free body diagrams.0079

Then, we write expressions for the net force in the x and y directions and set the net force = mass × acceleration since Newton's law s law tells us f = ma.0088

Finally, solve the resulting equations.0098

We will do lots of examples.0100

Let us start off with a block on the surface.0102

Two forces f1 to the left and f2 to the right are applied to a block on the frictionless horizontal surface as shown.0106

If the magnitude of the blocks acceleration is 2 m/s² find the mass of the block.0112

Since you get a bigger force to the left let us call to the left our x direction here.0119

If we look our net force in the x direction we have 12 in the positive direction - we have 2 in the negative x direction = mass × the acceleration in the x direction.0126

12 -2 is 10 N must equal our mass × acceleration 2 m/s².0141

Therefore, our mass must be 10 N / 2 m/s² or 5 kg.0149

Nice and straightforward.0160

How about if we have some concurrent forces?0162

A 25 N horizontal force northward in the 35 N horizontal for southward at concurrently at a 15 kg object on a frictionless surface.0165

What is the magnitude of the objects acceleration?0175

Let us draw that.0178

We have got 25 N to the north, we have 35 N that is our object.0179

35 N south what is the magnitude of the objects acceleration?0190

Why do not we call it down or south our y direction since we can see that is the bigger force.0200

I am going to write that the net force in the y direction is 35 N -25 N because that points in the -y direction and0207

that must be equal to mass × acceleration in the y direction or 10 N = 15 kg mass × acceleration.0218

Or acceleration then is equal to 10 N / 15 kg which is 0.67 m/s².0232

That is Newton’s second law.0246

As we talk about Newton’s second law an opportune time to talk about mass vs. weight.0249

If you recall mass is a measurement of an object’s inertia.0254

It tells you how much stuff something is made up of.0257

It is a constant.0259

Your mass should be the same unless somebody comes over and starts chopping off limbs and other things that are unpleasant.0261

Your mass is relatively constant.0267

Weight is the force of gravity on an object.0269

Weight which we write as mg can vary with gravitational field strength.0272

Your weight on earth would be different than your weight on the moon.0276

Your mass however would be the same.0281

Let us do an example with mass vs. Weight.0286

An astronaut weighs 1000 N on earth what is the weight of the astronaut on planet X where the gravitational field strength mg is 6 m/s².0289

Let us figure out the astronauts mass first.0299

The weight of the astronaut mg is 1000 N and therefore mass must be 1000 N / 10 m/s² here on earth or 100 kg.0304

On this planet X, however, the weight is m × gravitational field strength.0319

Let us call that g sub x for g on planet X which is 100 kg mass does not change × 6 m/s² or 600 N.0325

An alien on planet X weighs 400 N what is the mass of the alien?0341

The weight of the alien mg x is 400 N therefore the mass is 400 N ÷ gx which is 6 m/s² or 66.7 kg.0347

What does this have to do with Newton’s second law?0370

Here is the rule.0372

We have been writing Newton’s second law f = MA.0374

We have been writing the force of gravity the weight of an object as mg.0379

Force of gravity is just a specific type of force is mass × we called g the acceleration due to gravity.0384

Writing weight is mg is just an interpretation of Newton’s second law.0391

To shorthand that and we have been using the same step.0397

Already taking into account that we understand Newton’s second law.0399

Let us look at a translational equilibrium problem.0404

In the diagram a 20 N force due north and the 29 N due east act concurrently on an object in the same place of the same time.0407

What additional force is required to bring the object and equilibrium?0416

We need to have one force that is going to balance these two.0420

The way I do this is the first thing I'm going to do is figure out what the resultant of these two are.0424

Let us add them together to see what their net force would be.0430

If we got 20 N to the right and 20 N up I am going to line those up tip to tail so I can add them.0433

I am going to take this 20 N north and I'm going to move it here to the right to remember0439

we can slide vectors as long as we do not change their direction or length.0444

And then to find the resultant I will draw a line from the starting point in the first to the ending point of the last.0449

And I can use the Pythagorean Theorem to figure out its magnitude.0455

That will be the √20² + 20² which is about 28 N.0460

Right now we have the equivalent of a 28 N force NE 45°.0467

What force would be required to bring that in the equilibrium?0473

We have the exact opposite force.0475

We will start a force at the same point and have it go 28 N and down into the left.0478

What we call this force the force that required to bring an object in the equilibrium is known as the equilibrant.0487

Our answer would be 28 N southwest.0497

Another translational equilibrium problem.0503

A 3 N force in the 4 N force acting concurrently on a point.0506

Which force could not produce equilibrium with these forces?0510

Now let us see.0514

The 1 N force produce equilibrium with these.0515

We have a 3 N force, we have a 4 N force and have it go to the left.0518

It looks like the addition of a 1 N force could completely balance those out.0527

We get back to our starting point.0533

1 would work that could produce equilibrium.0535

I think 7 could work too.0539

Let us take a look if we have 3 N and now we line up the 4 N force there.0541

A 7 N force could bring this back to our starting point 7 N in the equilibrant that cannot be the answer.0546

However, there is no way that we can have a 3 N force if 1 N force and no matter what we do with the 9 N force,0555

no matter where we put it there is no way we get back to where we start where we get no net force.0565

It is going to be 9 N force, that is not going to work.0569

You may be wondering how you would do that with the 4 N force?0573

That would work 3 N probably something like that.0576

4 N and you can make a triangle out of it and you can get back to where you started.0583

But 9 N cannot produce equilibrium with a 3 and 4 N force.0588

A way you can test that is add up the 2 forces 3 + 4 = 7 and subtract the two forces.0593

4 -3 is 1.0601

Only forces that are between 1 and 7 could produce equilibrium with that.0602

Anything outside of that range not going to work.0607

Problem here we have got a 15 kg wagon pulled to the right across a surface by a tension of 100 N an angle of 30° above the horizontal.0615

A frictional force of 20 N to the left axis simultaneously.0623

What is the acceleration of the wagon?0627

Let us start off by drawing a free body diagram.0630

Let us get back to those.0632

Especially as our problems start to get a little more complex.0635

And as we look here it is pulled across the surface by a tension of 100 N and an angle of 30° above the horizontal.0644

There is 100 N an angle of 30°.0653

A frictional force of 20 N x to the left.0657

There is our frictional force.0662

What is the acceleration of the wagon?0665

Let us do our next step and break that 100 N force up into its components with a pseudo free body diagram.0668

We will draw our axis again.0679

We still have force of friction to the left the x component of our 100 N is going to be 100 cos 30.0686

That is going to be 100 cos 30 is 86.6 N.0696

In the y direction, we are going to have 100 sin 30 the y component which would be 50 N.0703

We can go to our Newton’s second law equation net force in the x direction is equal to0713

we have 86.6 N to the right - our force of friction which it tells us is 20 N to the left0721

and all of that must be equal to our mass × acceleration in the x direction.0729

Acceleration in the x is 86.6 -20 or 66.6 N ÷ our mass 15 kg which is about 4.44 m/s².0736

Alright, let us take a look at baseball problem.0757

A 0.15 kg baseball moving at 20 m/s is stopped by a player in 0.01 s.0761

What is the average force stopping the ball?0767

We are going to learn some other ways we can solve this two here in a few lessons.0770

But for now let us start by finding its acceleration.0773

Its acceleration is its change in velocity ÷ time which is its final velocity - its initial velocity ÷ time.0777

We can use our kinematic equations because it is a constant acceleration.0785

It is going to be 0 - 20 m/s / 0.01 s or -2000 m/s².0790

What is the negative mean?0802

We are calling its initial velocity positive so this means it is in the opposite direction of its initial velocity.0804

It is slowing down.0810

If we want to know the force, the force is going to be mass × acceleration with Newton’s second law0813

is going to be our mass 0.15 kg × acceleration -2000 m/s² or -300 N.0821

And again, all that negative means it is in the direction opposite what we called positive the direction the ball is going initially.0835

The force in the acceleration are always going to be in the same direction so they are both negative by Newton’s second law.0842

Let us take a look at some steel beams.0850

A steel beam of mass 100 kg is attached by cable to a 200 kg beam above it.0854

The two were raised upward with an acceleration of 1 m/s² by another cable attached to the top beam.0860

If we assume the cables are mass less find the tension in each cable.0866

Let us start.0870

We will call this a y direction and we will analyze beam 1 first.0872

Here is our free body diagram for beam 1.0878

We have tension 1 pulling up on it, we have tension 2 pulling down on it and we have its mass m1 g down.0880

There is beam 1.0892

We did the same thing with beam 2.0895

We will draw a dot for our object.0898

We have T2 pulling it up and only its weight m2 g pulling it down.0899

Let us take a look at applying Newton’s second law here.0906

For beam 1 write the net force in the y direction is equal to we have T1 - T2 – m1 g and all of that =m1 a or we could write that T1 = T2 + m1 g + m1 a.0909

Let us go down and will start here on number 2.0938

Net force in the y direction for 2 is going to be T2 – m2 g and that has to equal m2 a, Newton’s second law or T2 is going to be equal to m2 × g + a.0943

Since we know those we can solve right away that is going to be equal to m2 which is 100 kg × g10 + a1 or 11 m/s².0961

T2 = 1100 N.0975

We can come back to beam 1 where T1 = T2 + m1 + m1 a0980

T1 = T2 1100 N + we have m1 g 200 × 10 + m1 a 200 × 10989

So T1 therefore = 2000 + 200 + 1100 is 3300 N.1006

There are the tensions in our cables.1017

Let us take a look at another one here.1022

The system below is accelerated by applying a tension T1 to the right most cable,1025

Assuming the system is frictionless determine the tension T2 in terms of T1.1030

You can do this in a very detailed way or we can use a little bit of systems thinking to try and make it nice and simple.1035

The first thing I'm going to look at is I’m going to take a look and say you know what if all of these we consider one object for the moment.1042

We have T1 applying a force on that entire object.1050

We could write that T1 the net force on that big object is the total mass 7 × the acceleration a or a = T1 / 7.1054

If we come over here and we look at just of this block is an object is a 2 kg block there is our free body diagram.1069

We got a normal force, its weight, I will call this m2 I suppose and T2 acting on it.1079

The only unbalanced force is going to be T2.1088

Normal force and its weight have to exactly balance otherwise it spontaneously accelerate up off the table or go through it neither of which happen.1090

Looking in the x direction at f net x = MA x or T2 = m2 which is 2 kg × ax1100

If T2 = 2a and a= T1 / 7 T2 must equal 2 × a T1 / 71113

T2 = 2 / 7 T1.1126

Taking a look at some banked curves.1137

Curves can be banked at just the right angle that vehicles traveling a specific speed can stay on the road with no friction required.1140

It is very important if you live somewhere like I do where there is lots of ice and the snow on the roads in the winter.1146

Given a radius for the curve in a specific velocity at what angle should the bank of the curve be built.1152

Let us start by drawing a free body diagram for our system.1159

There is our y, there is our x.1164

Our object are car even though it is going to be on an incline of some sort, is not moving in the direction that inclined.1170

We are not going to tilt our axis here.1177

Before we draw our object let us say we put our angle of our ramp.1181

Let us make it really steep and dramatic here.1185

Let us have it look something like that.1189

There is the angle of our road just as a reference.1190

We will call this angle θ which means that if we draw on normal to it, a normal force on our car this would also be θ.1194

Here is our normal force.1210

Here is the weight of our object of our car mg.1212

Or if we want to do our pseudo free body diagram let us do that right beside it.1217

If we break that up in the components we still have mg straight down.1233

The normal force if we want the horizontal component that is going to be the adjacent.1239

As I look at this angle the x component is going to be the opposite side in this case.1245

That is going to be n sin θ here and the vertical piece is a side that right beside the angle that adjacent to it.1252

This piece is going to be n cos θ.1263

Let us write our Newton’s second law equations.1268

Net force we will look at the x direction first is also going to be what is causing the centripetal force as it goes around a circle.1273

That is what is causing the centripetal force to allow it to move in a circle to go around a curve of some1282

radius is going to be the horizontal component of the normal force n sin θ.1286

F net x which is also the net centripetal force is n sin θ which is ma by Newton’s second law.1294

Since it is a centripetal acceleration moving in a curve, in a circle, we could write that as mv² / r.1304

We will call that equation 1.1312

Analyzing in the y direction, net force in the y direction we have n cos θ - mg and1315

that has to be equal to 0 because there is no acceleration vertically.1325

Therefore we can say that n must be equal to mg / cos θ.1329

We will call that our second equation.1336

Now to start to solve this let us see if we can combine equations 1 and 2 to write n sin θ = mv² / r we will start with 1.1340

But we now know that n = mg / cos θ.1359

This implies then that we still have a sin θ.1366

We have an mg from our n up here ÷ cos θ must be equal to mv²/ r.1373

On the left inside there, sin/ cos is the tan function so we have mg tan θ = mv² / r.1392

I can divide a mass of both sides to say that g tan θ = V² / r.1406

Or θ if we want the angle is going to be the inverse tan of V² / gr.1415

If we know the speed we want the vehicles to travel on that curve.1428

We know the radius of the curve.1432

We can determine exactly what angle we should build that curve at so that the cars do not require a whole lot of friction on the road.1435

Taking a look at tension in cords.1445

Determine the tensions T1 and T2 in cables holding a 10 N weight stationary.1447

As we look at this first thing I think I am going to do again is draw a free body diagram for my object.1454

We will start to draw our axis here and I will draw the one for the pseudo free body diagram too1461

because I can already see am going to have a force and angle.1470

We will draw our x, that should do.1475

Starting with the free body diagram there is our object.1484

We have its weight which is given to us.1489

It is not telling us it is mass when it says it is 10 N.1491

That is its weight so we have 10 N down.1494

We have tension T1 strings cords can only pull so that must be our T1.1497

We have a T2 an angle of 60°.1505

I'm going to do my pseudo free body diagram breaking up T2 into its components.1512

I still have 10 N down.1517

T1 to the left.1523

The x component of T2 is going to be T2 cos 60°.1525

We have T2 cos 60° here in the y component is going to be T2 sin 60°.1532

I can start writing my Newton’s second law equations.1542

We will start with the x.1545

Net force in the x direction is going to be we have T2 cos 60 - T1 and all of that is going to have to be equal to 0.1547

There is no acceleration it is sitting there in equilibrium which implies then that T1 is going to be equal to T2 cos 60°.1563

Let us take a look at the y now.1577

Net force in the y direction is going to be T2 sin 60° -10 N.1580

Again, that has to be equal to 0.1594

Therefore, T2 must equal 10 Newton’s / sin 60° which is 11.5 N.1598

There is T2.1608

I can put that into my equation for T1.1611

Therefore T1 = 11.5 cos 60° or 5.77 N.1615

We found the tension in our two cables there.1628

Looking at this from a graphical perspective we will take an example where the velocity of1633

an airplane in flight is shown as a function of time T in the graph.1637

Velocity time.1641

At which time is the force exerted by the plane's engines on the plane the greatest?1643

The way I think about this one is we are going to have the greatest force when we have the greatest acceleration.1649

We flee on the big force we need a big acceleration and since acceleration I should say force is proportional to acceleration.1655

Acceleration is the slope or the derivative of the velocity time graph.1665

Where do we have the biggest slope?1670

It is probably occurring somewhere right around there which if I draw that down is right around 4.5 s.1672

We have the greatest slope at around 4.5 s that must be the time that which we have the greatest force.1683

Given the graph of velocity vs time for the particle shown below where V is proportional to T²1695

sketch a graph of the net force exerted on the object as a function of time.1701

If V is changing that way acceleration is the derivative of velocity.1707

We start off with 0 slope if I wanted to put acceleration on the same graph in red we start off at 0 acceleration.1712

As we go along we are going to have higher and higher, larger and larger amounts of acceleration.1720

It looks like our acceleration graph is probably going to be linear something of this shape.1726

If we have that shape for acceleration we have to have the same shape for our force time graph since force = mass × acceleration.1736

So I would draw our force graph with the shape like that when you are increasing up to the right.1746

Newton’s third law this is a pretty cool one as well.1757

All forces come in pairs, you never have just a single force, they are always pairs.1760

If object 1 exerts a force on object 2, object 2 exerts a force of equal magnitude but opposite in direction back on object 1.1765

Or if you wrote it mathematically the force of object 1 on object 2 is equal in magnitude but opposite in direction to the force of 2 on 1.1775

I should note that those are vectors of course, Newton’s third law.1785

If I punch you in the nose with a force of 100 N, your nose punches my fist with a force of 100 N back in the opposite direction.1790

Alright, some examples.1802

How does a cat run forward if the cat runs forward does not it push backwards on the ground?1803

It is the ground actually propels the cat forward?1810

Or if you want to swim forward which way do you push the water?1813

You push backward on the water applying a force on the water and the water’s reactionary force by Newton’s third law is what pushes you forward.1817

Or if you want to jump in the air which way do you push with your legs?1825

Do not you push down on the ground as hard as you can as quickly as you can?1828

You push down on the ground it is the opposite force of the ground pushing up on you that propels you up into the air.1832

When we have these two forces from different objects we call those action reaction pairs.1843

If a girl is kicking a soccer ball let us identify some of those pairs.1847

We have the force of the girl’s foot on the ball and the ball is actually applying a force back on the girl’s foot or a rocket ship in space.1851

The rocket ship is propelled by pushing gases out.1866

The reactionary force is that those gases push the rocket causing the acceleration.1874

How about gravity on you? gravity on me?1884

Right now the Earth's gravity is pulling me down.1887

Guess what though, I'm pulling the earth upward with the exact same force.1894

How come I have such a more dramatic effect?1904

If I jump off a cliff, I'm being pulled down toward the earth with the same force I'm pulling the earth up toward me,1906

why does it look like it is going to hurt the earth nearly as much as me?1913

I'm going to accelerate a lot more because for the same force I have a much smaller mass.1916

That same force applied to the Earth's huge mass gives a very little almost an immeasurable acceleration.1922

Earth's mass is approximately 81 × the mass of the moon.1933

If Earth exerts a gravitational force of magnitude f on the moon, the magnitude of the gravitational force of the moon on Earth is.1937

It is a straightforward application of Newton’s third law.1944

If the Earth is pulling on the moon with force f, the moon must pull on the earth with force f.1948

Forces come in pairs that are equal and opposite.1953

Or the sailboat question.1958

A 400 little girl standing on a dock exerts a force of 100 N on the 10,000 N sailboat as she pushes it away from the dock.1961

How much force does the sailboat exert on the girl?1970

In order to know the force the sailboat exerts on the girl we need to find the force that the girl exert on the sailboat.1973

She exerts a force of 100 N on the sailboat, the sailboat must have that exact same magnitude of force back on her.1980

This 10,000 N here that is just describing the weight of the sailboat not talking about the interaction of the girl and the sailboat.1987

Alright and another one, hammer and nail.1997