For more information, please see full course syllabus of AP Physics C: Mechanics

For more information, please see full course syllabus of AP Physics C: Mechanics

### Newton's Second & Third Laws of Motion

- Newton’s 2nd Law of Motion: The acceleration of an object is directly proportional to the force applied to the object and inversely proportional to the object’s inertial mass.
- Newton’s 3rd Law of Motion: All forces come in pairs. The force of object one on object two is equal in magnitude and opposite in direction to the force of object two on object one.

### Newton's Second & Third Laws of Motion

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Objectives
- Newton's 2nd Law of Motion
- Applying Newton's 2nd Law
- Example I: Block on a Surface
- Example II: Concurrent Forces
- Mass vs. Weight
- Example III: Mass vs. Weight
- Example IV: Translational Equilibrium
- Example V: Translational Equilibrium
- Example VI: Determining Acceleration
- Example VII: Stopping a Baseball
- Example VIII: Steel Beams
- Example IX: Tension Between Blocks
- Example X: Banked Curves
- Example XI: Tension in Cords
- Example XII: Graphical Interpretation
- Example XIII: Force from Velocity
- Newton's 3rd Law
- Examples - Newton's 3rd Law
- Action-Reaction Pairs
- Example XIV: Force of Gravity
- Example XV: Sailboat
- Example XVI: Hammer and Nail
- Example XVII: Net Force

- Intro 0:00
- Objectives 0:09
- Newton's 2nd Law of Motion 0:36
- Newton's 2nd Law of Motion
- Applying Newton's 2nd Law 1:12
- Step 1
- Step 2
- Step 3
- Step 4
- Example I: Block on a Surface 1:42
- Example II: Concurrent Forces 2:42
- Mass vs. Weight 4:09
- Mass
- Weight
- Example III: Mass vs. Weight 4:45
- Example IV: Translational Equilibrium 6:43
- Example V: Translational Equilibrium 8:23
- Example VI: Determining Acceleration 10:13
- Example VII: Stopping a Baseball 12:38
- Example VIII: Steel Beams 14:11
- Example IX: Tension Between Blocks 17:03
- Example X: Banked Curves 18:57
- Example XI: Tension in Cords 24:03
- Example XII: Graphical Interpretation 27:13
- Example XIII: Force from Velocity 28:12
- Newton's 3rd Law 29:16
- Newton's 3rd Law
- Examples - Newton's 3rd Law 30:01
- Examples - Newton's 3rd Law
- Action-Reaction Pairs 30:40
- Girl Kicking Soccer Ball
- Rocket Ship in Space
- Gravity on You
- Example XIV: Force of Gravity 32:11
- Example XV: Sailboat 32:38
- Example XVI: Hammer and Nail 33:18
- Example XVII: Net Force 33:47

### AP Physics C: Mechanics Online Course

I. Introduction | ||
---|---|---|

What is Physics? | 7:12 | |

Math Review | 1:00:51 | |

II. Kinematics | ||

Describing Motion I | 23:47 | |

Describing Motion II | 36:47 | |

Projectile Motion | 30:34 | |

Circular & Relative Motion | 30:24 | |

III. Dynamics | ||

Newton's First Law & Free Body Diagrams | 23:57 | |

Newton's Second & Third Laws of Motion | 23:57 | |

Friction | 20:41 | |

Retarding & Drag Forces | 32:10 | |

Ramps & Inclines | 20:31 | |

Atwood Machines | 24:58 | |

IV. Work, Energy, & Power | ||

Work | 37:34 | |

Energy & Conservative Forces | 28:04 | |

Conservation of Energy | 54:56 | |

Power | 16:44 | |

V. Momentum | ||

Momentum & Impulse | 13:09 | |

Conservation of Linear Momentum | 46:30 | |

Center of Mass | 28:26 | |

VI. Uniform Circular Motion | ||

Uniform Circular Motion | 21:36 | |

VII. Rotational Motion | ||

Rotational Kinematics | 32:52 | |

Moment of Inertia | 24:00 | |

Torque | 26:09 | |

Rotational Dynamics | 56:58 | |

Angular Momentum | 33:02 | |

VIII. Oscillations | ||

Oscillations | 1:01:12 | |

IX. Gravity & Orbits | ||

Gravity & Orbits | 34:59 | |

X. Sample AP Exam | ||

1998 AP Practice Exam: Multiple Choice | 28:11 | |

1998 AP Practice Exam: Free Response Questions (FRQ) | 28:11 |

### Transcription: 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 and*0207

*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 remember*0439

*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 to*0713

*we have 86.6 N to the right - our force of friction which it tells us is 20 N to the left*0721

*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 law*0813

*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 a*0980

*T1 = T2 1100 N + we have m1 g 200 × 10 + m1 a 200 × 1*0989

*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 × ax*1100

*If T2 = 2a and a= T1 / 7 T2 must equal 2 × a T1 / 7*1113

*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 some*1282

*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 and*1315

*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 too*1461

*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 of*1633

*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 T²*1695

*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

*A carpenter hits the nail with a hammer, compare to the magnitude of the force the hammer exerts on the nail.*2001

*The magnitude of the force the nail inserts on the hammer during contact is the exact same.*2006

*Why does the nail get propelled forward and the hammer not so much different?*2015

*The nail has a much smaller mass.*2019

*And our last question here.*2026

*If force is only come in pairs that are equal and opposite why do not all forces cancel each other?*2028

*The trick is these forces act on different objects.*2034

*They all act on the same object they would cancel out but if they do not they are acting on different objects.*2042

*Think about it in terms of free body diagram to be doing a free body diagram for each of these objects.*2048

*You are not going to have a free body diagram that has the initial force in the reactionary force or the paired force all on the same diagram.*2053

*You may have heard Newton’s third law stated as for every action there is an equal and opposite reaction.*2065

*Which is something of a restatement of the law but it also can be somewhat confusing*2070

*because you are not really talking about what actions and reactions are.*2075

*Instead, Newton’s third law the force of object 1 and object 2 is equal in magnitude and opposite direction to the force of object 2 on object 1.*2078

*Forces only come in pairs, you can never have a singular force.*2088

*Thank you for watching www.educator.com.*2093

*We will see you again real soon and make it a great day everybody.*2104

1 answer

Last reply by: Professor Dan Fullerton

Wed Oct 19, 2016 7:43 AM

Post by James Glass on October 19, 2016

Hello,

In example 10 (x), the problem with the banked curve, why doesn't (mg)sin(theta) somehow contribute to the force due to centripetal acceleration (m(v^2/r)? I think I understand why the y component of the centripetal force is (Fnormal)cos(theta).Thanks.

0 answers

Post by James Glass on October 19, 2016

Hello,

In example 10 (x), the problem with the banked curve, why doesn't (mg)sin(theta) somehow contribute to the force due to centripetal acceleration (m(v^2/r)? I think I understand why the y component of the centripetal force is (Fnormal)cos(theta).Thanks.

2 answers

Last reply by: Cathy Zhao

Sat Aug 13, 2016 8:51 PM

Post by Cathy Zhao on August 12, 2016

On example 9, why the acceleration of T2 equals to that of T1?

2 answers

Last reply by: Cathy Zhao

Sat Aug 13, 2016 8:51 PM

Post by Cathy Zhao on August 12, 2016

On example 6, why there is no acceleration in the y direction?

1 answer

Last reply by: Professor Dan Fullerton

Fri Aug 12, 2016 1:07 PM

Post by Dukaiwen Zhao on August 12, 2016

On example 5 (translational equilibrium), Answer choice D should be 5 N instead of 4 N right?

4 answers

Last reply by: Hemant Srivastava

Sat Jul 16, 2016 7:35 PM

Post by Hemant Srivastava on July 15, 2016

On example 8, shouldn't the gravity be -9.8 m/s^2?

1 answer

Last reply by: Professor Dan Fullerton

Mon Nov 23, 2015 7:16 AM

Post by Zhe Tian on November 19, 2015

Why is it on practice problem 9 you don't consider tension T2 for T1? Why isn't T1-T2=ma correct?

1 answer

Last reply by: Professor Dan Fullerton

Sun May 3, 2015 6:32 PM

Post by Nitin Prasad on May 3, 2015

In the block on a surface problem, why did you only find the horizontal component of acceleration? Shouldn't you also find the vertical component and take the vector sum?

2 answers

Last reply by: BRAD POOLE

Tue Mar 31, 2015 6:56 AM

Post by BRAD POOLE on March 29, 2015

In example 11. Does it mater what order you set up Fnet? You wrote

Fnet = T2 cos60 - T1 but can you write it as T1 - T2 cos 60? Your just trying to find the net so it shouldn't mater right? Thanks.