For more information, please see full course syllabus of AP Physics 1 & 2

For more information, please see full course syllabus of AP Physics 1 & 2

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## Study Guides

## Download Lecture Slides

## Table of Contents

## Transcription

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### Newton's 2nd Law of Motion

- FBDs are tools for visualizing forces on a single object and writing equations to represent a physical situation.
- The acceleration of an object is directly proportional to the net force experienced and inversely proportional to its inertial mass.
- The net force on an object is the vector sum of the individual forces.

### Newton's 2nd Law 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
- Objective
- Free Body Diagrams
- Drawing FBDs
- Example 1: Falling Elephant
- Example 2: Falling Elephant with Air Resistance
- Example 3: Soda on Table
- Example 4: Box in Equilibrium
- Example 5: Block on a Ramp
- Pseudo-FBDs
- Draw When Forces Don't Line Up with Axes
- Break Forces That Don’t Line Up with Axes into Components That Do
- Example 6: Objects on a Ramp
- Example 7: Car on a Banked Turn
- Newton's 2nd Law of Motion
- Newton's 1st Two Laws Compared
- Applying Newton's 2nd Law
- Example 8: Applying Newton's 2nd Law
- Example 9: Stopping a Baseball
- Example 10: Block on a Surface
- Example 11: Concurrent Forces
- Mass vs. Weight
- Example 12: Mass vs. Weight
- Translational Equilibrium
- Example 13: Translational Equilibrium
- Example 14: Translational Equilibrium
- Example 15: Determining Acceleration
- Example 16: Suspended Mass

- Intro 0:00
- Objective 0:07
- Free Body Diagrams 0:37
- Tools Used to Analyze Physical Situations
- Show All the Forces Acting on a Single Object
- Drawing FBDs 0:58
- Draw Object of Interest as a Dot
- Sketch a Coordinate System
- Example 1: Falling Elephant 1:18
- Example 2: Falling Elephant with Air Resistance 2:07
- Example 3: Soda on Table 3:00
- Example 4: Box in Equilibrium 4:25
- Example 5: Block on a Ramp 5:01
- Pseudo-FBDs 5:53
- Draw When Forces Don't Line Up with Axes
- Break Forces That Don’t Line Up with Axes into Components That Do
- Example 6: Objects on a Ramp 6:32
- Example 7: Car on a Banked Turn 10:23
- Newton's 2nd Law of Motion 12:56
- The Acceleration of an Object is in the Direction of the Directly Proportional to the Net Force Applied
- Newton's 1st Two Laws Compared 13:45
- Newton's 1st Law
- Newton's 2nd Law
- Applying Newton's 2nd Law 14:50
- Example 8: Applying Newton's 2nd Law 15:23
- Example 9: Stopping a Baseball 16:52
- Example 10: Block on a Surface 19:51
- Example 11: Concurrent Forces 21:16
- Mass vs. Weight 22:28
- Mass
- Weight
- Example 12: Mass vs. Weight 23:16
- Translational Equilibrium 24:47
- Occurs When There Is No Net Force on an Object
- Equilibrant
- Example 13: Translational Equilibrium 25:29
- Example 14: Translational Equilibrium 26:56
- Example 15: Determining Acceleration 28:05
- Example 16: Suspended Mass 31:03

### AP Physics 1 & 2 Exam Online Course

### Transcription: 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/s ^{2}.*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/s ^{2}.*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/s ^{2}, 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/s ^{2}.*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/s ^{2}.*1375

*On the surface of the moon, g is about 1.6 m/s ^{2}.*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/s ^{2} -- 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 2N*1667

*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/s ^{2}.*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

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1 answer

Last reply by: Professor Dan Fullerton

Thu Jun 23, 2016 10:20 AM

Post by Peter Ke on June 22, 2016

I really don't understand example #14. Would you mind explaining in detail why choice 1,2, and 4 are not the answer. I do however understand why choice 3 is the correct answer.

0 answers

Post by Saadman Elman on June 19, 2016

Your lecture is very effective. It helped me a lot. Thanks.

1 answer

Last reply by: Professor Dan Fullerton

Sun Sep 27, 2015 3:56 PM

Post by Bilbo Baggins on September 27, 2015

I looked up net force vs average force and they were different. Why is the answer to example nine the net force?

2 answers

Last reply by: Jim Tang

Fri Jul 24, 2015 7:35 PM

Post by Jim Tang on July 24, 2015

In Example #14, how does 4N produce equilibrium with 3N and 4N?

1 answer

Last reply by: Professor Dan Fullerton

Tue May 26, 2015 6:09 AM

Post by Ryan Rad on May 25, 2015

Hi quick question, at 12:00 how did u figure out which is your angle theta?? Im so confused, you mentioned by geometry we know, but could you elaborate please?? T

Thank you, oh and btw ur lectures are awesome! :)

1 answer

Last reply by: Professor Dan Fullerton

Tue May 5, 2015 7:23 PM

Post by Mutong Zhou on May 5, 2015

For example 4, why the answer choice D is not one of the answer?

1 answer

Last reply by: Professor Dan Fullerton

Fri Nov 7, 2014 6:20 PM

Post by mohammad mostafa on November 7, 2014

hi i have a question about example 10, is not the direction to the left so it should be negative,as the force of f1=12n > than f2

3 answers

Last reply by: Professor Dan Fullerton

Wed Feb 3, 2016 6:27 AM

Post by Caleb Martin on October 2, 2014

Hi,

I'm trying to firgure out what part of Newtons Second law applies to in this problem: "A force F with arrow applied to an object of mass m1 produces an acceleration of 3.60 m/s2. The same force applied to a second object of mass m2 produces an acceleration of 2.00 m/s2.

(a) What is the value of the ratio m1/m2?

ANS.:

(b) If m1 and m2 are combined into one object, find its acceleration under the action of the force F with arrow.

ANS.: m/s2

2 answers

Last reply by: Him Tam

Sun Jul 20, 2014 11:58 AM

Post by Him Tam on July 19, 2014

In the examples with the boxes on an incline, why is there still a normal force? Does air resistance still matter even if the object is not in the air?

1 answer

Last reply by: Professor Dan Fullerton

Mon Mar 24, 2014 3:37 PM

Post by sara bacellar on March 24, 2014

Hi, great lecture.However, I still need you help, I can't explain to myself how you did cancel out a 4 N and a 3 N and another 4N. How did you work this out? Thank you so much :)

1 answer

Last reply by: Professor Dan Fullerton

Sun Mar 16, 2014 5:17 PM

Post by John Parker on March 16, 2014

Thank you so much for these extremely helpful videos! One question: In example 11, does the object accelerate downward through the surface? I was surprised that the acceleration wasn't zero; I expected the normal force to increase along with the added 10 newtons South.

1 answer

Last reply by: Professor Dan Fullerton

Sat Mar 15, 2014 10:10 AM

Post by UMAIR TARIQ on March 14, 2014

Thank you, you are awesome. I had a quick question why were the x component in the banked car example Normal sin theta and in example 15 it is cos.

thank you again!!!

1 answer

Last reply by: Professor Dan Fullerton

Thu Jan 9, 2014 5:53 AM

Post by Hyun Cho on January 9, 2014

Hey i have another question about the last example. Sin30x49+sin60x85=mg. but since the tensions of the rope are facing up, they are positive (if up is positive and down is negative). Well then since the gravitational acceleration is going downward, shouldnt the g in mg be negative and therefore the m is negative as well?? I know its impossible to have negative mass, but its what i get

1 answer

Last reply by: Professor Dan Fullerton

Thu Jan 9, 2014 5:52 AM

Post by Hyun Cho on January 9, 2014

hi that was a great lecture but i dont get one thing. in example14, im fully aware why 9n cannot be equaled. but at the same time, i cant see how 4 is equaled either. could you show me how 4n can result?

1 answer

Last reply by: Professor Dan Fullerton

Thu Jan 9, 2014 5:50 AM

Post by Hyun Cho on January 8, 2014

hi i have a question. in example 6, you used the slanted plane as the x axis so that the normal force in perpendicular to the x axis, but in example, you just randomly made a x and y axis and drew the normal force and mg as they are.. how do i know what will be the x and y axis?

1 answer

Last reply by: Professor Dan Fullerton

Sat Nov 16, 2013 9:17 PM

Post by Constantin Ficiu on November 16, 2013

Great Lecture. I grasp the concepts of Forces and Newton's Laws in here, on the lectures taught by you, professor than the one's taught in my university.

Thank you.

1 answer

Last reply by: Professor Dan Fullerton

Fri Aug 23, 2013 12:17 PM

Post by Larry wang on August 23, 2013

Great lecture. Now I am going over this topic, and on example 14, however, I can't explain to myself on how you have derived 1N force could cancel out two concurrent forces (3N and 4N). Are those two concurrent forces somehow acting at an angle. If so what's the example. How do you work this out? Thank you very much

1 answer

Last reply by: Professor Dan Fullerton

Thu Jul 18, 2013 2:32 PM

Post by KyungYeop Kim on July 18, 2013

I have a question. If something is moving at a constant speed(no acceleration), then does it mean it doesn't have Force? but surely some force must be acting on it to move. I'm confused; when calculating force, what do the velocity and acceleration have to do with force(F)? (I know F=mxa, but it doesn't help)

1 answer

Last reply by: Professor Dan Fullerton

Sat Jun 1, 2013 3:18 PM

Post by Jude Nawlo on June 1, 2013

In example 7, how do you determine where angle theta should be with reference to the initial diagram? I can't tell where to put the angle?

1 answer

Last reply by: Professor Dan Fullerton

Wed May 22, 2013 6:01 AM

Post by kevin vaughn on May 21, 2013

for example 7 t= 12:17, why is the opposite of theta Fnsintheta. shouldn't that be the y component?

7 answers

Last reply by: Arshin Jain

Tue Apr 22, 2014 4:37 AM

Post by Nawaphan Jedjomnongkit on May 10, 2013

in ex14 how about choice 4N? How it can produce equilibrium with 3 and 4N ? Thank you

1 answer

Last reply by: Professor Dan Fullerton

Sat Apr 27, 2013 5:46 PM

Post by Edward Xavier on April 27, 2013

great lecture :D

1 answer

Last reply by: Professor Dan Fullerton

Mon Mar 25, 2013 5:42 AM

Post by John Smith on March 24, 2013

If Newton's first law really is a special case of the second then why are we taught the "three laws of motion" in school? Is it for historical reasons?