For more information, please see full course syllabus of High School Physics

For more information, please see full course syllabus of High School Physics

### Newton's 2nd Law: Multiple Dimensions

- With force in multiple dimensions, we switch to using vectors:
→F
_{net}= m · →a, ⇔ (F _{xnet}, F_{ynet})= (m ·a _{x}, m ·a_{y}). - We work with each component of the force vector separately. That also means we can break a force vector into its component vectors.
- The normal force is called
*normal*because that means (in math-speak) "perpendicular". A surface can not resist force that is parallel to it. - If something is on an incline, the portion of gravity that is perpendicular to the surface is cancelled out, but the portion of gravity that is parallel is not.
- Be careful when figuring out how much of the force is parallel! It is very easy to make a mistake and put an angle in the wrong place. Make a diagram and think carefully.

### Newton's 2nd Law: Multiple Dimensions

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
- Newton's 2nd Law in Multiple Dimensions
- Components
- Special Forces
- Normal Forces
- Why Do We Call It the Normal Forces?
- Normal Forces on a Flat Horizontal and Vertical Surface
- Normal Forces on an Incline
- Example 1: A 5kg Block is Pushed By a Force of 3N to the North and a Force of 4N to the East
- Example 2: A 20kg Block is On an Incline of 50Â° With a Rope Holding It In Place
- Example 3: A 10kg Block is On an Incline of 20Â° Attached By Rope to a Free-hanging Block of 5kg

- Intro 0:00
- Newton's 2nd Law in Multiple Dimensions 0:12
- Newton's 2nd Law in Multiple Dimensions
- Components 0:52
- Components
- Example: Force in Component Form
- Special Forces 2:39
- Review of Special Forces: Gravity, Normal Force, and Tension
- Normal Forces 3:35
- Why Do We Call It the Normal Forces?
- Normal Forces on a Flat Horizontal and Vertical Surface
- Normal Forces on an Incline
- Example 1: A 5kg Block is Pushed By a Force of 3N to the North and a Force of 4N to the East 10:22
- Example 2: A 20kg Block is On an Incline of 50° With a Rope Holding It In Place 16:08
- Example 3: A 10kg Block is On an Incline of 20° Attached By Rope to a Free-hanging Block of 5kg 20:50

### High School Physics Online Course

### Transcription: Newton's 2nd Law: Multiple Dimensions

*Hi, welcome back to educator.com, today we are going to be talking about Newton's second law and applying to multiple dimensions.*0000

*Last time we only talked about applying it in one dimension, today we are going to be talking about what happens in more than one dimension.*0006

*As in kinematics, each dimension works independently.*0011

*Acceleration in the y direction is completely unrelated to acceleration in x direction.*0016

*As far as Newtonian mechanics is concerned, we can consider them completely independently.*0020

*Now we just turn our formula from before into vector based ones.*0030

*Now we have got, F_{net} = ma, so we can consider them as the different components as separate pieces, x_{net}, y_{net} is going to have acceleration in the x and acceleration in the y, both going to be completely separate components.*0033

*It is possible to break a force vector into its components just as we did in kinematics to break velocity, acceleration, displacement into separate components.*0053

*If we had a force of 10 N acting on an object 30 degrees above the horizontal, what will that force be in component form?*0063

*First, we drop a perpendicular sown.*0068

*Now, we have a right triangle, we can easily apply trigonometry.*0072

*If we want to know what this side is, this is the side that is going to be connected to sin(30), we got the hypotenuse = 10, so it is going to be (10 N) × sin(30) , which comes from sin(30) = side opposite / hypotenuse, if you get used to doing stuff in component form, you will get really fast at this, it is just going to be sin(30) times whatever the full magnitude of the vector is.*0075

*Similarly on the other side, on the bottom part it is going to be, (10 N) × cos(30) .*0120

*Now we know what each side would be, so we just use basic trigonometry, and we get that from a table/calculator.*0132

*We get, the length (the width) is going to be 8.66, and the height is going to be 5, both in newtons, 8.66 N and 5 N in horizontal and vertical directions respectively.*0139

*Before we talked about some special forces, and they are all still there, the force of gravity (weight) is still there, now it is only going to pull down in one direction, it either gets canceled out, or works at the same time as the horizontal forces are working.*0159

*Normal force, force that holds you up, still there, if something is resting on the top of a surface, that surface is still going to support it against gravity.*0187

*Tension, force on whatever object is connecting things, still there, and it is going to pull in whatever direction that rope is angled at.*0195

*If you have a rope connected to something, and you pull like this, then it is going to cause a force of that, and if you pull like this, that is going to cause a force like that, it just have to do with what angle it is acting.*0203

*But, I have not explained why it is called the 'normal' force!*0220

*Why do we call it the normal force?*0228

*That is a great question.*0230

*The normal force, it is a fancy word for Math, it simply means perpendicular.*0233

*The word 'normal' comes from the Latin word 'norma', which is the word for a 'carpenter's square', which is a tool used for making right angles.*0240

*Normal just comes from a right angle, so it just means perpendicular.*0249

*Two normal segments would be any two that intersect with a right angle between them.*0255

*If you had a curve of some sort, you could have a number of right angles that is sprouted out of it.*0262

*Each one of them is a little bit different, but they are all normal even though they are attached to a curve, because what matters is that they are perpendicular at the point of intersection.*0279

*So normal is just another way of saying that it is perpendicular.*0289

*Normal force is the force that is perpendicular, the gravity the surface can cancel out, because the surface can only put up a perpendicular force.*0293

*Consider these two versions:*0301

*We have got a man standing on a perfectly flat plane, now what happens is, the force of gravity pulls down on him by a certain amount, and because he is standing on a perfectly flat plane, the normal force is going to be able to be exactly opposite, and the two are going to cancel one another out.*0304

*So that is great.*0319

*Now, what happens when force of gravity is pulling on a man, who stands on the side of a building.*0320

*He does not have anything connecting his feet to the building, so we are not going to think about how he managed to get there to his position where his feet are on the building for a brief second before he plummets down.*0327

*Does he have a normal force on him?*0338

*No, because the normal force would have to work by going this way, and that does not make sense because there is nothing pushing him against it.*0341

*So you cannot have that.*0349

*The normal force would only work if he were pushed against something, but he has nothing to be pushed against to, he is just going to plummet down to the ground.*0350

*What happens if you were on an incline?*0367

*So we looked at the two extreme cases, flat surface and vertical surface, either the normal force is going to be completely there, or it is going to have no effect.*0369

*If you were on an incline, them some of it is going to be able to be perpendicular.*0390

*We can now create a right triangle, that has this part perpendicular, and this part parallel to the top of the triangle.*0394

*There will be a parallel section and a perpendicular section.*0406

*The important part is that, this is a right angle, and this is a right angle, and this is parallel to this.*0418

*So we know this is a right angle, so we know how much is perpendicular to the surface, and we know how much is parallel to the surface.*0428

*Let us figure out what would be the angle up here.*0434

*If we look at this as a slightly larger version, here is angle θ , and here is a right angle, we can figure out from basic trigonometry what that angle is.*0441

*We know that across two intersections, we know that these have to equal angles.*0462

*So, this θ must be equal to this other angle over here, which means if we want to figure out what these two components are, we are going to get cos(θ) × F _{g} over here (because that is the side adjacent), and sin(θ) × F_{g} is going to be the parallel.*0468

*In general that is a rule to remember.*0506

*But it is important to make a diagram and understand yourself than slavishly fall over formula, if you do that you are liable to make mistake, the math can go wrong, it is up to us pay attention to what we are doing, and that is incredibly important.*0517

*But in general, if you are able to get this angle here in this right angle format, then we are going to have that sin(θ) is going to be the thing connected to the parallel part, and cos(θ) is going to be the thing connected to the perpendicular part, which makes sense,*0546

*because if θ was zero, if we were on a flat thing, then cos(0) would be 1, i.e. all of the force of gravity would be translated to the perpendicular and if it were 90 degree (perfectly vertical), then sin(90) = 1, which means all of the gravity would be translated into the parallel.*0566

*So, sin(θ) correlates to the parallel, cos(θ) correlates to the perpendicular, once we draw the diagram this way.*0590

*Depending on the diagram you use, you might wind up changing it out, or it might be that your problem works differently, so it is important for you to pay attention to what you are doing, and have it make sense, but this is a general structure that you can follow, also you would like somebody else to understand your work later.*0596

*First example:*0623

*We have got an object with mass of 5 kg sitting on a perfectly flat frictionless plane.*0624

*It is pushed by a force of 3 N to the North, and a force of 4 N to the East.*0629

*What is the acceleration of the object, what is the magnitude of that acceleration, and at what angle does the combined force move the object along at?*0634

*When we are saying it is on a perfectly flat frictionless plane, we imagine it like the middle of a frozen lake.*0647

*We are looking down from above.*0653

*So, we have got North, East, South, West like on a compass.*0655

*We make East-West the x positive direction.*0666

*That seems pretty reasonable, we could make North as x positive, but we want to make it something we are used to, not complicated, so we make East x positive and North y positive.*0670

*Some if we were to go South, that would be negative in the y component of our vector, and if West, that would be negative in the x component of our vector.*0685

*With that let us look at what our picture is, we have got our block sitting still at the moment, and it is pushed by a force of 3 N to the North, so 3 N 'up', and 4 N to the East, so 4 N to the 'right', 'up' and 'right' from the way we are visualizing it.*0692

*The important thing is that we have set up a new coordinate system to understand this.*0725

*So, 3 N positive y, and 4 N positive x.*0730

*That means now x component of our force is 4 N and y component of our force is 3 N .*0733

*So, if we want to find out what the acceleration of our object is, we are going to have to find out what the acceleration is for each of the components, in x and y, and together that will give us the acceleration vector, the combined acceleration.*0745

*We appeal to Newton's second law: F _{net} = ma, there is no other forces than this one force, actually two forces, but they do not interact since they are in different coordinates axes.*0760

*So we do not have to worry about any other forces, so F _{x} = ma_{x} , 4 N = (5 kg) × a_{x} , a_{x} = (4/5) m/s/s .*0772

*We do that for the y, F _{y} = ma_{y} , 3 N = (5 kg) × a_{y} , a_{y} = (3/5) m/s/s .*0800

*Now we have got our two pieces, we can put them together, we get, a = (4/5,3/5) m/s/s .*0831

*Here is our first answer.*0850

*If we want to find out what the magnitude of the vector is, that is just going to be the square root of each of the components squared.*0852

*So, square root of (4/5) ^{2} + (3/5)^{2} = square root of (16/25) +(9/25) = square root of (25/25) = 1 .*0859

*So, the magnitude of our acceleration vector is 1 m/s/s .*0883

*Finally, to figure out what the angle it is coming at, we want to set up a triangle so we can see the angle more easily.*0888

*So, here is a picture of what is going on.*0897

*We have got 4 N to the right, which has now become (4/5) m/s/s to the right, and also (3/5) m/s/s to the up.*0899

*If we want to figure out what angle that is, now we have got a triangle where we can set things up, we can see where the angle goes.*0919

*The angle above the horizontal, above the East, the angle North of East, we are going to able to figure that out by saying, tan(θ) = (y component)/(x component) = (3/5)/(4/5) = 3/4 .*0929

*Taking the arctan of that, θ = arctan(3/4) = 36.87 degrees .*0954

*Next example:*0967

*A 20 kg block is on an incline of 50 degrees with a rope holding in its place.*0968

*Assuming that the incline is frictionless, what would be the tension in the rope?*0973

*Then if we cut the rope, what acceleration would the block have?*0976

*First, what forces are at play here?*0980

*We got tension pulling this way, force of gravity pulling down.*0982

*Force of gravity = (20 kg) × (9.8 m/s/s) = 196 N .*0991

*So, how much of this is perpendicular, and how much of this is parallel?*1006

*Up here, we are able to get this as θ from the geometry, so this is going to be equal to 50 degrees.*1013

*So, force of gravity perpendicular = cos(50) × (196 N) = 126 N, and the force of gravity parallel = sin(50) × (196 N) = 150 N .*1028

*Now, there is one more force operating over here, and that is going to be the normal force.*1058

*So, the normal force, it can only work on what is operating perpendicular to the surface, how much is operating perpendicular to the surface?*1066

*The force normal = force of gravity perpendicular, which means that this force will cancel out this force, so the net force is only going to be the force of gravity parallel and the tension in the rope.*1073

*Now, is this thing moving?*1087

*We know that it is not, it is currently still, the rope is taut and it is staying in place.*1089

*So we know that the tension, we will make this the positive direction, Tension - force of gravity (parallel) = mass × acceleration = 0, because we got zero acceleration.*1094

*So, T = F _{g}(parallel) = 150 N .*1112

*So we have got the tension.*1121

*If we cut the rope, how much acceleration will the block have?*1124

*Now, it is not going to have anything pulling it away, it is only going to be moving parallel, because what we have done is, we have broken down it into the perpendicular and parallel.*1128

*The perpendicular is canceled out by the normal force, because that is how much force is translated to through the surface perpendicular into the object.*1136

*So the force of gravity that is perpendicular to the surface gets canceled out, but the force of gravity parallel to the surface will be able to accelerate the block along a parallel vector to the surface of the triangle, it will accelerate this way.*1143

*We do not know what 'a' is yet, but we can figure it out.*1161

*We had, T - F _{g}(parallel), but now there is no more T, so all we have is F_{g}(parallel) = (20 kg) × acceleration, (150 N)/(20 kg) = acceleration.*1164

*So our acceleration = 7.5 m/s/s .*1198

*What direction is that going?*1207

*It goes in the direction parallel to this.*1209

*This is just an acceleration, not a vector, we could figure out what the components are to the F _{g}(parallel), but that is going to take a lot of time, there will be some part x, some part y, and it will take a lot of effort, instead we can just know what it is off of our diagram, *1212

*because we do not have to say specifically what it is, we can just say it is working parallel to the top of the triangle, it is working parallel to the hypotenuse of the triangle, because that is really the most interesting thing, that gives us the useful information.*1227

*So, a =7.5 m/s/s, parallel and sloping down along the triangle.*1238

*Final example:*1252

*We have got a 10 kg block on an incline of 20 degrees, and it is attached by a rope to a free hanging block of 5 kg .*1254

*Assuming that it is frictionless, the rope and the pulley are mass-less, what is the acceleration of the blocks, and what is the tension in the rope?*1260

*As done before, we are going to look at this as a system.*1270

*It is a lot easier, although we could break it into the tensions and the gravities experienced by the blocks.*1274

*This one does not have any force normal operating on it, it is just going to have the force of gravity, which in its case, = 5 × 9.8 = 49 N .*1282

*How much gravity does this one have operating on it?*1298

*Its force of gravity = 10 × 9.8 = 98 N .*1301

*If we want to break this into its perpendicular and its parallel component, then here is our angle of 20 degrees.*1310

*So here we are going to have, cos(20) × F _{g}, so force of gravity (perpendicular) will be canceled out by the normal force, F_{g}(perpendicular) = cos(20) × 98 N = 33.5 N, but sorry!!, I gave you the wrong one there (parallel), but there is no need to worry since the normal and F_{g} (perpendicular) are going to cancel out, so we just have to worry about the parallel here.*1320

*We know the acceleration perpendicular to the surface of the triangle, because we know that they are going to get canceled out, because of the way the normal force works.*1386

*So, all we care about is, how much is parallel.*1394

*How much is parallel will be, sin(20) × F _{g} (parallel) = sin(20) × 33.5 N .*1396

*Like before, we are going to consider this as positive direction, so positive goes around the corner, goes down, we are going to consider it the same as we did in the last lesson when we worked out a problem similar to this.*1420

*The only thing left, we have not talked about these tensions working in the rope.*1432

*But, if we look at the system as a whole, the only thing pulling in an external fashion, are the two forces of gravity (the one that is going completely down), the tensions are going to be canceled out when we look it as a system, because the rope is going to be taut the entire time, so we can consider it to be a rigidly moving system, where the accelerations are going to be the same.*1440

*We have done a problem like this before, you should go back a lesson and look at the final example from that one, because that is very similar, we are just throwing in the incline to make the things little bit more difficult.*1469

*Now we know what our forces are, the only two forces we need to care about, are forces of gravity, for this one, = 49 N, we want to find out the acceleration and the tension in the rope.*1478

*For this one, force of gravity (parallel) = 33.5 N .*1496

*So, what is the mass over here?*1505

*This has a mass of 10 kg, this has a mass of 5 kg .*1507

*So, what are the forces acting on the system?*1513

*This is the positive direction, F _{net} (system) = mass (system) × acceleration, the accelerations being the same (since rigid).*1515

*So, 49 N - 33.5 N = (10 kg +5 kg) × a , 15.5 N = 15 × a, a = 1.03 m/s/s .*1536

*So there is going to be an acceleration this way, and an acceleration this way.*1563

*For this one it is going to be going directly down (for the smaller block), and for the larger one it is going to move parallel and up the surface of the triangle, and they are both going to be 1.03 m/s/s .*1568

*What is the tension?*1580

*For this, we just look at the free-body diagram.*1582

*Let us consider the smaller one.*1585

*It has some tension T, and it has some force of gravity pulling down.*1587

*We know what the force of gravity already is, it is 49 N .*1593

*What is the tension, we do not know that, but we do know that it has an acceleration = 1.03 m/s/s, and we know its mass = 5 kg .*1596

*We have got that, F _{net} = ma , F_{g} - T = 5 × 1.03 , 49 - (5 × 1.03) = T , T = 43.85 N , which makes a lot of sense because if the block is going to be able to fall, there need to be slightly less tension in the rope, because it has to have a net force that is stronger downwards than upwards, otherwise it will not accelerate down, it will accelerate up.*1610

*That finishes it for this lesson, hope you enjoyed it.*1661

1 answer

Last reply by: Professor Selhorst-Jones

Sun Mar 12, 2017 6:39 PM

Post by bienvenido villabroza on March 11 at 09:54:00 PM

For example 1, can my answer for the question "what is a for the object" be 1 m/s^2, 36.87 North of East? I understand why the answer in your slide is (4/5,3/5) m/s^2 but would my answer be acceptable as well?

2 answers

Last reply by: Kuthala Nathan

Sat Jun 6, 2015 3:34 AM

Post by Kuthala Nathan on June 6, 2015

Hi Professor,

I really enjoy your lecture and it is very easy to understand. Thank you!

In Example 3, why is the F net system 49N-33.5N and not 49N+33.5N ? Thanks in advance

2 answers

Last reply by: Tori Carroll

Sun Oct 5, 2014 12:25 PM

Post by Tori Carroll on October 4, 2014

How would the acceleration be found if the plane was not frictionless?

4 answers

Last reply by: Jack Wilshere

Sat Jan 11, 2014 12:44 AM

Post by Hoe Li En on January 4, 2014

hi, with regards to the diagram at 10 min 04, the hypotenuse of the triangle represents the force of gravity, while the (cosÃŽËœ)Fg represents the normal force...what about the (sinÃŽËœ)Fg ? what does it represent? the force of friction acting on the object?

1 answer

Last reply by: Professor Selhorst-Jones

Thu Oct 31, 2013 9:21 AM

Post by Benny Hui on October 31, 2013

Hi Professor,

This is my first time to see your lecture. It is clear and easy to understand. Thank you for making this video.

Going back to the example 3. There was a gravity acting on the 5Kg block. So, by F=ma, you deduced that there was a 49N downward force acting on this block. Using the same formula, you deduced that there was a 33.5N acting on the 10Kg block along the slide.

My question is this. Why can't we see the body as the whole system and say the tension between two blocks was 49N - 33.5N = 15.5?

Thanks

1 answer

Last reply by: Professor Selhorst-Jones

Mon Nov 19, 2012 9:06 AM

Post by varsha sharma on November 19, 2012

In example 1 why the vertical component is considered x direction and vice versa ? should'nt the vertical components be y and horizontal be x components.

1 answer

Last reply by: Professor Selhorst-Jones

Sun Oct 21, 2012 11:57 AM

Post by Ahmed Almushqab on October 21, 2012

In the slide "Example : force in componenet form"

how did you get the result (8.66 & 5 )

cause 10sin30 & 10cos30 dont give the result that you wrote sir

could you explain please ?