Hey, guys. So up until now, we've been dealing with Newton's first and second laws. Remember that the first law is the law of inertia, and the second law is f=ma. And we use f=ma. We're usually just looking at all the forces that act on one particular object. Now we're going to start to see what happens when objects exert forces on each other. So to do that, we're going to need Newton's third law. So let's go ahead and get into it. Newton's third law is otherwise known as the law of action reaction. And basically, what action reaction means or says is that every action or more specifically every force is going to result in a reaction force. And there's a couple of things you should know about this reaction. t's of equal magnitude, meaning it's the same size or the samestrength, but this force acts in opposite directions in the opposite direction. What do I mean by that? So imagine you're person A; you go up to a block, and you push on block B. So we're going to have a force here, and let's just call this force 50 Newtons. Now, because this is the block this is the force of you on the block, I'm going to label this F_{AB}. So this is an action This is a force. So what Newton's third law says is that there's an equal magnitude but oppositere opposite directionreaction force, meaning it's the same 50, but now it's going to point backwards. So how do we draw this Do we draw this like this And do we call this F_{BA} is equal to negative 50 because it points to the left Well, no. Because if every 50 newtons was countered by an equal and opposite 50 on the same object, then nothing could ever move, because everything would always be at equilibrium Every 50 would be countered by a negative 50. So what's really really importantabout Newton's third law is that all forces exist in action reaction force pairs, and these force pairs always act on different objects. Meaning, you, person A, push on the block with 50 Newtons. Thethe block pushes back on u with an equal but opposite 50 newtons So this is F_{BA} that acts on u, and this is going to be negative 50 Newtons. Thattakes careof the fact that itgoesbackwards. So in generalwhat this means is that F_{AB} is the negative of F_{BA}. That's how you'll see this written here in yourtextbooks. So what this means is that thisF_{AB} is the actionandthe reaction is the negative F_{BA}. There's a coupleof realworld examplesthat we'vebecome reallyfamiliarwith. For example, the normal forcewhich is aresponse toa surface push,isactually an action reaction pair. So if this block here is Aand the floor hereis Bthe weight forceof A██ pushes down on the floor. So the weightforce is the action And in response,the floor pushesback onblock A That'sthe reaction,and that'sthe normal force Nowthese thingsactuallydon't have to be touching Another exampleof this is the weight force. So the earthis goingto pull on youtowardsthe earthcenter This is the weight force Thisis mg. This is the action But you alsopullbackonthe earthwithan equalandoppositereaction force. So this is a reaction. So youactually pull on the earth just as hard as it pulls on youTheonlydifference is your mass compared to the earth is way, way smaller, so you'regoing to accelerate more than the earth does. Alright, So let's get into the example here. So you're at 80 kilograms. You stand on a frozen lake with a 40 kilogram ice block, and you push it with 20 newtons So I'm going to drawthis outreal quick This is80 andyou're upagainsta 40 kilogram block And you'regoingto push it We knowthisis Fwhich equalsthisis goingto be 20.AlrightSo what I'm going to do here is I'm going to call you A, the 80 kilogram person, and I'm going to call B the block. So what thatmeansis thatthisforce hereis actually F_{AB}, and that's 20 We want to doin thisfirst part is we want to figure out the forcethat the blockexertson youSo what does thatmean In this§first block in in thisfirstpartherewe want to╒figure out not F_{AB}. We want to figure out F_{BA}. So to do that, we're going to have to stick to the steps. We know the first thing we're going to do is draw a free body diagram. But now that we're talking about multiple objects, what you have to do is you're going to have to draw free body diagrams for each of these objects. Remember,we have 2 now in this problem. So I'm going to go ahead and do that over here.I've got the 40 kilogram block. I've got the weight force. That's what I check for. This is the weight Then I have any applied forces, which I know I have. This is my F_{AB}, and I know this is 20 And then I also have I don't have any tensions. So I've got a weight. I've got a apply force, but I have no tensions I do have a normal force because you're you're on the frozen lake This is the normal force And I've got no friction. So if these are the only 2 verticalorcexordinates,the only horizontal forceis F_{AB}. Now, what we want to do is we want to figure out the free body diagram for you. So you're going tobeoverhere.You have a weight as well. This is W equals mg. And so what we said from Newton's third law is that if you push on the block with_a force of 20newtons, thenthe actionthat'sthe action. Then the reaction is the block pushes back on you with an equal but opposite force. So that means that there's this forceoverherethat points to the leftThisis negative F_{BA}. And so what does that equal Well, if F_{AB} is 20,then that means that negative F_{BA} is negative 20. And so that's the answer. So I'm just going to finish off the free body diagram over here. So thatmeansthat F_{BA},by Newton's thirdlaw, is equal to negative 20 newtonsRight,Solets keep going now.Now we want tocalculate the acceleration of the block.So now we want to calcalculateA_{B}, so I'm goingto callthatSo what we want to do is if we want to figureout the acceleration of B of the block, we're going to have to use f=ma Butrememberwe're goingto do xm...

6. Intro to Forces (Dynamics)

Newton's Third Law & Action-Reaction Pairs

6. Intro to Forces (Dynamics)

# Newton's Third Law & Action-Reaction Pairs - Online Tutor, Practice Problems & Exam Prep

1

concept

### Newton's Third Law & Action-Reaction Pairs

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#### Video transcript

2

Problem

ProblemWhich of the options is NOT an action-reaction pair in the following situation? A book slides across the floor, slowing down due to friction.

A

Friction on the book from the floor & friction on the floor from the book

B

Weight of the book from the Earth & gravitational force on the Earth from the book

C

Weight of the book & normal force on the book

D

Normal force on the book from the floor & normal force on the floor from the book

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PRACTICE PROBLEMS AND ACTIVITIES (10)

- Boxes A and B are in contact on a horizontal, frictionless surface (Fig. E4.23). Box A has mass 20.0 kg and bo...
- World-class sprinters can accelerate out of the starting blocks with an acceleration that is nearly horizontal...
- A small car of mass 380 kg is pushing a large truck of mass 900 kg due east on a level road. The car exerts a ...
- An 85 kg cheerleader stands on a scale that reads in kg. b. What does the scale read if the 85 kg cheerleader...
- A house painter uses the chair-and-pulley arrangement of FIGURE P7.45 to lift himself up the side of a house. ...
- A 75 kg archer on ice skates is standing at rest on very smooth ice. He shoots a 450 g arrow horizontally. Whe...
- The foot of a 55 kg sprinter is on the ground for 0.25 s while her body accelerates from rest to 2.0 m/s. (b) ...
- Blocks with masses of 1 kg, 2 kg, and 3 kg are lined up in a row on a frictionless table. All three are pushed...
- A 1000 kg car is pushing an out-of-gear 2000 kg truck that has a dead battery. When the driver steps on the ac...
- (b) How much force does the astronaut exert on his chair while accelerating straight up at 10 m/s^2?