Ch 04: Intro to Forces (Dynamics)See all chapters

Sections | |||
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Intro to Forces + Newton's Laws | 16 mins | 0 completed | Learn |

Force Problems with Motion | 28 mins | 0 completed | Learn |

Forces with Multiple Objects | 26 mins | 0 completed | Learn |

Vertical Forces & Equilibrium | 34 mins | 0 completed | Learn |

More 1D Equilibrium | 16 mins | 0 completed | Learn |

Vertical Forces & Acceleration | 26 mins | 0 completed | Learn |

Landing & Jumping Problems | 19 mins | 0 completed | Learn |

Forces in 2D | 35 mins | 0 completed | Learn |

2D Equilibrium | 25 mins | 0 completed | Learn |

Concept #1: Forces with Multiple Objects

**Transcript**

Hey guys I now want to talk about problems force problems involving multiple objects Let's check it out. So the big thing to know the big thing to remember is that objects that are connected to objects acting as a system will have the same velocity and the same acceleration. If two or more objects are connected as a system and they move together they have to have the same velocity and they have to have the same acceleration otherwise they'll be getting closer or farther apart from each other. But instead of instead of calling these velocities let's say for two objects V-1 and V-2 that's two variables. I'm just going to call it V because it's the same so I have instead one variable. Same thing with A Instead of calling it A1 and A2 I'm just gonna call it A because it's the same that way I have one variable and now two variables. OK. The way to solve this is very similar to what we'll be doing with problems of just one object and we're going to draw a complete picture and very often you get that already drawn out for you and then for each object you get to do this as many times as you have objects you're going to draw the free body diagram. Free body diagram that means your F.B.D. You gonna write for each object sum of all forces equal ma and we're gonna solve. So if you have three objects that draw three diagrams and you gonna write F equals ma three times and combine those equations. These are the same exact steps as you would do for one object. It's just that if you have two objects you do that twice one for each. OK. So let's try this example here. Two blocks masses two and three. Are initially at rest on a horizontal frictionless surface connected to each other by a light string. So there's a string here so there's a tension, you pull horizontally on the three so this is your pull right there on the three with a force of F equals 15 Newtons. It's a constant force. We want to calculate the system's acceleration and the tension between the two blocks. OK. So I want to know what is (a). That is the entire system you have remember (a1) and (a2) are the same things so I can just call them (a) and I want to know what (a) is and I want to know what is the tension between the two right here. OK. Now I have two objects and where do I start my tip here pro tip number 1 is that you're going to begin with the simplest objects. Now what do you mean by simplest. You want the object that has the least amount of forces acting on it and that's usually the object that is at the ends of the problem. OK or at the edge of the problem. And in this case will be object number 2. Let me show you real quick, object number two has tension pulling to the right. Object three has tension pulling it to the left. By the way these two tensions are the same because it's the same strength. And it has this F here, so you can see this guy has two forces on it and this guy only has one force on it. So I'm going to start with this because it's simpler. OK. So sum of all forces equals ma, let me draw the free body diagram this is not a free body diagram . Free body diagram would look like this. So free body diagram for the two and the three is like this. OK. So the sum of all forces on the two equals the mass of the two and the acceleration of the second block of the two kilogram block. But all the accelerations are the same. So I can just call this a. I'm gonna do the same for the third object. Some of all forces on three is the mass of three and the acceleration of three which is just a. So two has only one force on it. So that force is T going to the right so I'm gonna call this plus T, mass two is two kilograms that's why we call that (m2) because it was two kilograms and acceleration we could call it (m1), (m2) if you want it. So that said there's nothing else I can do here. And notice that I'm stuck. I don't know T I don't a. So let's continue here with this next block. You're supposed to get stuck you're supposed to not know this and to write all the equations. So this block has T to the left so negative T and F to the right. Mass is three and acceleration I don't have. I'm going to move something around here so this is F is 15 minus T equals 3a. Again notice that just like here I had one equation two unknowns here I also have one equation two unknowns. But they're the same 2 unknows. So now I have two equations with the same two unknowns it's called the system of equations and I can solve this. Now the way you probably are used to doing this is you can let's say I can plug in 2a for this T over here it would be negative 2a because it's negative T and then I would be able to solve for a. Right so I can plug one isolate one variable one equation plug it into the other and all that but for these problems you're always going to be able to add the two equations and that's what I want to show you. So I'm gonna get this equation right here and stack it up over here and add it to the first equation. Now every time you do this tension the tensions are going to cancel because all the tensions are doing is holding this thing together so the tensions will cancel. And the you do this by the way you add everything that's on the left together. So all I get here is a fifteen that's why the T cancels because I was combining them and then you add everything on the right together that's gonna give me a 5a. Ok so you're always going to be able to do that. And this gives me a equals 15 over 5 that is 3. Once I have that to find T so that's part (a) a is 3. Once I have that to find T I just have to plug a back in there. Right. So I just plug a back over there and I know that T equals 2a so it's 2, 3, 6 Newtons. Alright so that's the final answer for these two.

So you're always going to be able to add up the equations even if you had three objects four objects. You're always going to be able to add it up. So let me show you whether you can do this a little bit easier, even simpler than this. Right. You can combine the objects to find the acceleration if you look at this you're pulling with 15 on the three but the two is connected to it. So if you think about it you're really pulling five kilograms total mass right. So I could combine the objects to find the acceleration so instead of two and a three I can just say that I'm pulling a five and with a force of 15. Now it's just got a whole lot easier. I could just right sum of all forces equals ma, their is a single object so the force is 15 and mass is of five so the acceleration is a three. And I did that in three lines instead of how many of these took. The problem with this is that I can't I cannot do this. I can no longer find T once I do this at least not with this diagram. And the reason is since I merge these objects here T is kind of like in here somewhere right. It's now an internal force. You wouldn't be able to see T it doesn't exist anymore basically so to be able to solve for T I'm going to have to go back to one of these two original diagrams right and I'm gonna pick the simplest of the two. Tension is touching the two kilograms and the three kilograms. They both have tension on them. So I can just pick one of the two of them I'm going to pick the second two kilogram because it's the simplest one. Right. Some will say here's a two kilogram it's being pulled by a T and I'm going to say some more forces and that two is the mass of the 2a. And the force is a tension positive the mass is two kilograms and the acceleration I just got over here is a three, six Newtows So this whole question got solved in five lines two of which are just F equals ma, ok so you can combine things. But then once you find the acceleration you're going to you're going to have to write the F equals ma for individual objects and plug your acceleration there. OK. So that's what happens when you have two objects that are connected together. Now a set up very similar to that applies also to these two other situations here. When you have a mass inside another mass. So mass moves with another mass inside like a passenger in a car. So imagine that this is a car right here it's a really crappy car and there is a seat in. It's terrible. And you're sitting on the seat. Ok hope you can do better than that. So in other situations if you have two objects stacked up on top of each other little m and Big M. Ok. So the idea is those they let's say the passenger is little m and the car is big M. The idea is there's a force. There's a force that the engine applies on the car. Or maybe it's a cable that's pulling the car whatever. And that force equals to, it's the force on the car. So it's the mass of the car and the acceleration of the whole system. There is also a force that's applied on the driver. The force of the seats on the driver when you're in a car. The reason why you go forward with the car is that the seat pushes you forward because the seat is fixed in the car. And this would be mass of the driver. Because it's the force on the driver and acceleration and these two accelerations are the same, the forces aren't but the accelerations are and by the way when you do this you can combine this whole thing as just one objects and you could use this to be the total mass of both. So I'm just kind of explaining this actually worked out this problem in example two. But for now I'm just give you a brief look, here let's say push here with pull here with an F pull of whatever that is.

That F Pull is going to be ma. Now if this object on top comes moves with moves with the object in the bottom it's because there's some friction here and we'll do this later. But just conceptually if they're both accelerating to the right together they both need to force these guys being pulled on. This guy is somehow again we will talk about being pushed as well. So F of little m on little m is little m acceleration and the accelerations are the same because they move together. So again we'll do this later when we deal with friction. But for now let's look at this guy inside the car. So it says one in 2940 Newton force accelerates a 900 kilograms car, how hard is this C push on an 80 kilogram driver. So we're gonna draw simple version of this. Here's the car mass and this guy is here in the back of the car being pushed forward little way and the force that's pulling the car, pushing the car forward doesn't matter is 2940 Newtons. All right. And I want to know how hard is the seat pushing on the driver. So this is going to cause some sort of acceleration to the car and the driver is also going to have that acceleration. So I want to know the force that the seat pushes on the driver. So I want to know since it's on the driver, sum of all forces on the driver is mass of the driver which is little m and a. The only force on the driver here is the force of the seat on driver. Which is what I want to calculate, his mass is 80 and the acceleration I don't have some I'm stuck. But guess what this acceleration is going to be acceleration of the car, acceleration of driver is the same as the acceleration of the car. So I can go back here and try to figure out the acceleration of the car. Now this force 2940 is pulling both the car and the guy the driver. Right so the driver is inside the car it's all one big system so I can write that this one force, sum of all forces on the system is the mass of the system and the acceleration of the system which is just a. The force is 2940. The mass is 900. Plus 80. And if you move everything around you you that a equals three meters per second square. Which I can now plug back in here and 80 times 3. The guy is being pushed forward with 240 Newtons. So the big thing to remember here guys is that whenever you accelerate whether you're getting faster or slower or whenever your velocity is changing there has to be a force acting on you. And if two objects are connected then they have to accelerate together. Very often you gonna have to find acceleration of one and plug it back into the other to be able to solve for what you're looking for. All right. So that's it for this one. If you have any questions then let me know.

Practice: A 0.5 kg rope pulls a 15 kg block across a frictionless horizontal surface. If the block moves with 2 m/s2 , how much force pulls the rope forward? (Hint: if the rope used to pull an object is NOT massless, it acts as an additional object).

Practice: Two blocks are initially at rest on a horizontal frictionless surface, as shown. You push on the 4 kg block with a constant horizontal force F. If the contact force between the two blocks is 12 N, calculate the magnitude of force F.

Practice: A 5,000-kg truck carrying a 300-kg crate on its horizontal flatbed comes to a complete stop from +20 m/s in 40 m. What force (use +/- to indicate direction) must the truck apply on the crate, so that it stops without slipping forward?

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Concept #1: Forces with Multiple Objects

Practice

Practice

Practice

Two boxes, one of mass M and one of mass m, lie next to each other. If a force F is applied on box M, as indicated in the figure, draw a free body diagram for each box, with the appropriate value for each force given in terms of known vairables (M, m, F and g). Assume there is no friction along the surface.

A 6,00-kg block is in contact with a 4.00-kg block on a horizontal frictionless surface as shown in the figure. The 6.00-kg block is being pushed by a horizontal 20.0-N force as shown. What is the magnitude of the force that the 6.00-kg block exerts on the 4.00-kg block?A) 8.00 NB) 20.0 NC) 10.0 ND) 6.00 NE) 4.00 N

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