Pearson+ LogoPearson+ Logo
Start typing, then use the up and down arrows to select an option from the list.


Learn the toughest concepts covered in Physics with step-by-step video tutorials and practice problems by world-class tutors

7. Friction, Inclines, Systems

Systems of Objects with Friction


Connected Objects With Friction

Play a video:
Was this helpful?
Hey, guys. So now that we understand friction a little bit better in some problems, you're gonna run into situations where you have systems of objects where you have these two objects or more that are connected. They could be touching each other or they have some cables or something like that. And there's also going to be friction. And the basic idea here is that whenever this happens and friction is not negligible, you're gonna have to consider the friction that acts on each individual objects. Really, the steps to solve these problems are gonna be the same. Just remember that connected objects always have the same velocity and acceleration. Basically, they always move together as a system. So let's go ahead and take a look at this problem here. We've got these two blocks. This a 5 kg block. I'll call a this one. I'll call B, and we have the coefficients of friction we want to figure out. First, we want to draw a free body diagrams for both objects and then find the acceleration of that system. So that means that we're not gonna take the shortcut here because we want to draw the free body diagrams for both objects, and we also have to consider the friction acting on each. Okay, so we've got the free body diagram for this one here. This is going to be M A G. And then we've got our attention force that acts this way. And then we've got a normal force. Now, the last thing is, we also have some friction. And whether this block is going to be moving to the right or it wants to move to the rights, the friction is going to act backwards. But actually, in this problem, we know exactly what kind of friction we're dealing with. We're told that the objects start moving, which means we know we're dealing with kinetic friction. So we're dealing with kinetic friction here. All right. And so now let's go ahead and take a look at our second block. Now we have the weight force MBG. Now we have our applied force. This F is equal to 90. Remember this block? The 10 kg block is the one that's being pulled with 90 newton. So that's our applying force here. And now, because of an action reaction pair, we also have some tension that's going backwards, and then we have some normal force. And then finally, some friction. So just like before, we have a friction force that points to the left. This is F K. All right, so those are free body diagrams, and we've actually already covered the next step as well. But determining the type of friction we're working with, remember, Sometimes the problem text will just literally tell you what kinds of kind of friction new you're working with here. All right, so basically, we're just gonna go ahead and start our f equals m a. And we're gonna start with the simplest or the fewest number of forces. So that's going to be blocked a over here. So if I want to figure out B, which is the acceleration, then I'm gonna look at Block A and I have to write out my F equals M A. So I'm looking for the acceleration here. All right, So what are my forces? Well, I'm gonna have to pick a direction of positive first. And because the blocks are being pulled to the rights, it's probably a safe bet that the system is going to accelerate to the rights, so that's gonna be my direction of positive. So when I expand my forces, I have attention minus f k equals M A. All right, So then what I can do here is I can actually expand and that can replace this f k with. It's an expression. Uh um u k times the normal. This is equal to m a. All right, remember, the normal force is actually, uh I have to go back into the diagram and figure that out. So here are normal force. If this blocks are only sliding horizontally, those have to cancel. So, basically are two vertical forces here have to cancel out some normals equal to mg. So what we can do here is tension minus mu K m G equals m A. And so finally, we can go ahead and start replacing some values. We know that the coefficient is 0.3 and then we've got five times 9.8 equals five a. So if you go ahead and work this out, you're going to get tension minus 14.7 equals five. Now, unfortunately, we can't go any further with this because we want to figure out the acceleration. But we can't because we don't have the tension force. So remember, whenever this happens, you're just going to go and start the other object and see if you can find an equation there. So here we got our forces equals mass B times the acceleration. So this actually was mass A. So we've got to write all over forces here. Our forces are going to be the F, which is the 90 minus your tension force, which is backwards. Remember, both of these are actually gonna be negative. So we're gonna have minus f k equals M B A. All right, so now we have this F minus tension minus u k times the normal equals M b A So f minus Tension minus u k. And now we're gonna use the normal force. Not for this object, though. We're going to use it for this object. And so remember, this normal force here is just gonna be equal to empty BG times G. So this is gonna be MB times G. This one is m eight times g. All right, so we have m b times g is equal to M B times A. Okay. And so now we've got here is We've got 90 minus tension minus, and this is going to be 0.3 times 10 times 9.8 is equal to 10 A. So you get 90 minus tension minus. And then this just turns into a number. This is 29.4. So this is equal to, um, 10 A. And so what happens is these two are numbers, right? You have 90 and the negative 29.4. So you can actually simplify one step further. And you can say this is 60.6. Minus T is equal to 10. All right, now we can't go any further. We have an acceleration in time. Uh, sorry. Acceleration. Intention. So I'm going to label these two equations here. This is my first equation here. My first equation of, you know, with two unknowns. And then this is the other one here. So this is like to now remember, with these kinds of problems, we're just gonna solve them using equation, addition or substitution. We're gonna pick addition here, So let's go ahead and do that. So we're gonna line them up this equation Number one, we've got tension minus 14.7 equals five a, and I'm gonna scoot this over just a little bit. Tension. And then finally, we have our 60.6 minus. Tension equals 10 A. So here's what happens is when we add these equations straight down, we add these equations like this, you're gonna get attention. That cancels with tension and basically 60.6 minus 14.7. So you have what this becomes is 45.9 equals 15 A. So now we just saw for the acceleration and it ends up being 3.6 And that's the answer. So let me know if you guys have any questions, really. Just, you know, the same steps that we've been working with so far? Let's move on to the next one.

Two blocks are connected by a cord over a pulley. Block A rests on a rough tabletop. Block B has mass mB=2kg and hangs over the edge of the table. The coefficients of friction between Block A and the tabletop are μs=0.6 and μk=0.4. What is the minimum mass Block A can have to keep the system from starting to move?