Stacked Blocks - Online Tutor, Practice Problems & Exam Prep

In stacked block problems, two blocks are connected without ropes, with friction between them. When pulling the bottom block on a frictionless surface, the maximum acceleration before they slide apart is determined by static friction. The equation for maximum acceleration is $a=\mu g$, where μ is the coefficient of static friction and g is the acceleration due to gravity. Understanding these concepts is crucial for analyzing motion and forces in connected systems.

Stacked Blocks

Hey, guys. So we've become really familiar with connected systems of objects and also friction. And in this video, I want to talk about a specific type of problem where you have objects that are connected not by ropes or cables, but they're basically just stacked on top of each other. I like to call these problems stacked blocks problems and there's some really interesting things that happen here. So let's go ahead and take a look. Now, we're gonna come back to this point in just a second here. We're gonna skip and start with the example because it's actually very easy to understand what's happening here. The idea is that we have these 2 blocks. 1 is 10 5 kilograms. I'll call this 1 a and this 1 b. So the idea here is that the floor is frictionless. There's no friction on the bottom surface, but between the two blocks, there is going to be friction.'

So what happens is I'm gonna pull the bottom block with some force, which means I'm gonna give it some acceleration. Notice how the two blocks stay together. What I wanna do is I wanna figure out the maximum acceleration that I can give the bottom block so that the two boxes remain moving together. Here's what this means. You pull, there's some acceleration but the boxes stay together. You pull harder, there's more acceleration, and the boxes still stay together. Eventually, you can pull hard enough, you can give the bottom block an acceleration fast enough that the boxes actually start sliding relative to each other, they're no longer moving together. I wanna figure out the maximum acceleration that I can do that with.

So basically, this is just gonna be like any other connected objects with systems with friction type problem. We're gonna draw the free body diagrams for both. Let's go ahead and get started here. So here I've got A and B. The free body diagram for A is gonna look like this. Actually, let me scoot this down a little bit. So I've got my weight force. This is mg, and then I have any applied forces or tensions. But there's no applied force that's acting on the top block. Remember, I'm only pulling on the bottom block here. So when I draw this free body diagram, there's no F. So I've got a normal force though because I've got these two surfaces in contact. One way you could think about this is that A, the weight force push down on B, so there's a reaction force to that surface push. So I'm gonna call this the normal between the two objects and B A.

Now finally, what happens is we're gonna look at friction. Right? So what happens is here, you've got this block and it's moving to the right. So there has to be a force that's acting on the top block to keep it accelerating and moving to the right as well. And that force is the force of friction. Remember that friction tries to stop or prevent velocity between the two surfaces. So we're pulling on the bottom block, it's moving, and friction actually keeps the top block moving as well. So when you have objects that are stacked on top of each other, the force that acts on the top object that causes it to move is the force of static friction.

The last thing I wanna talk about is we have remember when we talked about static and kinetic friction, we basically said moving versus not moving. We could be a little bit more specific here because now these two surfaces can possibly move relative to each other. So to be more specific, the friction is gonna be kinetic whenever the relative velocity between the two surfaces is not 0. Anytime you have these two surfaces that are sliding relative to each other, there's gonna be some kinetic friction. Friction is static on the other hand whenever the relative velocity between the two surfaces is equal to 0. Basically, anytime they are moving together like this, that's gonna be static friction.

So let's go ahead and get started here. We already know what the type of friction that we're dealing with. We actually have to go ahead and draw the free body diagram for B. So Let's go ahead and do that real quick. We've got the weight force that acts on B. And now we've got the applied force. This is my F. We have the normal force that's acting from the floor. This is from the floor onto B. I'm gonna call this N B, but there's also another normal force that's basically part of an action reaction pair. B pushes upward on A, so A pushes downwards on B because of action reaction. So this is a downwards force, and I'm gonna call this NAB. There's one more action reaction pair though as well. So remember, we said that there's friction between the two surfaces. There's gonna be a friction force to the right on the top block. Because of action reaction, there's an equal and opposite force that acts on the bottom block. So on the bottom block, there's another friction force that acts to the left. So really these two friction forces are actually going to be the same. These are the same fs's.

F = m a

If we're trying to figure out the acceleration, we're gonna start with the simplest object. So basically, we're gonna start with object A. So we got the sum of all forces, and really there's only one force to consider, and I'm gonna choose the right direction to be positive. So I've got this fs, the static friction here, and I'm trying to figure out the maximum acceleration. So if I figure out the maximum acceleration right before the objects start sliding relative to each other, then I'm not just going up against any old friction or static friction. This is actually going to be fs max. What I'm looking for here is the acceleration where this the static friction is going to be maximum. So we have an equation for this.

μ _s ∙ N BA = ma a _max

If you think about this, these two forces, the weight and the normal, have to cancel each other because the top block isn't accelerating vertically. So this NBA here is really just the weight force. So that means we have μ static times mag equals ma times a max. And if you can see here what happens is that the mass of block A actually cancels out from the equation. And this a max is just going to equal μ static times gravity. So it's really just going to be 0.7 times 9.8, and the acceleration maximum is going to be 6.86 meters per second squared. This is the maximum acceleration. Anything faster than 6.86, and the blocks now start sliding relative to each other, and the friction actually becomes kinetic.

I don't even have to go into the f equals m a for object B, and that's it for this one. So hopefully that made sense, and thanks for watching.

Stacked Block Tied to Wall

Everyone, welcome back. So in this practice problem, we have these two stacked blocks on top of each other in this diagram. Basically, we have block A that's sort of resting on top of block B, but I'm pulling block B to the right with a force of 45 Newtons. There's a twist to this one because instead of the whole system moving, what happens is that block A on the top is actually tied to the wall with some kind of a rope. We're supposed to figure out what the tension force is on block A. We are given some masses and some coefficients of friction. So let's just jump right in and stick to our steps. We're going to draw free body diagrams and choose our direction of positive, like we've done a bunch of times before.

Let's start out with block A. The free body diagram is going to be the weight force, so this is going to be m_Ag, and there's also a normal force because it's resting on top of B. In other words, that's the normal force from B on A. There's a tension force that's holding it to the wall, this T over here. If these were the only three forces acting, the block would accelerate in this direction. However, that doesn't make any sense because block A is on top of block B. As you pull block B to the right, why would block A go to the left? That makes no sense. So there must be some kind of a force that's sort of keeping it from moving off to the right, and that force is actually friction. As you pull B, you're going to cause these two surfaces to slip. So block A needs a force to prevent that slipping, which is going to be a force this way, that's the force of friction trying to prevent that motion.

There are no other forces on block A. Let's take a look at block B, which is a little bit more complicated. We start off with the weight force, which is going to be m_Bg. There's also an action-reaction pair from the normal force; B pushes up on A, so that means A has to push down on B. This is going to be the normal force from A on B. That means the normal force that's on B from the ground has to support both of those things, which is going to be N_B. Those are all the vertical forces. For the horizontal forces, we know that we're starting off with an applied force of 45 Newtons. There are actually two forces we have to consider on the left; block A on B is going to exert a friction force in this direction. This is going to be the friction force of B on A. But that means, because of action-reaction, there has to be another force that's acting from A on B. So, these are basically just the action-reaction pair of the friction force between the two surfaces.

Now, there's actually one more friction force to consider. We're told here that the coefficient of kinetic friction between all of the surfaces is 0.2. Thus, there's another friction force, which we'll denote as F_B, and it's really the force on block B from the ground. So there are two friction forces here. Now that we know that these frictions are all kinetic frictions, let's move on to the third step. We're going to write F = m a starting with block A and look at forces in the X-axis. The equation here turns out to be F_BAk - T = m_A × a. However, given that the rope is holding it, the acceleration is actually zero, meaning F_BAk = T. With this, knowing that the friction force is μ _k × N_BA, and that N_BA is equal to m_Ag, the weight force, the tension force can be found by substituting these values. Plugging in the numbers, T equals 0.2 × 5 kg × 9.8 m/s², which calculates to 9.8 Newtons. So, the tension force keeping block A attached to the wall is 9.8 Newtons.

Thank you for watching, and I'll see you in the next tutorial.

A 4kg block sits on top of a 6kg block which is on a frictionless surface. The coefficients of friction between the two blocks are μs=0.5 and μk=0.3. Calculate the maximum force you can pull on the bottom block with so that the objects move together.

• A. 49 N
• B. 19.6 N
• C. 3.27 N
• D. 4.9 N

Here’s what students ask on this topic:

What is the maximum acceleration in stacked block problems?

The maximum acceleration in stacked block problems is determined by the coefficient of static friction (μ_s) between the two blocks and the acceleration due to gravity (g). The formula for maximum acceleration (a_max) is given by:

$a=\mu$_sg

Here, μ_s is the coefficient of static friction, and g is the acceleration due to gravity, approximately 9.8 m/s². This equation helps determine the point at which the blocks will start to slide relative to each other.

How do you draw free body diagrams for stacked block problems?

To draw free body diagrams for stacked block problems, follow these steps:

1. Identify all forces acting on each block. For the top block (A), consider its weight (m_Ag), the normal force from the bottom block (N_BA), and the static friction force (f_s).
2. For the bottom block (B), include its weight (m_Bg), the normal force from the floor (N_B), the normal force from the top block (N_AB), the applied force (F), and the friction force (f_s).
3. Ensure that action-reaction pairs are correctly represented, such as N_BA and N_AB being equal and opposite.

These diagrams help visualize the forces and solve for acceleration and other quantities.

What role does static friction play in stacked block problems?

In stacked block problems, static friction plays a crucial role in keeping the blocks moving together without sliding relative to each other. Static friction acts at the interface between the two blocks and opposes the relative motion. The maximum static friction force (f_s,max) is given by:

$f$_s,max=μ_sN

Here, μ_s is the coefficient of static friction, and N is the normal force between the blocks. When the applied force exceeds this maximum static friction, the blocks start to slide, and kinetic friction takes over.

How do you calculate the normal force between two stacked blocks?

The normal force between two stacked blocks is determined by the weight of the top block. If block A is on top of block B, the normal force (N_BA) exerted by block B on block A is equal to the weight of block A:

$N$_BA=m_Ag

Here, m_A is the mass of the top block, and g is the acceleration due to gravity (approximately 9.8 m/s²). This normal force is crucial for calculating the static friction force between the blocks.

What happens when the applied force exceeds the maximum static friction in stacked block problems?

When the applied force exceeds the maximum static friction in stacked block problems, the blocks begin to slide relative to each other. At this point, the friction between the blocks transitions from static to kinetic. The kinetic friction force (f_k) is generally less than the maximum static friction force and is given by:

$f$_k=μ_kN

Here, μ_k is the coefficient of kinetic friction, and N is the normal force between the blocks. The blocks will then move with different accelerations, and the analysis must account for kinetic friction.