Systems of Objects with Friction - Online Tutor, Practice Problems & Exam Prep

In systems with connected objects experiencing friction, each object's friction must be considered. When analyzing such systems, remember that connected objects share the same velocity and acceleration. Begin by drawing free body diagrams for each object, identifying forces like tension, normal force, and kinetic friction. Use Newton's second law, F = m × a, to set up equations for each object, allowing you to solve for unknowns like acceleration and tension through substitution or addition of equations.

Connected Objects With Friction

So now that we understand friction a little bit better, in some problems, you're going to 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 have some cables or something like that, and there's also going to be friction. The basic idea here is that whenever this happens and friction is not negligible, you're going to have to consider the friction that acts on each individual object. Really, the steps to solve these problems are going to be the same. Just remember that connected objects always have the same velocity and acceleration. 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 kilogram block I'll call a, and this one I'll call b, and we have the coefficients of friction. First, we want to draw free body diagrams for both objects and then find the acceleration of the system. So that means that we're not going to take the shortcut here because we want to draw the free body diagrams for both objects and also have to consider the friction acting on each. Okay?

So, we've got the free body diagram for the first one here. This is going to be mag and then we've got our tension force that acts this way and then we've got our normal force. Now the last thing is we also have some friction, and whether this block is going to move to the right or wants to move to the right, 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 now let's 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, the 10 kilogram block is the one that's being pulled with 90 Newtons. So that's our applied force here. And now because of an action-reaction pair, we also have some tension that is going backwards, and then we have some normal force, and finally, some friction. So just like before, we have a friction force that points to the left.

So those are our free body diagrams. We've actually already covered the next step as well with determining the type of friction we're working with. Remember, sometimes the problem text will just literally tell you what kind of friction you're working with here. Alright. So basically, we're just going to go ahead and start our F = m a, and we're going to start with the simplest or the fewest number of forces. So that's going to be block a over here. So, if I want to figure out the acceleration, then I'm going to look at block a and I have to write out my F = m a. So, I am looking for the acceleration here.

So, what are my forces? Well, I'm going to have to pick a direction of positive first, and because the blocks are being pulled to the right, it's probably a safe bet that the system is going to accelerate to the right. So that's going to be my direction of positive. So when I expand my forces, I have tension minus f_k equals ma. I can replace this f_k with its expression, μ_k N, which is equal to ma. Remember, the normal force is actually going to have to go back into the diagram and figure that out. So here, our normal force, if these blocks are only sliding horizontally, those have to cancel. So basically, our two vertical forces here have to cancel out so normal is equal to mg. So what we can do here is tension minus μ_k mg equals ma. 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 5×9.8 equals 5a. So if you go ahead and work this out, you're going to get tension minus 14.7 equals 5a.

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, just go and start the other object and see if you can find an equation there. So here we got our forces equal to mass b times the acceleration. So this actually was mass a. So we got to write all our forces here; our forces are going to be the F which is the 90 minus your tension force, which is backwards. 가기. So now we have this, F - tension - μ_k × normal equals mb a. So F - tension - μ_k. And now we're going to 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 going to be equal to mb g times g. So this is going to be mb g. This one is ma g. Alright. So we have mb g is equal to mb a. Okay. And so now, what 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 10a. So you get 90 minus tension minus, and then this just turns into a number. This is 29.4. So this is equal to 10a. And so what happens is these two are numbers. 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 10a. So now we can't go any further. We have acceleration and tension. 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 my two. Now remember with these kinds of problems, we're just going to solve them using equation addition or substitution. And we're going to pick the addition here, so let's go ahead and do that. So we're going to line them up. So this is equation number 1. We've got tension minus 14.7 equals 5a, and I'm going to scoot this over just a little bit. This is tension. And then finally, we have our 60.6 minus tension equals 10a. So here what happens is when we add these equations straight down, we add these equations like this, you're going to get tension that cancels with tension, and basically 60.6 minus 14.7. So you have, what this becomes is 45.9 equals 15a. So now we just solve for the acceleration, and it ends up being 3.06, and that's the answer. Let me know if you have any questions. We're really just using the same steps that we've been working with so far. Let's move on to the next.

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?

• A. 3.33 kg
• B. 5 kg
• C. 32.7 kg

Here’s what students ask on this topic:

How do you draw free body diagrams for systems of objects with friction?

To draw free body diagrams for systems of objects with friction, follow these steps:

1. Identify each object in the system and draw them separately.

2. For each object, draw vectors representing all forces acting on it. These include:

• Gravitational force (${\mathrm{mg}}_i$), where $i$ is the object index.
• Normal force ($N_i$).
• Tension force ($T$), if objects are connected by a string or cable.
• Frictional force ($f_k$), which opposes the direction of motion.

3. Label each force clearly and indicate the direction of motion or applied forces.

4. Ensure that the forces are balanced according to Newton's second law, $F=ma$.

What is the role of kinetic friction in systems of connected objects?

Kinetic friction plays a crucial role in systems of connected objects by opposing the motion of each object. It is calculated using the formula:

$f_k={\mu }_k{N}_i$

where ${\mu }_k$ is the coefficient of kinetic friction and $N_i$ is the normal force on the object. Kinetic friction affects the net force acting on each object, influencing the system's overall acceleration. When solving problems, you must account for kinetic friction for each object to accurately determine the system's behavior.

How do you solve for acceleration in systems of objects with friction?

To solve for acceleration in systems of objects with friction, follow these steps:

1. Draw free body diagrams for each object, identifying all forces.

2. Write Newton's second law for each object: $F=ma$.

3. Include frictional forces using $f_k={\mu }_k{N}_i$.

4. Set up equations for each object, considering tension and other forces.

5. Solve the system of equations using substitution or addition to find the unknowns, such as acceleration and tension.

6. Ensure the acceleration is consistent across all objects, as they move together.

What are the common mistakes to avoid when analyzing systems of objects with friction?

Common mistakes to avoid include:

• Forgetting to include all forces in the free body diagrams, such as tension or friction.
• Incorrectly calculating the normal force, which affects the frictional force.
• Assuming static friction when the problem specifies kinetic friction.
• Not considering that connected objects share the same acceleration.
• Failing to correctly apply Newton's second law to each object.
• Ignoring the direction of forces, leading to sign errors in equations.

Carefully follow the steps and double-check your work to avoid these errors.

How do you determine the type of friction (static or kinetic) in a problem involving systems of objects?

To determine the type of friction in a problem involving systems of objects, consider the following:

1. **Problem Statement**: Check if the problem explicitly states whether the objects are moving or stationary.

2. **Motion Status**: If the objects are moving, kinetic friction is involved. If they are stationary but on the verge of moving, static friction applies.

3. **Force Analysis**: Compare the applied force to the maximum static friction force ($f_s={\mu }_s{N}_i$). If the applied force exceeds this value, the objects will move, and kinetic friction will take over.

4. **Given Coefficients**: Use the given coefficients of friction to identify the type, as problems often provide ${\mu }_s$ for static and ${\mu }_k$ for kinetic friction.