Sliding Door Force Calculation: Friction And Applied Force

Hey guys! Let's dive into a classic physics problem involving a sliding door, friction, and the force required to get it moving. This type of question often pops up in exams like the ENEM, so understanding the concepts here is super important. We'll break down the problem step-by-step, making sure everything is crystal clear.

Understanding the Problem: The Sliding Door Scenario

Okay, so picture this: we have an 810 N sliding door mounted on a horizontal track. Think of it like a super-sized version of a closet door. The key thing here is that the door isn't moving yet – it's static. This means we need to overcome static friction to get it going. The problem tells us the coefficients of static friction between the track and the door at two points, A and B. At point A, the coefficient is 0.20, and at point B, it's 0.30. These coefficients basically tell us how “sticky” the surfaces are – a higher coefficient means more friction. Our goal is to figure out the minimum horizontal force we need to apply to the door to make it slide. This involves some fundamental physics principles, primarily concerning forces, friction, and equilibrium.

Delving Deeper into Friction

Friction, that force that resists motion, is crucial in this scenario. Since the door is initially at rest, we're dealing with static friction. Static friction is a force that opposes the initiation of motion. Its magnitude can vary, but it has a maximum value, which is given by the product of the static friction coefficient (μs) and the normal force (N). The normal force is the force exerted by a surface that is supporting an object, acting perpendicular to the surface. In our case, the normal forces at points A and B will be equal to the vertical components of the forces supporting the door. It’s important to understand that the static friction force will increase up to its maximum value before the door starts to move. Until we apply a force that exceeds this maximum static friction, the door will remain at rest. The different coefficients at points A and B indicate varying levels of friction, adding another layer to our calculation. Remember, friction always opposes motion, so the direction of the frictional forces will be opposite to the direction we intend to move the door.

Forces in Play: A Closer Look

To solve this problem, we need to identify and analyze all the forces acting on the door. First off, we have the weight of the door, which is given as 810 N. This force acts vertically downwards due to gravity. Next, we have the normal forces acting at points A and B. These forces, which we'll call NA and NB, act upwards, supporting the door against gravity. We also have the static friction forces at A and B, which we'll call fA and fB. These forces act horizontally, opposing the direction we want to move the door. And finally, we have the horizontal force that we need to calculate – the force applied to the door to make it slide. This force will act in the direction we want the door to move. Visualizing these forces with a free-body diagram can be incredibly helpful. A free-body diagram is a simple sketch that shows an object and all the forces acting on it. By drawing such a diagram, we can clearly see the relationships between the forces and apply the laws of equilibrium to solve for the unknowns.

Solving the Problem: A Step-by-Step Approach

Alright, let's get down to business and solve this problem! We'll take a step-by-step approach to make it super clear. First, we need to consider the equilibrium conditions. Since the door is initially at rest, the sum of all forces acting on it in both the horizontal and vertical directions must be zero. Also, the sum of all torques (rotational forces) about any point must be zero. This gives us a set of equations that we can solve to find the unknown forces.

Step 1: Vertical Equilibrium

Let's start with the vertical forces. The door is in equilibrium in the vertical direction, which means the upward forces must balance the downward forces. The upward forces are the normal forces at points A (NA) and B (NB), and the downward force is the weight of the door (810 N). So, we can write the equation:

NA + NB = 810 N

This equation tells us that the sum of the normal forces at A and B must equal the weight of the door. However, we can't solve for NA and NB individually from this equation alone. We need another equation, which we'll get from considering the torques.

Step 2: Torque Equilibrium

Now, let's think about torques. Torque is a twisting force that can cause an object to rotate. For the door to be in equilibrium, the sum of the torques about any point must be zero. We can choose any point to calculate the torques, but a smart choice can simplify the calculations. Let's choose point A as our pivot point. This means the torque due to the normal force at A (NA) is zero, since the force acts at the pivot point. The torques are caused by the weight of the door and the normal force at B (NB). We will assume the center of gravity of the door is at its geometric center. If we denote the horizontal distance between A and B as d and assume that the center of gravity is located at d/2 from both supports, the torque equation will be:

NB * d - 810 N * (d/2) = 0

Notice how the weight of the door creates a clockwise torque, while the normal force at B creates a counter-clockwise torque. Solving this equation for NB, we get:

NB = 405 N

Now that we know NB, we can plug it back into the vertical equilibrium equation (NA + NB = 810 N) to find NA:

NA = 810 N - 405 N = 405 N

Step 3: Maximum Static Friction Forces

With the normal forces NA and NB known, we can calculate the maximum static friction forces at points A and B. Remember, the maximum static friction force is given by the product of the static friction coefficient and the normal force. So, the maximum static friction force at A (fA) is:

fA = μsA * NA = 0.20 * 405 N = 81 N

And the maximum static friction force at B (fB) is:

fB = μsB * NB = 0.30 * 405 N = 121.5 N

These values represent the maximum frictional forces that can be exerted at A and B before the door starts to move.

Step 4: Horizontal Equilibrium and Applied Force

Finally, we can consider the horizontal forces. To get the door moving, the applied horizontal force (F) must be equal to or greater than the sum of the static friction forces at A and B. So, for the minimum force required to initiate motion, we have:

F = fA + fB = 81 N + 121.5 N = 202.5 N

Therefore, the minimum horizontal force that must be applied to the door to make it slide is 202.5 N.

Key Takeaways and Exam Tips

So, there you have it! We've successfully calculated the horizontal force needed to move the sliding door. Let's recap the key concepts and some exam tips:

• Friction is Key: Remember the difference between static and kinetic friction. In this problem, we dealt with static friction because the door was initially at rest.
• Equilibrium is Your Friend: The concept of equilibrium (sum of forces and torques equals zero) is fundamental in solving these types of problems.
• Free-Body Diagrams are Lifesavers: Drawing a free-body diagram helps visualize all the forces acting on the object, making it easier to set up the equations.
• Choose Your Pivot Wisely: When calculating torques, choosing a strategic pivot point can simplify the calculations.
• Units Matter: Always pay attention to the units of measurement and make sure they are consistent throughout the problem.

For exams like the ENEM, practice is crucial. Work through similar problems, focusing on understanding the underlying principles rather than just memorizing formulas. Break down complex problems into smaller, manageable steps, and don't be afraid to draw diagrams and visualize the situation. You got this!

Conclusion

I hope this breakdown has helped you understand how to tackle sliding door friction problems! Remember, physics is all about understanding the forces and how they interact. By mastering these core concepts and practicing regularly, you'll be well-prepared to ace any similar question that comes your way. Keep practicing, guys, and you'll become physics pros in no time!