Friction: Slip or Tip?

Hello! Welcome to our final lesson in the Statics module.

Throughout this module, we have mastered the art of keeping things still. We've calculated the external forces and reactions on everything from simple beams to complex 3D structures, ensuring they are in static equilibrium. In our last lesson, we extended these principles to 3D, using six equilibrium equations to solve for reactions.

Today, we'll address one final, crucial aspect of static equilibrium: friction. Our learning outcome is to apply Coulomb's friction law to determine if objects will slip or tip under applied forces. This topic is a perfect capstone for statics, as it requires us to use both force and moment equilibrium equations together to solve a common and practical type of problem.

1. The Fundamentals of Dry Friction

Friction is the force that resists the relative motion or tendency of such motion between two surfaces in contact. In this course, we are concerned with dry friction, also known as Coulomb friction.

To begin, please read the first part of the following resource. It provides a clear definition of friction and explains the crucial difference between static and kinetic friction.

As you've just read, there are two key takeaways:

1. Static Friction ($F_f$): This is a reactionary force. It is only as large as it needs to be to prevent motion. It has a maximum possible value, beyond which it cannot hold the object in place. This is expressed as an inequality:
   $F_f\le {\mu }_{s}N$
   where ${\mu }_s$ is the coefficient of static friction and $N$ is the normal force between the surfaces.

2. Kinetic Friction ($F_k$): This force acts when the object is already sliding. It has a constant magnitude given by:
   $F_k={\mu }_{k}N$
   where ${\mu }_k$ is the coefficient of kinetic friction. Generally, ${\mu }_k<{\mu }_s$.

The moment just before an object begins to slide is called impending motion. At this point, the static friction force has reached its maximum value: $F_{f,max}={\mu }_{s}N$. This equation is the cornerstone of our slip analysis.

2. The Slip vs. Tip Problem

Now, consider pushing a tall, heavy object like a filing cabinet. If you push it near the bottom, it will likely slide. If you push it near the top, it might tip over instead. How do we determine which will happen?

The governing principle is simple: The event that requires the least amount of applied force is the one that will occur first.

Our task is to calculate two distinct forces:

1. The force required to make the object slip.
2. The force required to make the object tip.

We then compare them to see which is smaller.

3. A Systematic Approach to Analysis

To solve these problems, we analyze two separate "what-if" scenarios.

Scenario A: Assume Slipping is Impending

1. Condition: The object is on the verge of sliding. The friction force is at its maximum: $F_f={\mu }_{s}N$.
2. Analysis: Draw the Free Body Diagram (FBD). Apply the force equilibrium equations ($\sum F_x=0$ and $\sum F_y=0$) to solve for the pushing force, which we'll call $P_{slip}$.

Scenario B: Assume Tipping is Impending

1. Condition: The object is on the verge of rotating about a pivot point (usually a bottom corner). At this exact moment, the distributed normal force from the ground has shifted entirely to act at this single pivot point. The friction force $F_f$ is unknown and is just a reaction force; it is not necessarily at its maximum.
2. Analysis: Draw the FBD, showing the normal force $N$ acting at the pivot corner. Apply the moment equilibrium equation ($\sum M_{pivot}=0$) about that corner. This allows you to solve for the pushing force, which we'll call $P_{tip}$. Choosing the pivot point to sum moments is efficient, as it eliminates the unknown normal and friction forces from the moment equation.

The Verdict

Once you have both $P_{slip}$ and $P_{tip}$, you simply compare them:

• If $P_{slip}<P_{tip}$, the object will slip.
• If $P_{tip}<P_{slip}$, the object will tip.
• The force required to cause motion is $\min (P_{slip},P_{tip})$.

This entire methodology is summarized visually in the diagram below.

Test your understanding!

A box has a width of 1 meter and its center of mass is at its geometric center. The coefficient of static friction is ${\mu }_s$. A horizontal force $P$ is applied at a height h above the ground. If we want to guarantee that the box slips before it tips, should we apply the force $P$ at a high or low position on the box? Why?

Show answer

You should apply the force at a low position. A lower application height h creates a smaller moment that would cause tipping ($M=P\times h$). To cause tipping, this moment must overcome the restoring moment created by the box's weight. A smaller h means a larger force P is needed to cause tipping. By keeping h low, you make it more likely that the force required to slip ($P_{slip}={\mu }_{s}N$) will be reached before the force required to tip.

4. A Worked Example: Two Methods of Solution

There are two common ways to solve these problems. The video below demonstrates a slightly different, but very insightful, method. It starts by assuming slipping will occur and then verifies if that assumption is physically possible.

Watch this video to see a complete worked example.

This "assume and verify" method is a powerful tool. It is particularly useful when a problem asks "What happens if a specific force P is applied?" You can assume the object is in equilibrium and solve for the required friction force $F_f$ and the location of the normal force $N$. Then you check:

1. Is $F_f\le {\mu }_{s}N$? If not, it slips.
2. Is the location of $N$ within the object's base? If not, it tips.

5. Application and Practice

Friction is a fundamental consideration in countless engineering designs. For aerospace, understanding friction is critical for designing braking systems, analyzing the grip of rover wheels on planetary surfaces, and modeling the behavior of mechanisms in space. The analysis of whether a planetary rover will slip or tip over on a steep incline is a direct application of what we've learned today.

To solidify your understanding, I recommend working through the problems provided in the following resource. They come with full PDF solutions for you to check your work.

Conclusion

This lesson concludes our module on Statics. We've brought together the principles of force and moment equilibrium to analyze the practical problem of friction and stability.

Key Takeaways:

• Static friction is a variable reaction force with a maximum value of $F_{f,max}={\mu }_{s}N$, which occurs at impending motion.
• To determine if an object will slip or tip, you must find out which event occurs at a lower applied force.
• To find $P_{slip}$: Assume $F_f={\mu }_{s}N$ and solve the force equilibrium equations.
• To find $P_{tip}$: Assume the normal force $N$ acts at the pivot corner and solve the moment equilibrium equation about that corner.
• An alternative method is to assume equilibrium, solve for the required reaction forces, and check if they are physically possible (i.e., if $F_f\le {\mu }_{s}N$ and if $N$ is within the base).

Next Lesson Preview:

We have spent this entire module analyzing the external forces on rigid bodies. But in the real world, bodies are not perfectly rigid; they deform under load. The external forces we've been calculating create internal forces and deformations within a material.

In our next module, Mechanics of Materials, we will shift our focus from the external to the internal. Our very first lesson will introduce the foundational concepts of normal stress and strain, which describe how a material internally responds to being pulled or pushed. This will be our first step in understanding the strength and behavior of engineering materials.