Problem 25 - College Physics

Question

(a) How many uniform, identical textbooks of width \(25.0 \mathrm{~cm}\) can be stacked on top of each other on a level surface without the stack falling over if each successive book is displaced \(3.00 \mathrm{~cm}\) in width relative to the book below it? (b) If the books are \(5.00 \mathrm{~cm}\) thick, what will be the height of the center of mass of the stack above the level surface?

Step-by-Step Solution

Answer

3 books; center of mass height is 7.5 cm above the surface.

Step 1: Understanding the Problem

We have a series of textbooks each with width \(25.0 \text{ cm}\) and they are stacked such that each book is displaced by \(3.0 \text{ cm}\) from the one below it. Our goal is to determine how many books can be stacked without the pile toppling.

Step 2: Concept of Center of Mass for Stability

For the stack to be stable, the center of mass of the stack must remain above the base of the stack. If a subsequent book moves the center of mass beyond the base, the stack will topple over.

Step 3: Calculating the Center of Mass

Stacking \(n\) books with each displaced by \(3.0 \text{ cm}\) means the center of mass is at \(\frac{n(n-1)}{2} \times 3.0 \text{ cm}\) from the first book's center line. This must be less than or equal to half the width of a book (\(12.5 \text{ cm}\)) for stability.

Step 4: Setting Up the Equation for Stability

The inequality for stability is \(\frac{n(n-1)}{2} \times 3.0 \leq 12.5\). Solving this inequality will give us the maximum number of books \(n\).

Step 5: Solving for Maximum Books Without Toppling

Simplify and solve for \(n\):\[\frac{n(n-1)}{2} \leq \frac{12.5}{3} \\frac{n(n-1)}{2} \leq 4.1667 (n-1) \leq 8.3334\].Trying \(n=4\): \(4(4-1)=12\), therefore \(n=3\) is the maximum.

Step 6: Calculating Height of Center of Mass

If \(n=3\) books are stacked and each book is \(5.0 \text{ cm}\) thick, the height of the books is \(3 \times 5.0 = 15.0 \text{ cm}\). The center of mass of the stack is the height of the center of mass of a uniform stack: \(\frac{15.0}{2} = 7.5 \text{ cm}\).

Key Concepts

Center of Mass

The center of mass is a crucial concept in physics, especially when discussing stability. It is the point where we can think of the entire mass of an object being concentrated. This helps when predicting movements and balancing objects.
In the exercise, the aim is to keep the center of mass of the textbook stack directly above the base. If the center of mass shifts from this spot, the stack will become unbalanced and fall over.
To find the center of mass of a stack of books, we use the formula: \[\text{Center of mass position} = \frac{n(n-1)}{2} \times 3.0 \text{ cm}\] where each book is displaced by \(3.0 \text{ cm}\). The correct positioning of the center of mass ensures the books don't topple.
Thus, maintaining a stable center of mass is fundamental in stacking objects without them falling.

Stability in Physics

Stability in physics often refers to an object's ability to remain balanced without falling. In the textbook problem, the key to stability lies in the position of the center of mass.
For the stack to remain stable, its center of mass must stay within the confines of the book directly beneath it. If it shifts too much, gravity will take over, and the stack will fall.
To ensure stability, physicists calculate whether the center of mass will overhang the base. For our textbooks, the calculation involves stacking books such that their center remains centralized:

• The maximum number of books \(n\) can be calculated by ensuring that: \[ \frac{n(n-1)}{2} \times 3.0 \leq 12.5 \text{ cm} \]

By solving this, we ensure the center of mass does not exceed the halfway point of the bottom book's width, maintaining stability.

Physics Calculations

Physics often requires precise calculations to solve real-world problems. In our exercise, calculations help determine the number of books one can stack without them falling over.
Calculating begins with understanding equations such as:

• For the stability: \[ \frac{n(n-1)}{2} \leq 4.1667 \]
• For center of mass height: \(\text{height} = \frac{\text{total height}}{2} \)

Solving these equations provides the solution.
For example, solving \(\frac{n(n-1)}{2} \leq 4.1667\) tells us the maximum number \(n\) of books that can be safely stacked. Using trial and error, with \(n=4\) proving unstable and \(n=3\) being stable.

Problem Solving in Physics

Problem solving in physics is about breaking down complex situations into simpler, manageable parts.
The exercise about stacking textbooks guides you through this process. First, identify the main challenge: ensuring the stack doesn't topple. Then, use fundamental concepts like center of mass and stability to address this.
Breaking problems into steps, as shown:

• Understand the conditions for stability and center of mass.
• Calculate mathematically whether these conditions are met using known formulas.
• Verify results with practical assumptions like trial-and-error to confirm stability.

Approaching physics problems this way helps in structuring the solutions effectively and ensures you thoroughly understand each concept involved.