Question

consider the two configurations of books on the edge of a table shown below. which of the following is clearly true? a. case 1 will tip b. case 2 will tip c. neither will tip d. both will tip

Explanation:

Step1: Analyze Case 1

For Case 1, assume the length of each book is \(L = 1\) unit. The center - of - mass of the lower book (relative to the edge of the table) is at \(x_1=\frac{1}{4}\) (to the right of the table edge). The center - of - mass of the upper book relative to the edge of the lower book is \(\frac{1}{2}\). The combined center - of - mass of the two - book system \(x_{cm1}\) (using the formula \(x_{cm}=\frac{m_1x_1 + m_2x_2}{m_1 + m_2}\), assuming \(m_1=m_2 = m\)):

$$x_{cm1}=\frac{m\times\frac{1}{4}+m\times(\frac{1}{4}+\frac{1}{2})}{2m}=\frac{m\times\frac{1}{4}+m\times\frac{3}{4}}{2m}=\frac{\frac{1 + 3}{4}m}{2m}=\frac{1}{2}$$

The system will tip if the center - of - mass is to the right of the table edge. Here, the combined center - of - mass is at the edge of the table.

Step2: Analyze Case 2

For Case 2, the center - of - mass of the lower book (relative to the table edge) is \(x_1=\frac{1}{2}\) (to the right of the table edge). The center - of - mass of the upper book relative to the edge of the lower book is \(\frac{1}{4}\). The combined center - of - mass of the two - book system \(x_{cm2}\) (using \(x_{cm}=\frac{m_1x_1 + m_2x_2}{m_1 + m_2}\), \(m_1 = m_2=m\)):

$$x_{cm2}=\frac{m\times\frac{1}{2}+m\times(\frac{1}{2}+\frac{1}{4})}{2m}=\frac{m\times\frac{1}{2}+m\times\frac{3}{4}}{2m}=\frac{\frac{2+3}{4}m}{2m}=\frac{5}{8}>\frac{1}{2}$$

Answer:

B. Case 2 will tip