Infinite numbers of blocks are placed on a ta | vedclass

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(C) Let the mass of the $1^{st}$ block be $m_1 = m$ and its length be $x_1 = x$. The mass of the $n^{th}$ block is $m_n = m/4^{n-1}$ and its length is

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$3x/7$

Vedclass · Answer

(C) Let the mass of the $1^{st}$ block be $m_1 = m$ and its length be $x_1 = x$. The mass of the $n^{th}$ block is $m_n = m/4^{n-1}$ and its length is $x_n = x/2^{n-1}$.To find the center of mass of the system relative to the left edge (axis $AA'$),we calculate the weighted average of the center of mass of each block.The center of mass of the $n^{th}$ block from the axis $AA'$ is $x_{cm,n} = \frac{1}{2} \cdot \frac{x}{2^{n-1}} = \frac{x}{2^n}$.The total mass of the system is $M = \sum_{n=1}^{\infty} \frac{m}{4^{n-1}} = m \left( 1 + \frac{1}{4} + \frac{1}{16} + \dots \right) = m \left( \frac{1}{1 - 1/4} \right) = \frac{4m}{3}$.The position of the center of mass of the entire system is $X_{CM} = \frac{1}{M} \sum_{n=1}^{\infty} m_n x_{cm,n} = \frac{3}{4m} \sum_{n=1}^{\infty} \left( \frac{m}{4^{n-1}} \cdot \frac{x}{2^n} \right) = \frac{3x}{4} \sum_{n=1}^{\infty} \frac{1}{2^n \cdot 4^{n-1}} = \frac{3x}{4} \sum_{n=1}^{\infty} \frac{1}{2^n \cdot 2^{2n-2}} = \frac{3x}{4} \sum_{n=1}^{\infty} \frac{1}{2^{3n-2}} = \frac{3x}{4} \left( \frac{1/2}{1 - 1/8} \right) = \frac{3x}{4} \left( \frac{1/2}{7/8} \right) = \frac{3x}{4} \cdot \frac{4}{7} = \frac{3x}{7}$.For the system to be on the verge of toppling,the center of mass must lie exactly at the edge of the table,so $L = X_{CM} = 3x/7$.

Infinite numbers of blocks are placed on a table edge as shown in the diagram. What is the minimum value of $L$ so that the blocks are just going to topple over the table?

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