Weights of 1,g, 2,g, dots, 100,g are suspende | vedclass

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(C) The system is in equilibrium when supported at its center of mass $(COM)$.The masses are $m_i = i$ (in grams) and their positions are $x_i = i$ (i

Vedclass · Accepted Answer

$67$

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(C) The system is in equilibrium when supported at its center of mass $(COM)$.The masses are $m_i = i$ (in grams) and their positions are $x_i = i$ (in cm) for $i = 1, 2, \dots, 100$.The center of mass is given by:$COM = \frac{\sum_{i=1}^{100} m_i x_i}{\sum_{i=1}^{100} m_i} = \frac{\sum_{i=1}^{100} i^2}{\sum_{i=1}^{100} i}$Using the summation formulas:$\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$ and $\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$For $n = 100$:$COM = \frac{\frac{100(101)(201)}{6}}{\frac{100(101)}{2}} = \frac{201}{3} = 67\,cm$.Thus,the system should be supported at the $67\,cm$ mark.

Weights of $1\,g, 2\,g, \dots, 100\,g$ are suspended from the $1\,cm, 2\,cm, \dots, 100\,cm$ marks respectively of a light metre scale. Where should it be supported for the system to be in equilibrium?

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