Calling the mass of a single brick \(m\), then the mass of the \(N-1\) bricks is \((N-1)m\), and the center of mass of the bottom brick is \(\frac{1}{2}L\) farther from the origin than the center of g...Calling the mass of a single brick \(m\), then the mass of the \(N-1\) bricks is \((N-1)m\), and the center of mass of the bottom brick is \(\frac{1}{2}L\) farther from the origin than the center of gravity of the \(N-1\) bricks. [For a reminder of how to find the center of mass of two extended objects with their own centers of mass, go here.]
