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(C) Let the left edge of the bottom book (book $1$) be at $x = 0$. The length of each book is $L$.For book $1$,the center of mass is at $x_1 = L/2$.Fo

Vedclass · Accepted Answer

$0.875\ L$

Vedclass · Answer

(C) Let the left edge of the bottom book (book $1$) be at $x = 0$. The length of each book is $L$.For book $1$,the center of mass is at $x_1 = L/2$.For book $2$,the left edge is at $x = L/4$,so the center of mass is at $x_2 = L/4 + L/2 = 3L/4$.For book $3$,the left edge is at $x = L/4 + L/4 = L/2$,so the center of mass is at $x_3 = L/2 + L/2 = L$.For book $4$,the left edge is at $x = L/4 + L/4 + L/4 = 3L/4$,so the center of mass is at $x_4 = 3L/4 + L/2 = 5L/4$.Since all books have equal mass $m$,the $x$-coordinate of the center of mass of the system is:$x_{cm} = \frac{m(x_1 + x_2 + x_3 + x_4)}{4m} = \frac{L/2 + 3L/4 + L + 5L/4}{4}$$x_{cm} = \frac{(2L + 3L + 4L + 5L)/4}{4} = \frac{14L/4}{4} = \frac{14L}{16} = \frac{7L}{8} = 0.875\ L$.

Find the $x$ coordinate of the centre of mass of the books shown in the figure. Every book has mass $m$.

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