3 identical books, each with length L, are stacked over the edge of a table. The top book overhangs the middle book by 1/2L and the middle book overhangs the bottom book by 1/4L. How much can the bottom book be allowed to overhang the edge of the table without the books the (other) books falling?
The Attempt at a Solution
I approached this problem assuming that in order for the three books to not fall, their center of gravity must be at least over the table so that there is normal force to counterbalance the books' gravitational force. And so the net torque of the book mass must 0. I placed the pivot point for the torque at the left end of the bottom book
So far, I have \tau=-[(mL/2)+m(L-xL)+m(L+L/4)]/3m * 3mg
x= the fraction of the bottom book's length allowed to overhang the table.
but I have no idea what to do with the normal force and what the normal force value is.
I read from another thread in which people said that as long as the center of mass of the 2nd book is above the edge of the bottom book and the same for the bottom book over the table, the problem is solved. (https://www.physicsforums.com/archive/index.php/t-73791.html)
I tried that on mastering physics with answer being 1/4L and tried again with 1/2L. neither worked...can anyone help me?