Place a rectangular book with a length L in parallel with the end line of the desk and push it a little by little. How long can a book go out of the desk without falling off? (use the statics and equilibrium concepts to explain this)

Logically, it can only go until it has traveled at least #L//2# from the end of the desk, assuming that the end of the book begins at the edge of the desk and the pushing is done such that the book's static equilibrium is maintained (the energy exchange with the desk is reversible).

This can be understood by drawing a free-body diagram. Take the initial position to be:

#" "" "" "" "" "" "" "color(white)(.)overbrace(barul|" "" "|)^("Length "L)#
#bar(" "" "" "" "" "" "" "" "" "" ")# #" "" "" "" "bb((1))#

and an intermediate position to be:

#" "" "" "" "" "" "" "" "color(white)(//)overbrace(barul|" "" "|)^("Length "L)#
#bar(" "" "" "" "" "" "" "" "" "" ")# #" "" "" "" "bb((2))#
#" "" "" "" "" "" "" "" "" "underbrace(color(white)(.//))_("Length "L//2)#

• In #(1)#, we have the normal force #vecF_N# from the table pushing up, and the force due to gravity #vecF_g# pushing down on the entire length of the book.

Those balance out, and (obviously) the book does not fall through the table as a result.

The sum of the vertical forces is:

#sum_i vecF_(y,i) = vecF_g - vecF_N = 0#

where up is negative and down is positive, but the negative is in the minus operation. This is chosen so that the torque later on will be positive for clockwise rotation.

• In #(2)#, #vecF_N# and #vecF_g# balance out on the left half of the book, but only #vecF_g# acts on the right half of the book.

If in #(2)# we allow #vecF_g# to act on the right half to rotate the book clockwise, #vecF_g# on the left half counterbalances it to rotate it counterclockwise because equal lengths of the book are on and off the table.

We say that the clockwise/counterclockwise torque #tau# on both halves are equal, and the book is in static equilibrium with respect to rotation:

#sum_i tau_i = tau_R - tau_L#

#= overbrace(vecL/2 xx vecF_g)^"right half of book" - overbrace(vecL/2 xx vecF_g)^"left half of book"#

#= 0#

Past this intermediate position #(2)# (to the right), the torque #tau# introduced on the right half of the book rotates the book clockwise off the table, as it becomes larger with larger lever arm length #r#, disturbing the static equilibrium in #(2)#.

#sum_i tau_i = tau_R - tau_L#

#= vecr_R xx vecF_g - vecr_L xx vecF_g#

#= {(0,r_R = L//2"),(Ialpha,r_R > L//2):}#

where #I# is the moment of inertia of the book and #alpha# is its angular acceleration, analogous to #vecF_"linear" = mveca#.

(Here I labeled #r_L# to be the length of the book on the table, and #r_R# to be the length of the book off the table.)