p13_034.pdf

34. We locate the origin of the x axis at the edge of the table and choose rightwards positive. The criterion

(in part (a)) is that the center of mass of the block above another must be no further than the edge of

the one below; the criterion in part (b) is more subtle and is discussed below. Since the edge of the table

corresponds to x =0 then the total center of mass of the blocks must be zero.

(a) We treat this as three items: one on the upper left (composed of two bricks, one directly on top of

the other) of mass 2 m whose center is above the left edge of the bottom brick; a single brick at the

upper right of mass m which necessarily has its center over the right edge of the bottom brick (so

a 1 = L/ 2 trivially); and, the bottom brick of mass m . The total center of mass is

(2 m )( a 2 −

L )+ ma 2 + m ( a 2 −

L/ 2)

4 m

L/ 2. The middle layer consists of two bricks, and

we note that it is possible for each of their centers of mass to be beyond the respective edges of the

bottom one! This is due to the fact that the top brick is exerting downward forces (each equal to

half its weight) on the middle blocks – and in the extreme case, this may be thought of as a pair of

concentrated forces exerted at the innermost edges of the middle bricks. Also, in the extreme

case, the support force (upward) exerted

on a middle block (by the bottom one)

may be thought of as a concentrated force

located at the edge of the bottom block

(which is the point about which we com-

putetorques, inthefollowing). If(asindi-

catedinoursketch,where F top has mag-

nitude mg/ 2) we consider equilibrium of

torques on the rightmost brick, we obtain

F top

F bottom

mg b 1 −

2 L = mg

2 ( L

−

b 1 )

which leads to b 1 =2 L/ 3. Once we conclude from symmetry that b 2 = L/ 2 then we also arrive at

h = b 2 + b 1 =7 L/ 6.

which leads to a 2 =5 L/ 8. Consequently, h = a 2 + a 1 =9 L/ 8.

(b) We have four bricks (each of mass m ) where the center of mass of the top and the center of mass of

the bottom one have the same value x cm = b 2 −