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concentrated force located at the edge of the bottom block (which is the point about
which we compute torques, in the following).
If (as indicated in our sketch, where
G
F
top
has magnitude
mg
/2)
we consider equilibrium of torques on the rightmost brick, we
obtain
mg b
L
mg
Lb
11
1
22
−
F
H
G
I
K
J
=−
()
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.
75. 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
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Center Of Mass, Mass

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