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318 CHAPTER 10 Systems of Particles

(a) This fraction of the total (b)
mass, dm/M, is the same a
as this fraction of the total y = x
b
1
area, y dx/( ab). 1 b
2 3
y y
dm
a
dx y CM
1 a
3
x x
b
x CM
FIGURE 10.14 (a) A right triangle, with mass element dm of height y and width dx. (b) The center of mass is
one-third of the distance from the right angle along sides a and b.

We integrate this in Eq. (10.25) for x and sum the contributions from x 0
CM
to x b:

1 1 b 2x
x CM x dm xM 2 dx
M M 0 b
2 b 2 1 3 b
2 x dx 2 x `
2
b 0 b 3 0
2 1 3 2
(b 0) b
2
b 3 3
So the center of mass is two-thirds of the distance toward the right angle.
1
Performing a similar calculation for y yields y a .Thus each of x and
CM CM 3 CM
y is a distance away from the right angle equal to one-third of the length of the
CM
corresponding side (see Fig. 10.14b).

The Great Pyramid at Giza (see Fig. 10.15) has a height of
EXAMPLE 7
147 m and a square base. Assuming that the entire volume is
completely filled with stone of uniform density, find its center of mass.
SOLUTION: Because of symmetry, the center of mass must be on the vertical line
FIGURE 10.15 The Great Pyramid. through the apex. For convenience, we place the y axis along this line, and we
arrange this axis downward, with origin at the apex. We must
then find where the center of mass is on this y axis. Figure
10.16a shows a cross section through the pyramid, looking par-
allel to two sides.The half-angle at the apex is . By examina-
tion of the colored triangle, we see that at a height y (measured
from the apex) the half-width is x y tan and the full width
is 2x 2y tan . A horizontal slice through the pyramid at this
height is a square measuring 2x 2x (see Fig. 10.16b). The
volume of a horizontal slab of thickness dy at this height y is
2
2
therefore dV (2x) dy (2y tan ) dy. If we represent the uni-
form density of the stone by (the Greek letter rho), the pro-
portionality between mass and volume can be written
dm dV