Question

Help! This is problem 10P of Chapter 9 of the 'Fundamentals of Physics 5th edition'

The Great Pyramid of Cheops at El Gizeh, Egypt had height H = 147 m before its topmost stone fell. Its base is a square with edge length L = 230 m. Its volume V is equal to L^2H/3. Assume that the pyramid has a uniform density of 1.8*10^3 kg/m^3

Find:
(a) the original height of its center of mass above the base

(b) the work required to lift the blocks into place from the base level.

Answer 1

It seems that the book only has answers for odd numbered problems, only. Can someone verify my work/my calculation?

Answer 2

Here is how I did it.

Answer 3

First, find the mass of the pyramid by multiplying density and volume, getting 4665780000.

Answer 4

Now, using the integral definition of the center of mass,\[z _{cm}=\frac{ 1 }{ V }\int\limits_{}^{}z dV\]

Answer 5

From the diagram above, I got \[dV=l^2 dz\]

Answer 6

From that diagram, \[\frac{ 147 }{ \frac{ 230 }{ 2 } }=\frac{ 147-h }{ \frac{ l }{ 2 } }\]
Simplify to get \[l=\frac{ 230 }{ 147 }(147-h)\]

Answer 7

Now, substitute that in the integral to get\[\frac{ 1 }{ 4665780000 }\int\limits_{0}^{147}z(\frac{ 230 }{ 147 }(147-h))^2dz\]

Answer 8

Oh wait...
never mind