Question

The great pyramid of Giza consits of approximately 2 million stones, each weighing 1.5 tons ( 3000 lbs). The pyramid is 450 ft high and has a square base measuring 750 ft. Find the work rquired to lift the stones into place from ground level.

Would I use the integral of w=f*d; w=3000*x from 0 to 450? Do I need to do something with the square base and its volume?

Answer is 675,000,000,000 ft-lb

Answer 1

Here we need to calculate the work done in each horizontal square layer of the pyramid. For this, it is needed to integrate in the height of the pyramid each square layer.
First of all, we will need the mass density of the pyramid, which is easily found since the volume V of the pyramid is base^2 x height /3. In this case, V = 750 * 750 * 450/3 = 84*10^6 ft^3. The mass of the pyramid is 2*10^6* 3*10^3 lbs = 6*10^9 lbs. Therefore, the mass density d = mass/V = 6*10^9/(84*10^6) lbs/ft^3 = 71.42 lbs/ft^3.
With this, each layer of a base b and differential height dh, located at a height h, will need a work
dW = potential energy = mass*height*g = density*volume*height*g = d*b^2*dh*h*g
In order to find the base at a height h, we do a cross multiplication using that we know the height and the base of the total pyramid. It is satisfied that
b/h = 750 /450 = 1.666,    ===>   b = 1.666*h.
Therefore, the differential work becomes
dW = 1.666^2*d*h^3*g*dh = 1.666^2* 71.42 lbs/ft^3*32.174 ft/s^2*h^3*dh,
dW =6377.85*h^3*dh.
We integrate this over all the height, obtaining

I can see that the result obtained through this method (it is the exact way of obtaining the correct answer) is around the same order of magnitude of your answer (are you sure that you are not missing a zero decimal?)