Energy of Destruction of Plunging Excavator down Giant Shaft

The blogger’s note: 

On the evening of October 14, 2020 at a Hong Kong Highway Department construction site of the new 4.7km Central Kowloon Route at Ho Man Tin, an excavator accidentally broke loose from a tower crane as it was being lifted and plunged down the giant shaft.

The incident captured in a viral video showed that the excavator was being lifted by the crane, then slowly turned upside down before breaking free of its rigging and tumbling down the shaft. Several workers near the shaft could be seen running away as it happened. Fortunately, no one was injured.

An excavator lies at the bottom of a deep shaft after breaking free from the crane that was lowering it at a Hong Kong highway construction site (taken from SCMP)

A look down at the large ventilation shaft that an excavator fell to the bottom of on Wednesday Oct. 14, 2020 evening in Hong Kong (taken from SCMP)

After I received a copy of that video from a friend and posted into Dr. JB Lim’s chat group, he wrote the following and requested me to post into my blog (with minor editing by me)..  

 

Date: Oct. 18, 2020 @ 9.43AM

 

Thank you Sifu Ir. Lau for the video on the crane accident in Hong Kong.

I am now thinking on my own the energy of destruction generated by the falling excavator.

Let us start by using Newton’s laws of motion:

Let,

1.   v = u + at
2.  s = ut + 0.5 at²

3  s = 0.5 (u + v) t
4.  v² = u²+ 2as
5.  s = vt - 0.5 at²

where,

v = final velocity
u = initial velocity
a = acceleration due to gravity (9.81 m/s²)
t = time taken

s = displacement (assumed to be 100 metres down the shaft)

Let us do some simple calculations using Newton’s equations of motion.

We can clearly see there are two very important data we do not know.

First, we do not have any idea how long the excavator took to plunge down to the bottom of the shaft.

Second, we have no idea the height of the shaft through which the excavator fell.

My brother-in-law who is a retired Senior Civil Engineer in Singapore estimated the excavator to weigh 10 tons (10,000 kg).

Let us also assume the height of the shaft was about 100 metres.

Of course, the above data are only assumptions for the sake of this calculation.

Out of the 5 equations above, we shall use equation 4 to determine the final velocity.

initial velocity, u = 0

final velocity, v = √(0 + 2 x 9.81 x 100) = 44.29 m/s

Now that we know its final velocity as 44.29 m/s, we can calculate the kinetic energy (1/2 mv²) generated at the bottom of the shaft

= 0.5 x 10,000 kg x 44.29 x 44.29 = 9,808, 020 Joules

Wow!  That's a lot of energy!

Suppose now we use nuclear energy to lift the excavator up again to the top. How much mass of uranium using fission would be needed?

E = mc²
m =  E / c²

= 9, 808,020 / square of speed of light at 299,792,458 m/s
= 1.09 x 10^-10 gm
= 0. 0 000 000 002 gm of matter

Wow! Such a tiny, tiny amount of uranium is needed through nuclear fission to lift up the crane again.

That's the beauty of physics and mathematics whereby we can imagine many thoughts.

Later we shall calculate the hydroelectric power needed, and after that, how much water is needed to be lifted up by the Sun as clouds expressed in volume, and to what height to get the same amount of potential energy needed to restore the excavator back to its original position.

But let me rest first as I have only a small hand phone to do these calculations while lying in a very painful and uncomfortable position on my hospital bed.

By the way, the potential energy of the excavator was mgh = 10 000 kg × 9.81 m/s² x 100 m = 9,810, 000 Joules which is the same as its kinetic energy when the excavator dropped down, where m = mass, g = acceleration due to gravity, and h = height.

This confirms the calculation was correct.

The force of impact (F = ma) was 10,000 kg x 9.81 m/s² = 98,100 newton

Sorry. I have nothing else to do here lying on the bed 24 hrs a day but just thinking and wondering after watching a video on an accident involving excavator falling down a shaft in Hong Kong sent to me by my Sifu Ir TO Lau friend.

I wrote this just for the fun of wondering. I am not a physicist or an engineer like my Sifu Ir TO Lau.

I hope Ir. TO Lau can check the calculations for me as he is my Sifu since it was very difficult for me to calculate and type using a small handphone.

Thank you, Sifu.

Jb lim