What Was the Acceleration of the 1852 Crampton Coal-Fired Train Engine?

Click For Summary

Homework Help Overview

The discussion revolves around calculating the acceleration of the 1852 Crampton coal-fired train engine, given its mass, force capability, and frictional forces. The subject area includes classical mechanics, specifically dynamics involving forces and acceleration.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the acceleration using the net force and mass, but expresses uncertainty about the reasonableness of the result. Some participants question the expectations of acceleration for a train from that era.

Discussion Status

There is a mix of validation and questioning in the discussion. One participant confirms the original poster's procedure while another points out a potential misunderstanding in the notation of the acceleration value. The conversation reflects an exploration of historical context and expectations regarding train performance.

Contextual Notes

Participants discuss the historical capabilities of trains in the 1850s, suggesting that expectations for acceleration may need to be adjusted based on the technology of the time.

thua
Messages
9
Reaction score
0
When the Crampton coal-fired train engine was built in 1852, its mass was 48.3 t (1.0 t = 1.0 x 10^3 kg) and its force capability was rated at 22.4 kN. Assuming it was pulling train cars whose total mass doubled its own mass and the total friction on the engine and cars was 10.1 kN, what was the magnitude of the acceleration of the train?
What I have: (not sure if it's right though)
m = 48.3 t = (48.3 x 10^3 kg) x 3 = 144.9 x 10^3 kg
net force = 22.4kN - 10.1 kN = 12.3 kN = 12300 N
acceleration = ?
acceleration = net force / mass
= 12300 N / 144.9 x 10^3 kg
acceleration = 0.0849 m/s^2
acceleration = 8.49 x 10^3 kg
Is this right? because 0.0849 m/s^2 doesn't seem very reasonable to me. If it's wrong, could someone please help?
 
Physics news on Phys.org
Your procedure looks perfectly fine to me.
 
Were you expecting a train in 1852 to take off really fast?

Notice that in only 100 seconds (1 min 40 seconds), the train will be traveling over 8 m/s. Now that's a pretty good pace!

By the way:
"acceleration = 0.0849 m/s^2" is correct.

"acceleration = 8.49 x 10^3 kg" is nonsense. I presume you meant
"8.49 x 10^(-2) m/s^2".
 
Yah, trains back then could go... 40mph at the most probably back then.
 

Similar threads

Replies
17
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 21 ·
Replies
21
Views
3K
Replies
15
Views
2K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
5K
Replies
44
Views
4K
  • · Replies 33 ·
2
Replies
33
Views
3K
  • · Replies 8 ·
Replies
8
Views
8K