Force made by a mass falling on the floor

Click For Summary

Homework Help Overview

The original poster seeks assistance in calculating the force exerted by a mass (100 kg) falling from a height of 0.5 meters onto a bicycle crankarm, with the intention of determining the stress in the crankarm. The poster has proposed a method involving the impulse-momentum principle but is uncertain about the time variable to use in their calculations.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the original poster's approach to calculating force and question the appropriateness of the time interval chosen for the calculation. Suggestions are made regarding the spring constant of the bicycle's tires and the overall dynamics of the bicycle as a spring-damper system. There is also a consideration of the impact of the rider's knee bending on the force experienced.

Discussion Status

The discussion is ongoing, with various participants providing insights and suggestions. Some have raised questions about measuring the spring constant of the bicycle and the implications of the rider's posture on the impact force. There is no clear consensus yet, as multiple interpretations and approaches are being explored.

Contextual Notes

The original poster notes the complexity of the problem due to the simplifications involved, such as neglecting vibrations and the rider's potential bouncing. There is also an acknowledgment of the need for a maximum force estimate for simulation purposes in Solidworks.

vv3
Messages
2
Reaction score
0

Homework Statement


Hello, I need help with a simple task.
I want to calculate the force which appears when a mass falls on the floor without bouncing.
I have modeled a bicycle crankarm, and I want to calculate the stress in it. In order to do that I need a force.
So I want to know the force made by a man (100kg) falling from 0,5 meter directly on the pedals.

Homework Equations


I figured it could be calculated this way:
F=I=m∙a
I=m∙Δv/t; Δv=√(2gh);
But I do not know what time should I use.

The Attempt at a Solution


I was thinking of using 0,1 second for the time, but I am really not sure if that would be correct. I know this is a simplification of a complex model (it probably vibrates and the rider probably bounces a bit).

Do you think this is a good way of calculating the force? Do you have any other suggestions?
Please help me.
Thank you, Vid
 
Physics news on Phys.org
Can you determine the spring constant of the tires?
 
Hey,

thanks for the answer. It really got me thinking. Actually the whole bicycle is a spring and a damper.
I guess it is going to be hard to estimate its spring and damping coefficient. I will try to google it.
Well I am making a simulation of the crankarm in Solidworks and that is why I need a maximum force... Do you have a suggestion how I could estimate that in a different way?

Thank you guys,

Vid
 
You problem is complcated by the fact that the riders knees will bend somewhat. That also lessens the impact.

Is there any way you can measure the spring constant of the bicycle by applying loads and measurements from the pedal to the ground?
 
The man has become an engine, and you make the same error often made by people talking about how much torque or power the engine produces. It produces none at all. The load produces power/torque and the drive train transmits it back to the engine shaft. Look at this problem in the opposite way. How much power can the load take? This for me was at a near stall condition, on a steep grade, lowest gear, and a hundred pound pack on the bike. I decided to approximate this as a full stall with the front wheel against a wall. So now all I had to do was to find the stall torque of my engine, which was me plus the crank. A little testing in the gym showed that I could apply as much as 1500 lb in a sudden burst. I rounded that up to 2000 for safety and multiplied it by crank length. That gave me a max load that the rest of the system had to be designed for. After checking that against all the components, I found that the parts manufactures had all assumed a customer much stronger than me.
 

Similar threads

  • · Replies 21 ·
Replies
21
Views
2K
  • · Replies 23 ·
Replies
23
Views
3K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 25 ·
Replies
25
Views
2K
  • · Replies 20 ·
Replies
20
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 11 ·
Replies
11
Views
5K
Replies
1
Views
1K
Replies
7
Views
11K
  • · Replies 41 ·
2
Replies
41
Views
5K