Dimensional Analysis: Solving Confusing Steps

Click For Summary

Homework Help Overview

The discussion revolves around a dimensional analysis problem related to the force exerted by ice on buildings, specifically examining the relationship between force, surface area, and material properties. The original poster attempts to analyze the dimensions involved in the problem, particularly focusing on the independence of the force per unit area from the surface area.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the dimensional relationships between force, area, and material properties, questioning the independence of the derived quantities. There is a focus on the number of independent dimensions and the implications for dimensional analysis.

Discussion Status

The discussion is ongoing, with participants providing differing views on the independence of dimensions and the correct interpretation of the parameters involved. Some participants offer clarifications regarding the dimensional analysis, while others express confusion about the original poster's calculations.

Contextual Notes

There is mention of a potential error in the original poster's understanding of the dimensions involved, particularly regarding the independence of mass and time in the context of force. The thread has been locked due to the original poster's intent to revise their question.

promise899
Messages
2
Reaction score
1
Homework Statement
I have a question which has two part described below:

Let the force P to be described by the surface area A between the ice and the buildings. The paramater for ice is comp. str. Y, which has the dimesion of stress. i) Proove the max. force per unit area P/A is independent of A. ii) Ice is a brittle material. That suggest an alternative model, that relevant materia parameter might not be comp stress Y but might instead be fracture toughness K(FL^-3/2). Show that in that case P/A is not independent of A and find how it depends on A?
Relevant Equations
i )pi(1) = P^X1 * A^Y1 * Y^Z1
I tried to use dimensional analysis, there is variable for part i) m= P,A and Y also parameter is used in analysis is n=3(M,L,T). So m-n=0 number of dimensionless analysis group. I am confused at this step however I did this calculation to reach solution:

i) P=MLT-2 (Ice Force) m-n=4-3=1(number of dimensionless group)
γ =ML-2T-2 (Specific Weight) A=L2 (Contact Surface Area)

pi(1) = P^X1 * A^Y1 * Y^Z1

pi(1) =( P/A*Y)

How it shows P/A is not dependent of A? I could skip some points in this question also I stuck at part ii. Thanks for help!
 
  • Like
Likes   Reactions: Delta2
Physics news on Phys.org
promise899 said:
Homework Statement:: I have a question which has two part described below:

Let the force P to be described by the surface area A between the ice and the buildings. The parameter for ice is comp. str. Y, which has the dimesion of stress. i) Proove the max. force per unit area P/A is independent of A. ii) Ice is a brittle material. That suggest an alternative model, that relevant materia parameter might not be comp stress Y but might instead be fracture toughness K(FL^-3/2). Show that in that case P/A is not independent of A and find how it depends on A?
Relevant Equations:: i )pi(1) = P^X1 * A^Y1 * Y^Z1

is n=3(M,L,T). So m-n=0 number of dimensionless analysis group.
This is not correct as T and M are not independent dimensions here (ie, they always appear in the same combination, which is M/T^2). Hence, in this case, n=2.
 
Orodruin said:
This is not correct as T and M are not independent dimensions here (ie, they always appear in the same combination, which is M/T^2). Hence, in this case, n=2.
Force equal to P=ML/T^2 so we use M,L,T . In this way I think n=3
 
promise899 said:
Force equal to P=ML/T^2 so we use M,L,T . In this way I think n=3
Again, no - you think wrong. M only appears along with 1/T^2. If you know M appears to the power of k in a dimension you know that T will appear to the power -2k. That it also appears with an L in force is irrelevant. You only have two independent dimensions. This is also clear from the simple fact that you can construct a dimensionless quantity.
 
OP made an error in the question and will do a new post. Thread now locked.

Thanks
Bill
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 38 ·
2
Replies
38
Views
5K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
710
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
12K