Dimensional Analysis: Solving Confusing Steps

In summary, the conversation is discussing a question with two parts. The first part involves using dimensional analysis to prove that the maximum force per unit area, P/A, is independent of the surface area A. The second part considers an alternative model for the material parameter of ice, fracture toughness K, and how it affects the dependence of P/A on A. There is some confusion about the dimensions being used and how to approach this question.
  • #1
promise899
2
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!
 
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  • #2
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 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

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.
 
  • #3
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
 
  • #4
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.
 
  • #5
OP made an error in the question and will do a new post. Thread now locked.

Thanks
Bill
 

FAQ: Dimensional Analysis: Solving Confusing Steps

What is dimensional analysis and why is it important in science?

Dimensional analysis is a mathematical method used in science to convert units of measurement and ensure that equations and calculations are accurate. It involves using conversion factors and canceling out units to arrive at a desired unit. It is important because it helps scientists make precise and consistent measurements, which are crucial in understanding and explaining natural phenomena.

How do I know which units to use in dimensional analysis?

The units used in dimensional analysis should always be consistent with the units given in the problem or equation. It is important to pay attention to the units given and make sure they are the same on both sides of the equation. If necessary, use conversion factors to change units to the desired ones.

Can dimensional analysis be used for any type of measurement?

Yes, dimensional analysis can be used for any type of measurement as long as the units are consistent. It is commonly used in chemistry, physics, and other sciences to convert units of length, mass, volume, time, and other physical quantities.

What are some common mistakes to avoid when using dimensional analysis?

One common mistake is using the wrong conversion factor, which can lead to incorrect results. It is also important to pay attention to the units and make sure they are consistent throughout the calculation. Another mistake is forgetting to cancel out units, which can also result in incorrect answers.

How can I practice and improve my skills in using dimensional analysis?

The best way to practice dimensional analysis is by doing practice problems and exercises. You can also use online resources and tutorials to learn more about the method and its applications. As you practice, pay attention to the units and make sure they are consistent, and try to identify any mistakes or errors in your calculations.

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