Largest mass for Ultimate Tensile Strength

AI Thread Summary
The discussion focuses on determining the maximum radius and mass of a steel ball that a steel wire can support, given the wire's dimensions and the ultimate tensile strength of steel. Participants emphasize the importance of understanding the relationship between stress, strain, and material properties, particularly the ultimate strength as a measure of failure. There is a realization that the ultimate strength represents the stress at which failure occurs, clarifying previous misconceptions about the relationship between stress and strain. Suggestions include using free body diagrams for visualization and working backward from known material properties to find the required mass. The conversation concludes with a consensus on the correct interpretation of ultimate strength in solving the problem.
AJKing
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Vibrations and Waves, A.P. French, questions 3-9
Please /do not/ provide full solutions. Just suggestions.

Homework Statement



A solid steel ball is hung at the bottom of a steel wire of length 2m and radius 1mm, the ultimate strength of steel is 1.1E9 N/m2. What are the radius and mass of the biggest ball the wire can bear?

Homework Equations



Y = \frac{\sigma}{\epsilon} (line)

U = \frac{\sigma_u}{\epsilon_u} (point below line)

m = \frac{4}{3}\pi*r^3 \rho

\rho \approx 7850 Kg/m^3 (Internet)

The Attempt at a Solution



For the sake of sanity, I arrange eq.2 by knowns and unknowns.

\frac{U*A}{g*l_0} = \frac{m}{\Delta l}

I must find Δ l

I consider two points on the Y line:

Y = \frac{\sigma_u}{\epsilon_1}

Y = \frac{\sigma_2}{\epsilon_u}

both can be written as

Y = C\frac{\sigma_u}{\epsilon_u} = C U

C = Y/U

Then I stop and think about how this will lead me in circles.

------

I have a suspicion that the solution to my problem requires differential equations. Anyone second that notion?
 
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DEs not required. Do you know what Y represents?
 
You know the ultimate stress the wire is capable of sustaining. You also know the size of the wire. Can you use this information to work back and figure out the amount of mass required to produce this stress? Draw a free body diagram if it helps you to visualize the situation.
 
SteamKing said:
You know the ultimate stress the wire is capable of sustaining. You also know the size of the wire. Can you use this information to work back and figure out the amount of mass required to produce this stress? Draw a free body diagram if it helps you to visualize the situation.

Y is available in the text.

[strike]The ultimate stress σu. Do I really know this without knowing the ultimate strain εu?[/strike]

Have I been wrong this entire time? Is ultimate strength just the stress associated with failure?
Suggesting the following is incorrect:

U = \frac{\sigma_u}{\epsilon_u}
 
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Holy hell.

Yes. I was incorrect about ultimate strength!
Break through in understanding here folks.

Ultimate strength is the stress of failure.

Question solved correctly.

Our discussion in the other forum makes more sense now SteamKing - cheers :).
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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