Find Work Done with Increasing Tension of a Wire

Click For Summary
SUMMARY

The discussion focuses on calculating the work done when the tension in a wire increases quasi-statically from F1 to F2, utilizing Young's Modulus (Y), cross-sectional area (A), and length (L). The work done (W) is derived using the equation W = YA ln(L2/L), where the integral accounts for the change in length. A critical point raised is the necessity to consider changes in cross-sectional area, suggesting a potential need for a double integral if volume remains constant. The calculation assumes F1 equals zero, which is a valid approach for this scenario.

PREREQUISITES
  • Understanding of Young's Modulus (Y) in material science
  • Knowledge of integral calculus for evaluating work done
  • Familiarity with the relationship between tension, cross-sectional area, and length in elastic materials
  • Concept of volume conservation in deformable bodies
NEXT STEPS
  • Explore the derivation of work done in elastic materials using double integrals
  • Research the implications of volume conservation on cross-sectional area changes in wires
  • Study advanced applications of Young's Modulus in different materials
  • Learn about the effects of varying tension on the mechanical properties of materials
USEFUL FOR

Students in engineering and physics, particularly those studying material mechanics and elasticity, as well as professionals involved in structural analysis and design.

James Wine
Messages
8
Reaction score
0

Homework Statement


Let a wire's tension increase quasi-statically from [tex]F_{1}[/tex][tex]\rightarrow[/tex][tex]F_{2}[/tex], with Young's Modulus [tex]Y[/tex], Cross-sectional Area [tex]A[/tex], and Length [tex]L[/tex]. Find the work done


Homework Equations



[tex]F=\frac{YA}{L}[/tex] *Delta L, tex having issues
W= Integral from L to L_2

The Attempt at a Solution



W = YA*Integral[dL/L]
W = YA ln(L_2/L)

Question is if the length changes then the cross-sectional area should also change and would require this to be a double integral, 2nd question would be if I even thought about the first integral in the right way.
 
Physics news on Phys.org
If you really need to take into account the effect of A changing, you could use the approximation that the total volume remains constant, or you could use ratio[/url].

Note your calculation assumes that F1=0 because you used L, the initial length of the wire, for the lower limit of the integral. Your work looks fine otherwise.
 
Last edited by a moderator:

Similar threads

Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
4
Views
2K
  • · Replies 16 ·
Replies
16
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 10 ·
Replies
10
Views
1K