How Is the Spring Constant Derived from Young's Modulus, Length, and Area?

Click For Summary
SUMMARY

The spring constant k of a metal bar can be derived using Young's modulus Y, initial length L, and cross-sectional area A. The correct formula for the spring constant is k = Y * A / L. This derivation involves applying Hooke's Law and rearranging the equations to express force in terms of the material properties and dimensions of the bar. The initial misunderstanding in the discussion was related to the incorrect formulation of the relationship between force and elongation.

PREREQUISITES
  • Understanding of Hooke's Law
  • Familiarity with Young's modulus (Y)
  • Knowledge of tensile strain concepts
  • Basic algebra for rearranging equations
NEXT STEPS
  • Study the derivation of Hooke's Law in detail
  • Explore applications of Young's modulus in material science
  • Investigate the relationship between tensile strain and stress
  • Learn about the mechanical properties of different materials
USEFUL FOR

Students studying physics or engineering, particularly those focusing on material mechanics and elasticity, will benefit from this discussion.

nickb145
Messages
68
Reaction score
0

Homework Statement



Consider a metal bar of initial length L and cross-sectional area A. The Young's modulus of the material of the bar is Y. Find the "spring constant" k of such a bar for low values of tensile strain.


Express in terms of Y, L, and A.

Homework Equations




I know the hooks law equation and the modulus but i can't figure out the answer it wants me ot get

Maybe I'm doing it wrong but i am getting Y/(L*A)...

The Attempt at a Solution

 
Physics news on Phys.org
Spring constant = F/ΔL .....(1)

Youngs Mod. Y = F/ΔL * L/A ......(2)

Rearrange (2) to give an equation for F/ΔL

Substitute into 1
 
im getting F=(Y*A)/L
 
You are close but the answer isn't the force F = (Y*A)/L

The problem asks you to work out an equation for the spring constant k.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
3
Views
3K
  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 23 ·
Replies
23
Views
5K
Replies
10
Views
7K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K