How to calculate wire extension from tension and Young's modulus?

[lemon]

Homework Statement

A wire of diameter 5.0 mm supports a 2.8kg load.
(a) Determine the tension in the wire
(b) The original length of the wire was 2.0m Calculate its extension when supporting the load.

Homework Equations

Young's modulus for the material of the wire = 2.0 x 10^7N m^-2

The Attempt at a Solution

a) Tension - Weight = mass x acceleration
acceleration is 0
Therefore
Tension = Weight
W=mg=2.8 x 10
=28N

b) I have the equation E=(force(F)/area(A)) / (extension(ex)/original length(l))
Can I:
(F x ex) = (A x l)
ex= (A x l)/F?

Which would give ((2.5x10^-3)x2.0) / 28 = 0.18x10^-3m (2s.f.)

Please let me know if my method is good?
Thank you

 

[PhanthomJay]

Homework Statement

A wire of diameter 5.0 mm supports a 2.8kg load.
(a) Determine the tension in the wire
(b) The original length of the wire was 2.0m Calculate its extension when supporting the load.

Homework Equations

Young's modulus for the material of the wire = 2.0 x 10^7N m^-2

The Attempt at a Solution

a) Tension - Weight = mass x acceleration
acceleration is 0
Therefore
Tension = Weight
W=mg=2.8 x 10
=28N

Yes, good.

b) I have the equation E=(force(F)/area(A)) / (extension(ex)/original length(l))

Yes. since (F/A)/(ex/l) is Fl/A(ex) , you might want to rewrite it as E = Fl/A(ex)

Can I:
(F x ex) = (A x l)
ex= (A x l)/F?

your algebra is off. Use the formula I gave you, multiply both sides by (ex), and then divide both sides by E , and see what you get for ex = ? You should also check your units to see if you are getting (ex) in length units; if you are not, then your equation is wrong.