# Elastic deformation

A copper cylinder and a brass cylinder are stacked end to end, as in the drawing. Each cylinder has a radius of 0.22 cm. A compressive force of F = 6450 N is applied to the right end of the brass cylinder. Find the amount by which the length of the stack decreases.

Is F=Y(change in L/L0)A the equation I would use? I know that Y for brass is 9.0 x 10^10 and Y for copper is 1.1 x 10^11 so would I add those together or what?

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Length by which stack decreases= Decrease in length of copper+Decrease in length of brass.

BJ

apparently i'm still doing it wrong:
for copper i got:
6450=1.1*10^11(change in L/3).0022m = 7.9*10^-5

and for brass:
6450=9.0*10^10(change in L/5).0022m = 1.63*10^-4

when i added these together i got 2.43*10^-4

what am i doing wrong?

Astronuc
Staff Emeritus
Stress = Force /area

Strain = Stress/E, where E = Elastic (Young's) Modulus.

What is the meaning of strain in terms of change in length?

FredGarvin
1) You need to calculate the area of the cylinders, not just use the radii.

2) What are the units of your constants?

3) You didn't include a picture, so I am assuming that you left out that the cylinders are 3 m and 5 m in length?

yeah the length for the copper rod is 3 cm and for the brass rod its 5 cm..

for copper i got:
6450N=1.1*10^11N/m^2(change in L/3cm).22cm = 7.99*10^-7

and for brass:
6450N=9.0*10^10N/m^2(change in L/5cm).22cm = 1.62*10^-6

when i added these together i got 2.43*10^-6 cm

i still don't think this is right?

FredGarvin
$$6450 N = (1.1 x 10^11 \frac{N}{m^2})(\frac{\Delta L}{.03 m})(\pi * (.022 m)^2)$$
You can solve for $$\Delta L$$ which will be in meters for each metal.