Elastic deformation of stacked copper and brass cylinders under compression

[texasgrl05]

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?

 

[Dr.Brain]

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

BJ

 

[texasgrl05]

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]

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?

 

[texasgrl05]

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]

You need to use the proper units. Since you are dealing in Newtons, that is broken down into kg*m/sec^2. Sooooo...you need to convert all of your distances from centimeters into meters. Also, the "A" in the equation is for AREA. You keep using the radius of the cylinder in stead of the AREA.

Your first equation should look like:
$$6450 N = (1.1 x 10^11 \frac{N}{m^2})(\frac{\Delta L}{.03 m})(\pi * (.022 m)^2)$$

You can solve for $$\Delta L$$ which will be in meters for each metal.