Thermal Deformation: Find Change in Cable Length at -15°C

[Jim_Earle]

Hey everyone I am wondering if someone can just double check my work and formulas to see if I did this correctly. Thanks!
A 1.0-m rigid horizontal support is hung by two cables as shown. One cable is brass and the other is high density polyethylene plastic. At room temperature (21° C) the cables measure 3.50 m vertically. When the temperature drops to -15°C find the following.
The change in length in each cable.
Per my table I found Copper has a thermal expansion of coefficient α that equals .00018 m/m-C°
My L₀ = 3.50 m
∆T = TFinal – T orginal = -15°C – (21°C) = -36°C
∆ Lthermal = α L₀(∆T) = (.000018 m/m-C°)(3.50m)(-36°) = -.002268 m
Am I correct with the way I did this?

how do you get all the special symbols to insert into the forums also?

 

[AlephZero]

Your method is correct.

But you said the value of α is .00018/°C and then used .000018/°C, so check which value is the right one!