Heat expansions using simultaneous equations.

[lagwagon555]

Hi all, this isn't homework, just doing a bit of revision and I've come to a question I'm stuck on. Thanks for any help!

Homework Statement

A steel ball is resting on a hole in a brass surface, at 15 degrees celsius. The bearing has a diameter of 30mm, and the hole has a diameter at 29.994mm. What temperature do you have to heat both the ball and surface up to, in order for the ball to hit through the hole? The temperature will be the same for both the bearing and the hole.

a for brass = 1.9x10^-5
a for steel = 1.1x10^-5

Homework Equations

(Sorry, I haven't worked out how to use the proper method for inputting equations)
(change in)L = a.L(initial).(change in)T

The Attempt at a Solution

(change in)L = 3.3x10^-4.(change in)T for the bearing
(change in)L = 5.6x10^-4.(change in)T for the brass


I have no clue how to set up the equations from here. Any help would be greatly appreciated! Thanks.

 

[LowlyPion]

First of all help yourself to a Δ.

(And use it well.)

ΔL = a *L * ΔT

So set up 2 equations. One for ΔL1 and the other for ΔL2 in which you know ultimately that you want ΔL2 = ΔL1 + .006

Then solve for ΔT.

 

[J Hill]

Well, you what you need is:
L_{B0}+ \Delta L_{B} < L_{R0} +\Delta L_{R}
All that should be necessary is to substitute in the expression for thermal expansion. Where the B values are for the steel ball, and the R for the brass ring/surface

 

[lagwagon555]

First of all help yourself to a Δ.

(And use it well.)

ΔL = a *L * ΔT

So set up 2 equations. One for ΔL1 and the other for ΔL2 in which you know ultimately that you want ΔL2 = ΔL1 + .006

Then solve for ΔT.

I shall treasure it for the rest of my life!

And sorry to ask for even more details, I'm just useless with maths at the moment. So, if I set up my two equations, it should look like:

ΔL1 = 1.1x10^-5 *30 * ΔT (I assume you can keep the units in mm here)
ΔL2 = 1.9x10^-5 *29.994 * ΔT

How do you go about getting about solving for ΔT? I tried setting the ΔL2 equation to the ΔL1 equation plus .006, but when I solved for ΔT I got 8.25x10^-3, which is horribly wrong.