- #1
lagwagon555
- 60
- 1
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!
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
(Sorry, I haven't worked out how to use the proper method for inputting equations)
(change in)L = a.L(initial).(change in)T
(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.
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.