Thermal Expansion: Glass rectangle

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Homework Help Overview

The problem involves calculating the dimensions of a glass plate that will fit into a frame, considering thermal expansion. The glass plate is initially cut at a different temperature than the frame's installation temperature, and the linear expansivity of glass is provided.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the initial area of the glass plate and how to manipulate the area expansion formula. There are questions about the accuracy of calculated values and the impact of rounding. Some suggest checking results using linear expansion/contraction directly.

Discussion Status

Participants are actively comparing their calculations and discussing potential discrepancies in values. There is an acknowledgment of different approaches to verify results, but no consensus has been reached on the correctness of the answers.

Contextual Notes

Participants are working under the constraints of specific temperature conditions and the properties of glass, with an emphasis on ensuring calculations align with thermal expansion principles.

paulie
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Homework Statement


A rectangular windshield is to be assembled by installing a glass plate on a 3 ft by 1 ft frame at 60°C. The glass plate is cut at 68°F in such a way that its length is three times its width. The linear expansivity of glass is 5 x 10-6 /C°. (a) What area of the glass plate at 68°F will exactly fit the frame at 60°C? (b) What length of the glass plate at 68°F will exactly fit the frame at 60°C? (c) What width of the glass plate at 68°F will exactly fit the frame at 60°C?

Homework Equations


Af = Ao(1+2αΔT)
Where:
Af = Final Area
Ao = Initial Area
α = coefficient of expansion
ΔT = Change in Temperature

L = 3W
Where:
L = Length
W = Width

The Attempt at a Solution


From what I understand, I am suppose to find the initial area of the glass plate to be fitted in the frame.
First, manipulate the area expansion formula to find Ao:
Ao = Af / (1+2αΔT)
Substituting the values, I will arrive with:
Ao = 2.9988 ft2 → (a)
Next, find the length of the initial area by using L=3W, where:
Ao = L*W
Ao = L*L/3
L = √(3*Ao)
Substituting the values, I will arrive with:
L = 2.9944 ft → (b)
Finally, using L=3W to find the width:
L=3W
W=L/3
W=0.9998 ft → (c)

Is this the right answer? The initial values seems too close to the final.
 
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Hi paulie,

Welcome to Physics Forums!

I see a slightly different value for L, perhaps due to rounding or truncation? How many decimal places did you maintain for intermediate values in your calculations?

Another approach you might try in order to check your results is to go directly for the length and width values. You know that the length and width at 60°C are 3 ft and 1 ft respectively. What are those values at the cutting temperature? This would be a linear contraction, not an area.
 
Did you get 2.9994 ft for the length? My solution paper gives that value, it seems I typed it incorrectly.

Tried checking using linear expansion/contraction and it gives the same value, never knew I can use this to check.
 
paulie said:
Did you get 2.9994 ft for the length? My solution paper gives that value, it seems I typed it incorrectly.
Yes, that's what I got.
Tried checking using linear expansion/contraction and it gives the same value, never knew I can use this to check.
:smile: Now you know!
 
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Likes   Reactions: paulie
Thanks for you help! :)
 

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