Thermal Expansion: Glass rectangle

AI Thread Summary
A rectangular windshield is being assembled with a glass plate that expands from 68°F to fit a 3 ft by 1 ft frame at 60°C. The linear expansivity of glass is 5 x 10-6 /C°, and calculations show that the initial area of the glass plate should be approximately 2.9988 ft². The length of the glass plate is determined to be about 2.9944 ft, while the width is calculated to be around 0.9998 ft. Participants in the discussion confirm the calculations and suggest using linear expansion to verify results. The conversation emphasizes the importance of precision in calculations and alternative methods for checking answers.
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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Thanks for you help! :)
 
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