Physics for Scientists and Engineers
Physics for Scientists and Engineers
10th Edition
ISBN: 9781337553278
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Textbook Question
Chapter 18, Problem 31AP

Two metal bars are made of invar and a third bar is made of aluminum. At 0°C, each of the three bars is drilled with two holes 40.0 cm apart. Pins are put through the holes to assemble the bars into an equilateral triangle as in Figure P18.31. (a) First ignore the expansion of the invar. Find the angle between the invar bars as a function of Celsius temperature. (b) Is your answer accurate for negative as well as positive temperatures? (c) Is it accurate for 0°C? (d) Solve the problem again, including the expansion of the invar. Aluminum melts at 660°C and invar at 1 427°C. Assume the tabulated expansion coefficients are constant. What are (e) the greatest and (f) the smallest attainable angles between the invar bars?

Figure P18.31

Chapter 18, Problem 31AP, Two metal bars are made of invar and a third bar is made of aluminum. At 0C, each of the three bars

(a)

Expert Solution
Check Mark
To determine
The angle between the invar bars as a function of Celsius temperature.

Answer to Problem 31AP

The angle between the invar bars as a function of Celsius temperature is 2sin1(1+αAlTC2) .

Explanation of Solution

Given info: The distance between two holes drilled at 0°C is 40.0cm , the melting temperature of the aluminum is 660°C and the melting temperature of invar is 1427°C .

In the given diagram consider the right triangle that each invar bar makes with one half of the aluminum bar.

Then, the angle is,

sin(θ2)=L0(1+αAlΔT)2L0=L0(1+αAlΔT)2L0=(1+αAlTC)2

Here,

L0 is the length of each bar at 0°C .

αAl is the average coefficient of linear expansion of the aluminum.

ΔT is the temperature difference.

TC is the Celsius temperature.

Rearrange the above equation for θ .

sin(θ2)=(1+αAlTC)2θ=2sin1(1+αAlTC2)

Conclusion:

Therefore, the angle between the invar bars as a function of Celsius temperature is 2sin1(1+αAlTC2) .

(b)

Expert Solution
Check Mark
To determine
Whether the answer is accurate for negative as well as positive temperature or not.

Answer to Problem 31AP

Yes the answer is accurate for negative as well as positive temperature.

Explanation of Solution

Given info: The distance between two holes drilled at 0°C is 40.0cm , the melting temperature of the aluminum is 660°C and the melting temperature of invar is 1427°C .

If the temperature drops, the Celsius temperature becomes negative. The negative value of the Celsius temperature describes the contraction of the bars. So the answer is accurate for the negative temperature same as the positive temperature.

Conclusion:

Therefore, yes the answer is accurate for negative as well as positive temperature.

(c)

Expert Solution
Check Mark
To determine
Whether the answer is accurate at 0°C or not.

Answer to Problem 31AP

Yes, the answer is accurate at 0°C .

Explanation of Solution

Given info: The distance between two holes drilled at 0°C is 40.0cm , the melting temperature of the aluminum is 660°C and the melting temperature of invar is 1427°C .

The expression for the angle between the invar bars is,

θ=2sin1(1+αAlTC2)

Substitute 0°C for TC in above equation.

θ=2sin1(1+αAl(0°C)2)=2sin1(12)=60°

The value of each angle of the right angle triangle is 60° so this is accurate.

Conclusion:

Therefore, Yes, the answer is accurate at 0°C .

(d)

Expert Solution
Check Mark
To determine
The angle between the invar bars and the aluminum bar as a function of Celsius temperature.

Answer to Problem 31AP

The angle between the invar bars and the aluminum bar as a function of Celsius temperature is 2sin1(1+αAlTC2(1+αinvarTC)) .

Explanation of Solution

Given info: The distance between two holes drilled at 0°C is 40.0cm , the melting temperature of the aluminum is 660°C and the melting temperature of invar is 1427°C .

In the given diagram consider the right triangle that each invar bar makes with one half of the aluminum bar.

Then, the angle between invar bars and aluminum bar is,

sin(θ2)=L0(1+αAlΔT)2L0(1+αinvarΔT)=L0(1+αAlΔT)2L0(1+αinvarΔT)=(1+αAlTC)2(1+αinvarTC)

Here,

αinvar is the average coefficient of linear expansion of the invar.

Rearrange the above equation for θ .

sin(θ2)=(1+αAlTC)2(1+αinvarTC)θ=2sin1(1+αAlTC2(1+αinvarTC))

Conclusion:

Therefore, the angle between the invar bars and the aluminum bar as a function of Celsius temperature is 2sin1(1+αAlTC2(1+αinvarTC)) .

(e)

Expert Solution
Check Mark
To determine
The greatest attainable angle between the invar bars.

Answer to Problem 31AP

The greatest attainable angle between the invar bars is 61.0° .

Explanation of Solution

Given info: The distance between two holes drilled at 0°C is 40.0cm , the melting temperature of the aluminum is 660°C and the melting temperature of invar is 1427°C .

The average coefficient of linear expansion of the aluminum is 24×106°C1 .

The average coefficient of linear expansion of the invar is 0.9×106°C1 .

The greatest angle is occur at 660°C .

The equation for the angel between the invar bars is,

θ=2sin1(1+αAlTC2(1+αinvarTC))

Substitute 24×106°C1 for αAl , 0.9×106°C1 for αinvar and 660°C for TC in above equation to find θ .

θ=2sin1(1+(24×106°C1)(660°C)2(1+(0.9×106°C1)(660°C)))=2sin1(1.015842.001188)=2sin1(0.5076)=61.0°

Conclusion:

Therefore, the greatest attainable angle between the invar bars is 61.0° .

(f)

Expert Solution
Check Mark
To determine
The smallest attainable angle between the invar bars.

Answer to Problem 31AP

The smallest attainable angle between the invar bars is 59.6° .

Explanation of Solution

Given info: The distance between two holes drilled at 0°C is 40.0cm , the melting temperature of the aluminum is 660°C and the melting temperature of invar is 1427°C .

The average coefficient of linear expansion of the aluminum is 24×106°C1 .

The average coefficient of linear expansion of the invar is 0.9×106°C1 .

The smallest angle is occur at 273°C .

The equation for the angel between the invar bars is,

θ=2sin1(1+αAlTC2(1+αinvarTC))

Substitute 24×106°C1 for αAl , 0.9×106°C1 for αinvar and 273°C for TC in above equation to find θ .

θ=2sin1(1+(24×106°C1)(273°C)2(1+(0.9×106°C1)(273°C)))=2sin1(0.99341.9995)=2sin1(0.497)=59.6°

Conclusion:

Therefore, the smallest attainable angle between the invar bars is 59.6° .

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Chapter 18 Solutions

Physics for Scientists and Engineers

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