A cylindrical wire of radius R elongates when subjected to a tensile force F. Let L 0 represent the initial length of the wire and let L 1 represent the final length. Young’s modulus for the material is given by Y = F L 0 π R 2 ( L 1 − L 0 ) Assume that F = 800 ± 1 N, R = 0.75 ± 0.1 mm, L 0 = 25.0 ± 0.1 mm, and L 1 = 30.0 ± 0.1 mm. a. Estimate Y , and find the uncertainty in the estimate. b. Of the uncertainties in F , R , L 0 , and L 1 , only one has a non-negligible effect on the uncertainty in Y. Which one is it?

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Answer 1

Textbook Question

Chapter 3.4, Problem 15E

A cylindrical wire of radius R elongates when subjected to a tensile force F. Let L₀ represent the initial length of the wire and let L₁ represent the final length. Young’s modulus for the material is given by

$Y = F L 0 π R 2 ( L 1 − L 0 )$

Assume that F = 800 ± 1 N, R = 0.75 ± 0.1 mm, L₀ = 25.0 ± 0.1 mm, and L₁ = 30.0 ± 0.1 mm.

a. Estimate Y, and find the uncertainty in the estimate.
b. Of the uncertainties in F, R, L₀, and L₁, only one has a non-negligible effect on the uncertainty in Y. Which one is it?

a.

Expert Solution

To determine

Find the estimate of Young’s modulus of material.

Find the uncertainty in the Young’s modulus of material.

Answer to Problem 15E

The estimate of the Young’s modulus of material is $Y=2,264±608Nmm2_$ .

The uncertainty in the Young’s modulus of material is $σY=608Nmm2_$ .

Explanation of Solution

Given info:

The tensile force of cylindrical material is measured to be $F=800±1 N$ , the radius is measured to be $R=0.75±0.1 mm$ , initial length of the wire is measured to be $L0=25.0±0.1 mm$ and the final length of the wire is measured to be $L1=30.0±0.1 mm$ .

Calculation:

The form of the measurements of a process is,

$Measured value(μ)± Standard deviation(σ)$ .

Here, for a random sample the measured value will be sample mean and the population standard deviation will be sample standard deviation.

The form of the measurements of tensile force is $F=800±1 N$ .

Here, the measured value of tensile force is $F=800 N$ and the uncertainty in the tensile force is $σF=1 N$ .

The form of the measurements of radius is $R=0.75±0.1 mm$ .

Here, the measured value of radius is $R=0.75mm$ and the uncertainty in radius is $σR=0.1 mm$ .

The form of the measurements of initial length of the wire is $L0=25.0±0.1 mm$ .

Here, the measured value of initial length is $L0=25.0 mm$ and the uncertainty in initial length is $σL0=0.1 mm$ .

The form of the measurements of final length of the wire is $L1=30.0±0.1 mm$ .

Here, the measured value of final length is $L1=30.0mm$ and the uncertainty in final length is $σL1=0.1 mm$ .

Measured value of Young’s modulus of material:

The formula for Young’s modulus of material is $Y=FL0πR2(L1−L0)$ .

Here, $F=800 N$ , $R=0.75mm$ , $L0=25.0 mm$ and $L1=30.0mm$ .

The measured value of Young’s modulus of material is obtained as follows:

$Y=FL0πR2(L1−L0)=800×253.14159×0.752×(30−25)=2,264$

Thus, the measured value of Young’s modulus of material is $Y=2,264Nmm2_$ .

Uncertainty:

The uncertainty of a process is determined by the standard deviation of the measurements. In other words it can be said that, measure of variability of a process is known as uncertainty of the process.

Therefore, it can be said that uncertainty is simply $(σ)$ .

Standard deviation:

The standard deviation is based on how much each observation deviates from a central point represented by the mean. In general, the greater the distances between the individual observations and the mean, the greater the variability of the data set.

The general formula for standard deviation is,

$s=∑xi2−(∑xi)2nn−1$ .

From the properties of uncertainties for functions of one measurement it is known that,

If $X1,X2,...,Xn$ are independent measurements with uncertainties $σX1,σX2,...,σXn$ and if $U=U(X1,X2,..,Xn)$ is a function of $X1,X2,...,Xn$ then the uncertainty in the variable U is $σU=(∂U∂X1)2σX12+(∂U∂X2)2σX22+....+(∂U∂Xn)2σXn2$ .

Here, tensile force, radius, initial length and final length of the cylindrical wire are not constants. The Young’s modulus of material is a function of tensile force, radius, initial length and final length.

The uncertainty in the Young’s modulus of material is,

$σY=FL0πR2(L1−L0)=(∂Y∂F)2σF2+(∂Y∂R)2σR2+(∂Y∂L0)2σL02+(∂Y∂L1)2σL12=((∂(FL0πR2(L1−L0))∂F)2σF2+(∂(FL0πR2(L1−L0))∂R)2σR2+(∂(FL0πR2(L1−L0))∂L0)2σL02+(∂(FL0πR2(L1−L0))∂L1)2σL12)=((L0πR2(L1−L0))2×σF2+(−2FL0πR3(L1−L0))2×σR2+(FL1πR2(L1−L0))2×σL02+(−FL0πR2(L1−L0))2×σL12)=((2.82942)2×σF2+(−6,036.1)2×σR2+(543.249)2×σL02+(−452.707)2×σL12)$

$=((2.82942)2×(1)2+(−6,036.1)2×(0.1)2+(543.249)2×(0.1)2+(−452.707)2×(0.1)2)=608$

Thus, the uncertainty in the Young’s modulus of material is $σY=608Nmm2_$ .

Estimate of the Young’s modulus of material:

The estimate of the measurement of a process is,

$Measured value(μ)± Standard deviation(σ)$ .

Here, for a random sample the measured value will be sample mean and the population standard deviation will be sample standard deviation.

The estimate of Young’s modulus of material is,

$Y=Measured value of Y±σY=,264±608Nmm2$

Thus, the estimate of Young’s modulus of material is $Y=2,264±608Nmm2_$ .

b.

Expert Solution

To determine

Find the variable with non-negligible effect on the uncertainty in Y.

Answer to Problem 15E

The variable R has the non-negligible effect on the uncertainty in Y.

Explanation of Solution

Calculation:

From part (a), the uncertainty in the tensile force is $σF=1 N$ , uncertainty in radius is $σR=0.1 mm$ , uncertainty in initial length is $σL0=0.1 mm$ and the uncertainty in final length is $σL1=0.1 mm$ .

Uncertainty:

From the properties of uncertainties for functions of one measurement it is known that,

If $X1,X2,...,Xn$ are independent measurements with uncertainties $σX1,σX2,...,σXn$ and if $U=U(X1,X2,..,Xn)$ is a function of $X1,X2,...,Xn$ then the uncertainty in the variable U is $σU=(∂U∂X1)2σX12+(∂U∂X2)2σX22+....+(∂U∂Xn)2σXn2$ .

Here, tensile force, radius, initial length and final length of the cylindrical wire are not constants. The Young’s modulus of material is a function of tensile force, radius, initial length and final length.

The uncertainty in the Young’s modulus of material is,

$σY=FL0πR2(L1−L0)=(∂Y∂F)2σF2+(∂Y∂R)2σR2+(∂Y∂L0)2σL02+(∂Y∂L1)2σL12=((∂(FL0πR2(L1−L0))∂F)2σF2+(∂(FL0πR2(L1−L0))∂R)2σR2+(∂(FL0πR2(L1−L0))∂L0)2σL02+(∂(FL0πR2(L1−L0))∂L1)2σL12)=((L0πR2(L1−L0))2×σF2+(−2FL0πR3(L1−L0))2×σR2+(FL1πR2(L1−L0))2×σL02+(−FL0πR2(L1−L0))2×σL12)=((2.82942)2×σF2+(−6,036.1)2×σR2+(543.249)2×σL02+(−452.707)2×σL12)$

Uncertainty in Young’s modulus of material with negligible uncertainty in F:

Here, $σF=0 N$ , $σR=0.1 mm$ , $σL0=0.1 mm$ and $σL1=0.1 mm$ .

The uncertainty in Young’s modulus of material with negligible uncertainty in F is,

$σY=(2.82942)2×σF2+(−6,036.1)2×σR2+(543.249)2×σL02+(−452.707)2×σL12=(2.82942)2×(0)2+(−6,036.1)2×(0.1)2+(543.249)2×(0.1)2+(−452.707)2×(0.1)2=608$

Thus, the uncertainty in Young’s modulus of material with negligible uncertainty in F is $σY=608Nmm2_$ .

Uncertainty in Young’s modulus of material with negligible uncertainty in R:

Here, $σF=1 N$ , $σR=0 mm$ , $σL0=0.1 mm$ and $σL1=0.1 mm$ .

The uncertainty in Young’s modulus of material with negligible uncertainty in R is,

$σY=(2.82942)2×σF2+(−6,036.1)2×σR2+(543.249)2×σL02+(−452.707)2×σL12=(2.82942)2×(1)2+(−6,036.1)2×(0)2+(543.249)2×(0.1)2+(−452.707)2×(0.1)2=71$

Thus, the uncertainty in Young’s modulus of material with negligible uncertainty in R is $σY=71Nmm2_$ .

Uncertainty in Young’s modulus of material with negligible uncertainty in $L0$ :

Here, $σF=1 N$ , $σR=0.1 mm$ , $σL0=0mm$ and $σL1=0.1 mm$ .

The uncertainty in Young’s modulus of material with negligible uncertainty in $L0$ is,

$σY=(2.82942)2×σF2+(−6,036.1)2×σR2+(543.249)2×σL02+(−452.707)2×σL12=(2.82942)2×(1)2+(−6,036.1)2×(0.1)2+(543.249)2×(0)2+(−452.707)2×(0.1)2=605$

Thus, the uncertainty in Young’s modulus of material with negligible uncertainty in $L0$ is $σY=605Nmm2_$ .

Uncertainty in Young’s modulus of material with negligible uncertainty in $L1$ :

Here, $σF=1 N$ , $σR=0.1 mm$ , $σL0=0.1 mm$ and $σL1=0 mm$ .

The uncertainty in Young’s modulus of material with negligible uncertainty in $L1$ is,

$σY=(2.82942)2×σF2+(−6,036.1)2×σR2+(543.249)2×σL02+(−452.707)2×σL12=(2.82942)2×(1)2+(−6,036.1)2×(0.1)2+(543.249)2×(0.1)2+(−452.707)2×(0)2=606$

Thus, the uncertainty in Young’s modulus of material with negligible uncertainty in $L1$ is $σY=606Nmm2_$ .

Comparison:

The uncertainty in Young’s modulus of material with negligible uncertainty in F is $σY=608Nmm2_$ .

The uncertainty in Young’s modulus of material with negligible uncertainty in R is $σY=71Nmm2_$ .

The uncertainty in Young’s modulus of material with negligible uncertainty in $L0$ is $σY=605Nmm2_$ .

The uncertainty in Young’s modulus of material with negligible uncertainty in $L1$ is $σY=606Nmm2_$ .

From part(a), the uncertainty in the Young’s modulus of material is $σY=608Nmm2_$ .

By comparing all these uncertainties, the uncertainty in Y has drastic effect when the uncertainty in R is reduced to “0” and there is not much difference in the uncertainty in Y with the negligible uncertainty in F, $L0$ and $L1$ .

That is, there is typical difference in the values of 71 and 608.

Therefore, the uncertainty in R is not negligible.

Thus, the variable R has the non-negligible effect on the uncertainty in Y.

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Answer 2

Textbook Question

Chapter 3.4, Problem 15E

A cylindrical wire of radius R elongates when subjected to a tensile force F. Let L₀ represent the initial length of the wire and let L₁ represent the final length. Young’s modulus for the material is given by

$Y = F L 0 π R 2 ( L 1 − L 0 )$

Assume that F = 800 ± 1 N, R = 0.75 ± 0.1 mm, L₀ = 25.0 ± 0.1 mm, and L₁ = 30.0 ± 0.1 mm.

a. Estimate Y, and find the uncertainty in the estimate.
b. Of the uncertainties in F, R, L₀, and L₁, only one has a non-negligible effect on the uncertainty in Y. Which one is it?

a.

Expert Solution

To determine

Find the estimate of Young’s modulus of material.

Find the uncertainty in the Young’s modulus of material.

Answer to Problem 15E

The estimate of the Young’s modulus of material is $Y=2,264±608Nmm2_$ .

The uncertainty in the Young’s modulus of material is $σY=608Nmm2_$ .

Explanation of Solution

Given info:

The tensile force of cylindrical material is measured to be $F=800±1 N$ , the radius is measured to be $R=0.75±0.1 mm$ , initial length of the wire is measured to be $L0=25.0±0.1 mm$ and the final length of the wire is measured to be $L1=30.0±0.1 mm$ .

Calculation:

The form of the measurements of a process is,

$Measured value(μ)± Standard deviation(σ)$ .

Here, for a random sample the measured value will be sample mean and the population standard deviation will be sample standard deviation.

The form of the measurements of tensile force is $F=800±1 N$ .

Here, the measured value of tensile force is $F=800 N$ and the uncertainty in the tensile force is $σF=1 N$ .

The form of the measurements of radius is $R=0.75±0.1 mm$ .

Here, the measured value of radius is $R=0.75mm$ and the uncertainty in radius is $σR=0.1 mm$ .

The form of the measurements of initial length of the wire is $L0=25.0±0.1 mm$ .

Here, the measured value of initial length is $L0=25.0 mm$ and the uncertainty in initial length is $σL0=0.1 mm$ .

The form of the measurements of final length of the wire is $L1=30.0±0.1 mm$ .

Here, the measured value of final length is $L1=30.0mm$ and the uncertainty in final length is $σL1=0.1 mm$ .

Measured value of Young’s modulus of material:

The formula for Young’s modulus of material is $Y=FL0πR2(L1−L0)$ .

Here, $F=800 N$ , $R=0.75mm$ , $L0=25.0 mm$ and $L1=30.0mm$ .

The measured value of Young’s modulus of material is obtained as follows:

$Y=FL0πR2(L1−L0)=800×253.14159×0.752×(30−25)=2,264$

Thus, the measured value of Young’s modulus of material is $Y=2,264Nmm2_$ .

Uncertainty:

The uncertainty of a process is determined by the standard deviation of the measurements. In other words it can be said that, measure of variability of a process is known as uncertainty of the process.

Therefore, it can be said that uncertainty is simply $(σ)$ .

Standard deviation:

The standard deviation is based on how much each observation deviates from a central point represented by the mean. In general, the greater the distances between the individual observations and the mean, the greater the variability of the data set.

The general formula for standard deviation is,

$s=∑xi2−(∑xi)2nn−1$ .

From the properties of uncertainties for functions of one measurement it is known that,

If $X1,X2,...,Xn$ are independent measurements with uncertainties $σX1,σX2,...,σXn$ and if $U=U(X1,X2,..,Xn)$ is a function of $X1,X2,...,Xn$ then the uncertainty in the variable U is $σU=(∂U∂X1)2σX12+(∂U∂X2)2σX22+....+(∂U∂Xn)2σXn2$ .

Here, tensile force, radius, initial length and final length of the cylindrical wire are not constants. The Young’s modulus of material is a function of tensile force, radius, initial length and final length.

The uncertainty in the Young’s modulus of material is,

$σY=FL0πR2(L1−L0)=(∂Y∂F)2σF2+(∂Y∂R)2σR2+(∂Y∂L0)2σL02+(∂Y∂L1)2σL12=((∂(FL0πR2(L1−L0))∂F)2σF2+(∂(FL0πR2(L1−L0))∂R)2σR2+(∂(FL0πR2(L1−L0))∂L0)2σL02+(∂(FL0πR2(L1−L0))∂L1)2σL12)=((L0πR2(L1−L0))2×σF2+(−2FL0πR3(L1−L0))2×σR2+(FL1πR2(L1−L0))2×σL02+(−FL0πR2(L1−L0))2×σL12)=((2.82942)2×σF2+(−6,036.1)2×σR2+(543.249)2×σL02+(−452.707)2×σL12)$

$=((2.82942)2×(1)2+(−6,036.1)2×(0.1)2+(543.249)2×(0.1)2+(−452.707)2×(0.1)2)=608$

Thus, the uncertainty in the Young’s modulus of material is $σY=608Nmm2_$ .

Estimate of the Young’s modulus of material:

The estimate of the measurement of a process is,

$Measured value(μ)± Standard deviation(σ)$ .

Here, for a random sample the measured value will be sample mean and the population standard deviation will be sample standard deviation.

The estimate of Young’s modulus of material is,

$Y=Measured value of Y±σY=,264±608Nmm2$

Thus, the estimate of Young’s modulus of material is $Y=2,264±608Nmm2_$ .

b.

Expert Solution

To determine

Find the variable with non-negligible effect on the uncertainty in Y.

Answer to Problem 15E

The variable R has the non-negligible effect on the uncertainty in Y.

Explanation of Solution

Calculation:

From part (a), the uncertainty in the tensile force is $σF=1 N$ , uncertainty in radius is $σR=0.1 mm$ , uncertainty in initial length is $σL0=0.1 mm$ and the uncertainty in final length is $σL1=0.1 mm$ .

Uncertainty:

From the properties of uncertainties for functions of one measurement it is known that,

If $X1,X2,...,Xn$ are independent measurements with uncertainties $σX1,σX2,...,σXn$ and if $U=U(X1,X2,..,Xn)$ is a function of $X1,X2,...,Xn$ then the uncertainty in the variable U is $σU=(∂U∂X1)2σX12+(∂U∂X2)2σX22+....+(∂U∂Xn)2σXn2$ .