Chapter 19, Problem 78CP

(a)

To determine

To show: The stationary line is at distance of $\frac{L}{2}\left(1-\frac{\tan \theta }{{\mu }_{\text{k}}}\right)$.

Answer to Problem 78CP

The distance at which stationary line lie is $\frac{L}{2}\left(1-\frac{\tan \theta }{{\mu }_{\text{k}}}\right)$.

Explanation of Solution

Given info: The angle made by the roof with the horizontal plane is $\theta$, the initial temperature is $T_{\text{c}}$, the final temperature is $T_{\text{c}}$, the coefficient of the thermal expansion is ${\alpha }_1$, the expansion coefficient of the metal plate is ${\alpha }_2$, the length of the plate is $L$ and the coefficient of kinetic friction between the plate and the roof is ${\mu }_{\text{k}}$.

Consider the figure given below.

Figure 1

Consider that $xL$ represent the distance of the stationary line below the top edge of the plate.

The normal force on the lower part of the plane is,

$N=mg\left(1-x\right)\cos \theta$

Here,

$m$ is the mass of the plate.

$g$ is the acceleration due to gravity.

$x$ is the distance.

$\theta$ is the angle between plate and the roof.

The force due to gravity is,

$F_{\text{g}}=mg\sin \theta$

The equation for the kinematic friction force is,

$f_{\text{k}}=mg{\mu }_{\text{k}}\left(1-x\right)\cos \theta$

The equation for the downward force is,

$F=mg{\mu }_{\text{k}}x\cos \theta$

The force equation for the plate is,

$\begin{array}{c}{\displaystyle \sum F_{\text{x}}}=0\\ -mg{\mu }_{\text{k}}x\cos \theta +mg{\mu }_{\text{k}}\left(1-x\right)\cos \theta -mg\sin \theta =0\\ -2mg{\mu }_{\text{k}}x\cos \theta =mg\sin \theta -mg{\mu }_{\text{k}}\cos \theta \\ 2{\mu }_{\text{k}}x={\mu }_{\text{k}}-\tan \theta \end{array}$

Further, solve for $x$.

$\begin{array}{c}2{\mu }_{\text{k}}x={\mu }_{\text{k}}-\tan \theta \\ x=\frac{1}{2}-\frac{\tan \theta }{2{\mu }_{\text{k}}}\end{array}$

The distance of the stationary line below the top edge is,

$\begin{array}{c}xL=\frac{L}{2}-\frac{L\tan \theta }{2{\mu }_{\text{k}}}\\ =\frac{L}{2}\left(1-\frac{\tan \theta }{{\mu }_{\text{k}}}\right)\end{array}$

Conclusion:

Therefore, the distance at which stationary line lie is $\frac{L}{2}\left(1-\frac{\tan \theta }{{\mu }_{\text{k}}}\right)$.

(b)

To determine

To show: The stationary line is at that same distance above the bottom edge of the plate.

Answer to Problem 78CP

The stationary line is at that same distance above the bottom edge of the plate.

Explanation of Solution

Given info: The angle made by the roof with the horizontal plane is $\theta$, the initial temperature is $T_{\text{c}}$, the final temperature is $T_{\text{c}}$, the coefficient of the thermal expansion is ${\alpha }_1$, the expansion coefficient of the metal plate is ${\alpha }_2$, the length of the plate is $L$ and the coefficient of kinetic friction between the plate and the roof is ${\mu }_{\text{k}}$.

Consider the figure given below.

Figure 2

With the temperature falling, the plate contracts faster than the roof. The upper part slides down and feels an upward frictional force, $mg{\mu }_{\text{k}}\left(1-x\right)\cos \theta$. The lower part slides up and feels downward frictional force, $mg{\mu }_{\text{k}}x\cos \theta$.

Then the force equation remains same as in part (a) and the stationary line is above the bottom edge by,

$xL=\frac{L}{2}\left(1-\frac{\tan \theta }{{\mu }_{\text{k}}}\right)$

Conclusion:

Therefore, it is proved that the stationary line is at that same distance above the bottom edge of the plate.

(c)

To determine

To show: The plate steps down the roof like an inchworm moving each day by the distance $\frac{L}{{\mu }_{\text{k}}}\left({\alpha }_2-{\alpha }_1\right)\left(T_{\text{h}}-T_{\text{c}}\right)\tan \theta$.

Answer to Problem 78CP

The distance by which the plate steps down the roof like an inchworm moving each day is $\frac{L}{{\mu }_{\text{k}}}\left({\alpha }_2-{\alpha }_1\right)\left(T_{\text{h}}-T_{\text{c}}\right)\tan \theta$.

Explanation of Solution

Given info: The angle made by the roof with the horizontal plane is $\theta$, the initial temperature is $T_{\text{c}}$, the final temperature is $T_{\text{c}}$, the coefficient of the thermal expansion is ${\alpha }_1$, the expansion coefficient of the metal plate is ${\alpha }_2$, the length of the plate is $L$ and the coefficient of kinetic friction between the plate and the roof is ${\mu }_{\text{k}}$.

Consider the figure given below.

Figure 3

Consider the plate at dawn, as the temperature starts to rise. As in part (a), a line at distance $xL$ below the top edge of the plate stays stationary relative to the roof as long as the temperature rises.

In the above figure, the point $P$ on the plate at distance $xL$ above the bottom edge is destined to become the fixed point when the temperature starts falling. As the temperature rises, point $P$ on the plate slides down the roof relative to the upper fixed line from $\left(L-xL-xL\right)$ to $\left(L-xL-xL\right)\left(1+{\alpha }_2\Delta T\right)$.

The change in the length of the plate is,

$\Delta L_{\text{plate}}=\left(L-xL-xL\right){\alpha }_2\Delta T$

The change in the length of the roof is,

$\Delta L_{\text{roof}}=\left(L-xL-xL\right){\alpha }_1\Delta T$

The point on the roof originally under point $P$ on the plate slides down the roof relative to the upper fixed line from $\left(L-xL-xL\right)$ to $\left(L-xL-xL\right)\left(1+{\alpha }_1\Delta T\right)$ relative to the upper fixed line, a change of $\Delta L_{\text{roof}}$.

When the temperature drops, point $P$ remains stationary on the roof while the roof contracts, pulling point $P$ back by approximately $\Delta L_{\text{roof}}$. Thus relative to the upper fixed line, point $P$ is moved down the roof $\Delta L_{\text{plate}}-\Delta L_{\text{roof}}$.

The displacement for a day is,

$\begin{array}{c}\Delta L_{\text{plate}}-\Delta L_{\text{roof}}=\left({\alpha }_2-{\alpha }_1\right)\left(L-xL-xL\right)\Delta T\\ =\left({\alpha }_2-{\alpha }_1\right)\left(L-2xL\right)\Delta T\end{array}$

Substitute $\frac{L}{2}\left(1-\frac{\tan \theta }{{\mu }_{\text{k}}}\right)$ for $xL$ in the above expression.

$\begin{array}{c}\Delta L_{\text{plate}}-\Delta L_{\text{roof}}=\left({\alpha }_2-{\alpha }_1\right)\left(L-2\left(\frac{L}{2}\right)\left(1-\frac{\tan \theta }{{\mu }_{\text{k}}}\right)\right)\Delta T\\ =\left({\alpha }_2-{\alpha }_1\right)\left(L-L\left(1-\frac{\tan \theta }{{\mu }_{\text{k}}}\right)\right)\left(T_{\text{h}}-T_{\text{c}}\right)\\ =\left({\alpha }_2-{\alpha }_1\right)\left(L-L+\frac{L\tan \theta }{{\mu }_{\text{k}}}\right)\left(T_{\text{h}}-T_{\text{c}}\right)\\ =\frac{L}{{\mu }_{\text{k}}}\left({\alpha }_2-{\alpha }_1\right)\left(T_{\text{h}}-T_{\text{c}}\right)\tan \theta \end{array}$

Conclusion:

Therefore, the distance by which the plate steps down the roof like an inchworm moving each day is $\frac{L}{{\mu }_{\text{k}}}\left({\alpha }_2-{\alpha }_1\right)\left(T_{\text{h}}-T_{\text{c}}\right)\tan \theta$.

(d)

To determine

The distance an aluminum plate moves each day.

Answer to Problem 78CP

The distance an aluminum plate moves each day is $\text{0}\text{.275}\text{mm}$.

Explanation of Solution

Given info: The angle made by the roof with the horizontal plane is $\theta$, the initial temperature is $T_{\text{c}}$, the final temperature is $T_{\text{c}}$, the coefficient of the thermal expansion is ${\alpha }_1$, the expansion coefficient of the metal plate is ${\alpha }_2$, the length of the plate is $L$ and the coefficient of kinetic friction between the plate and the roof is ${\mu }_{\text{k}}$.

The length of the plate is $1.20\text{m}$, the initial temperature is $4.00°\text{C}$, the final temperature is $36.0°\text{C}$, the angle made by the roof with the horizontal plane is $18.5°$, the coefficient of linear expansion is $1.50\times {10}^{-5}°{\text{C}}^{-1}$ and the coefficient of kinetic friction is $0.420$.

The coefficient of linear expansion for aluminum is $24\times {10}^{-6}°{\text{C}}^{-1}$.

The formula for the displacement for a day is,

$\Delta L_{\text{plate}}-\Delta L_{\text{roof}}=\frac{L}{{\mu }_{\text{k}}}\left({\alpha }_2-{\alpha }_1\right)\left(T_{\text{h}}-T_{\text{c}}\right)\tan \theta$

Substitute $1.20\text{m}$ for $L$, $0.420$ for ${\mu }_{\text{k}}$, $1.50\times {10}^{-5}°{\text{C}}^{-1}$ for ${\alpha }_1$, $24\times {10}^{-6}°{\text{C}}^{-1}$ for ${\alpha }_2$, $4.00°\text{C}$ for $T_{\text{c}}$, $36.0°\text{C}$ for $T_{\text{h}}$ and $18.5°$ for $\theta$ in above equation.

$\begin{array}{c}\Delta L_{\text{plate}}-\Delta L_{\text{roof}}=\left(\frac{1.20\text{m}}{0.420}\right)\left(24\times {10}^{-6}°{\text{C}}^{-1}-1.50\times {10}^{-5}°{\text{C}}^{-1}\right)\left(36.0°\text{C}-4.00°\text{C}\right)\tan \left(18.5°\right)\\ =2.75\times {10}^{-4}\text{m}\\ =\text{0}\text{.275}\text{mm}\end{array}$

Conclusion:

Therefore, the distance an aluminum plate moves each day is $\text{0}\text{.275}\text{mm}$.

(e)

To determine

To explain: The effect on the plate if the expansion coefficient of the plate is less than the expansion coefficient of the roof.

Answer to Problem 78CP

The plate creeps down the roof each day by an amount given by $\frac{L}{{\mu }_{\text{k}}}\left({\alpha }_1-{\alpha }_2\right)\left(T_{\text{c}}-T_{\text{h}}\right)\tan \theta$.

Explanation of Solution

Given info: The angle made by the roof with the horizontal plane is $\theta$, the initial temperature is $T_{\text{c}}$, the final temperature is $T_{\text{c}}$, the coefficient of the thermal expansion is ${\alpha }_1$, the expansion coefficient of the metal plate is ${\alpha }_2$, the length of the plate is $L$ and the coefficient of kinetic friction between the plate and the roof is ${\mu }_{\text{k}}$.

If ${\alpha }_2<{\alpha }_1$, the forces of the friction reverse direction relative to part (a) and part (b) because the roof expands more than the plate as the temperature rises and less as the temperature falls.

The figure I, applies to the temperature falling and figure II applies to temperature rising. A point on the plate $xL$ from the top of the plate moves upward from lower fixed line by $\Delta L_{\text{plate}}$, and when the temperature drops, the upper fixed line of the plates is carried down the roof by $\Delta L_{\text{roof}}$. So the net change in the plate’s position is $\Delta L_{\text{roof}}-\Delta L_{\text{plate}}$, where $\Delta L_{\text{roof}}>\Delta L_{\text{plate}}$.

The plate creeps down the roof each day by an amount given by,

$\begin{array}{c}\Delta L_{\text{roof}}-\Delta L_{\text{plate}}=\left({\alpha }_1-{\alpha }_2\right)\left(L-L\left(1-\frac{\tan \theta }{{\mu }_{\text{k}}}\right)\right)\left(T_{\text{c}}-T_{\text{h}}\right)\\ =\left({\alpha }_1-{\alpha }_2\right)\left(L-L+\frac{L\tan \theta }{{\mu }_{\text{k}}}\right)\left(T_{\text{c}}-T_{\text{h}}\right)\\ =\frac{L}{{\mu }_{\text{k}}}\left({\alpha }_1-{\alpha }_2\right)\left(T_{\text{c}}-T_{\text{h}}\right)\tan \theta \end{array}$

Conclusion:

Therefore, the plate creeps down the roof each day by an amount given by $\frac{L}{{\mu }_{\text{k}}}\left({\alpha }_1-{\alpha }_2\right)\left(T_{\text{c}}-T_{\text{h}}\right)\tan \theta$.