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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Chapter 18, Problem 44CP

Review. A house roof is a perfectly flat plane that makes an angle θ with the horizontal. When its temperature changes, between Ti before dawn each day and Tk in the middle of each afternoon, the roof expands and contracts uniformly with a coefficient of thermal expansion α1. Resting on the roof is a flat, rectangular metal plate with expansion coefficient α2, greater than α1. The length of the plate is L, measured along the slope of the roof. The component of the plate’s weight perpendicular to the roof is supported by a normal force uniformly distributed over the area of the plate. The coefficient of kinetic friction between the plate and the roof is μk. The plate is always at the same temperature as the roof, so we assume its temperature is continuously changing. Because of the difference in expansion coefficients, each bit of the plate is moving relative to the roof below it, except for points along a certain horizontal line running across the plate called the stationary line. If the temperature is rising, parts of the plate below the stationary line are moving down relative to the roof and feel a force of kinetic friction acting up the roof. Elements of area above the stationary line are sliding up the roof, and on them kinetic friction acts downward parallel to the roof. The stationary line occupies no area, so we assume no force of static friction acts on the plate while the temperature is changing. The plate as a whole is very nearly in equilibrium, so the net friction force on it must be equal to the component of its weight acting down the incline. (a) Prove that the stationary line is at a distance of

L 2 ( 1 tan θ μ k )

below the top edge of the plate. (b) Analyze the forces that act on the plate when the temperature is falling and prove that the stationary line is at that same distance above the bottom edge of the plate. (c) Show that the plate steps down the roof like an inchworm, moving each day by the distance

L μ k ( α 2 α 1 ) ( T h T c ) tan θ

(d) Evaluate the distance an aluminum plate moves each day if its length is 1.20 m, the temperature cycles between 4.00°C and 36.0°C, and if the roof has slope 18.5°, coefficient of linear expansion 1.50 × 10−5 (°C) −1, and coefficient of friction 0.420 with the plate. (e) What If? What if the expansion coefficient of the plate is less than that of the roof? Will the plate creep up the roof?

(a)

Expert Solution
Check Mark
To determine

To show: The stationary line is at distance of L2(1tanθμk) .

Answer to Problem 44CP

The distance at which stationary line lie is L2(1tanθμk) .

Explanation of Solution

Given info: The angle made by the roof with the horizontal plane is θ , the initial temperature is Tc , the final temperature is Tc , the coefficient of the thermal expansion is α1 , the expansion coefficient of the metal plate is α2 , the length of the plate is L and the coefficient of kinetic friction between the plate and the roof is μk .

Consider the figure given below.

Physics for Scientists and Engineers, Chapter 18, Problem 44CP , additional homework tip  1

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(1x)cosθ

Here,

m is the mass of the plate.

g is the acceleration due to gravity.

x is the distance.

θ is the angle between plate and the roof.

The force due to gravity is,

Fg=mgsinθ

The equation for the kinematic friction force is,

fk=mgμk(1x)cosθ

The equation for the downward force is,

F=mgμkxcosθ

The force equation for the plate is,

Fx=0mgμkxcosθ+mgμk(1x)cosθmgsinθ=02mgμkxcosθ=mgsinθmgμkcosθ2μkx=μktanθ

Further, solve for x .

2μkx=μktanθx=12tanθ2μk

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

xL=L2Ltanθ2μk=L2(1tanθμk)

Conclusion:

Therefore, the distance at which stationary line lie is L2(1tanθμk) .

(b)

Expert Solution
Check Mark
To determine

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

Answer to Problem 44CP

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 θ , the initial temperature is Tc , the final temperature is Tc , the coefficient of the thermal expansion is α1 , the expansion coefficient of the metal plate is α2 , the length of the plate is L and the coefficient of kinetic friction between the plate and the roof is μk .

Consider the figure given below.

Physics for Scientists and Engineers, Chapter 18, Problem 44CP , additional homework tip  2

Figure 2

With the temperature falling, the plate contracts faster than the roof. The upper part slides down and feels an upward frictional force, mgμk(1x)cosθ . The lower part slides up and feels downward frictional force, mgμkxcosθ .

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

xL=L2(1tanθμk)

Conclusion:

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

(c)

Expert Solution
Check Mark
To determine

To show: The plate steps down the roof like an inchworm moving each day by the distance Lμk(α2α1)(ThTc)tanθ .

Answer to Problem 44CP

The distance by which the plate steps down the roof like an inchworm moving each day is Lμk(α2α1)(ThTc)tanθ .

Explanation of Solution

Given info: The angle made by the roof with the horizontal plane is θ , the initial temperature is Tc , the final temperature is Tc , the coefficient of the thermal expansion is α1 , the expansion coefficient of the metal plate is α2 , the length of the plate is L and the coefficient of kinetic friction between the plate and the roof is μk .

Consider the figure given below.

Physics for Scientists and Engineers, Chapter 18, Problem 44CP , additional homework tip  3

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 (LxLxL) to (LxLxL)(1+α2ΔT) .

The change in the length of the plate is,

ΔLplate=(LxLxL)α2ΔT

The change in the length of the roof is,

ΔLroof=(LxLxL)α1ΔT

The point on the roof originally under point P on the plate slides down the roof relative to the upper fixed line from (LxLxL) to (LxLxL)(1+α1ΔT) relative to the upper fixed line, a change of ΔLroof .

When the temperature drops, point P remains stationary on the roof while the roof contracts, pulling point P back by approximately ΔLroof . Thus relative to the upper fixed line, point P is moved down the roof ΔLplateΔLroof .

The displacement for a day is,

ΔLplateΔLroof=(α2α1)(LxLxL)ΔT=(α2α1)(L2xL)ΔT

Substitute L2(1tanθμk) for xL in the above expression.

ΔLplateΔLroof=(α2α1)(L2(L2)(1tanθμk))ΔT=(α2α1)(LL(1tanθμk))(ThTc)=(α2α1)(LL+Ltanθμk)(ThTc)=Lμk(α2α1)(ThTc)tanθ

Conclusion:

Therefore, the distance by which the plate steps down the roof like an inchworm moving each day is Lμk(α2α1)(ThTc)tanθ .

(d)

Expert Solution
Check Mark
To determine
The distance an aluminum plate moves each day.

Answer to Problem 44CP

The distance an aluminum plate moves each day is 0.275mm .

Explanation of Solution

Given info: The angle made by the roof with the horizontal plane is θ , the initial temperature is Tc , the final temperature is Tc , the coefficient of the thermal expansion is α1 , the expansion coefficient of the metal plate is α2 , the length of the plate is L and the coefficient of kinetic friction between the plate and the roof is μk .

The length of the plate is 1.20m , the initial temperature is 4.00°C , the final temperature is 36.0°C , the angle made by the roof with the horizontal plane is 18.5° , the coefficient of linear expansion is 1.50×105°C1 and the coefficient of kinetic friction is 0.420 .

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

The formula for the displacement for a day is,

ΔLplateΔLroof=Lμk(α2α1)(ThTc)tanθ

Substitute 1.20m for L , 0.420 for μk , 1.50×105°C1 for α1 , 24×106°C1 for α2 , 4.00°C for Tc , 36.0°C for Th and 18.5° for θ in above equation.

ΔLplateΔLroof=(1.20m0.420)(24×106°C11.50×105°C1)(36.0°C4.00°C)tan(18.5°)=2.75×104m=0.275mm

Conclusion:

Therefore, the distance an aluminum plate moves each day is 0.275mm .

(e)

Expert Solution
Check Mark
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 44CP

The plate creeps down the roof each day by an amount given by Lμk(α1α2)(TcTh)tanθ .

Explanation of Solution

Given info: The angle made by the roof with the horizontal plane is θ , the initial temperature is Tc , the final temperature is Tc , the coefficient of the thermal expansion is α1 , the expansion coefficient of the metal plate is α2 , the length of the plate is L and the coefficient of kinetic friction between the plate and the roof is μk .

If α2<α1 , 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 ΔLplate , and when the temperature drops, the upper fixed line of the plates is carried down the roof by ΔLroof . So the net change in the plate’s position is ΔLroofΔLplate , where ΔLroof>ΔLplate .

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

ΔLroofΔLplate=(α1α2)(LL(1tanθμk))(TcTh)=(α1α2)(LL+Ltanθμk)(TcTh)=Lμk(α1α2)(TcTh)tanθ

Conclusion:

Therefore, the plate creeps down the roof each day by an amount given by Lμk(α1α2)(TcTh)tanθ .

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Chapter 18 Solutions

Physics for Scientists and Engineers

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