(a) A luggage carousel at an airport has the form of a section of a large cone, steadily rotating about its vertical axis. Its metallic surface slopes downward toward the outside, making an angle of 20.0° with the horizontal. A piece of luggage having mass 30.0 kg is placed on the carousel at a position 7.46 m measured horizontally from the axis of rotation. The travel bag goes around once in 38.0 s. Calculate the force of staticfriction exerted by the carousel on the bag. (b) The drive motor is shifted to turn the carousel at a higher constant rate of rotation, and the piece of luggage is bumped to another position, 7.94 m from the axis of rotation. Now going around once in every 34.0 s, the bag is on the verge of slipping down the sloped surface. Calculate the coefficient of static friction between the bag and the carousel.

Question

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Answer 1

Textbook Question

Chapter 6, Problem 6.47AP

(a) A luggage carousel at an airport has the form of a section of a large cone, steadily rotating about its vertical axis. Its metallic surface slopes downward toward the outside, making an angle of 20.0° with the horizontal. A piece of luggage having mass 30.0 kg is placed on the carousel at a position 7.46 m measured horizontally from the axis of rotation. The travel bag goes around once in 38.0 s. Calculate the force of staticfriction exerted by the carousel on the bag. (b) The drive motor is shifted to turn the carousel at a higher constant rate of rotation, and the piece of luggage is bumped to another position, 7.94 m from the axis of rotation. Now going around once in every 34.0 s, the bag is on the verge of slipping down the sloped surface. Calculate the coefficient of static friction between the bag and the carousel.

(a)

Expert Solution

To determine

The force of static friction exerted by the carousel on the bag.

Answer to Problem 6.47AP

The force of static friction exerted by the carousel on the bag is $106 N$ up the incline.

Explanation of Solution

Given info: The metallic surface of a luggage carousel slopes downward toward outside at an angle $20.0°$ with the horizontal. The mass of luggage is $30.0 kg$ . The luggage is placed at the position $7.46 m$ measured horizontally from the axis of rotation. The travel bag goes around once in $38.0 s$ .

Formula to calculate the centripetal force is,

$Fc=mv2R$

The horizontal force acting on the bag in horizontal direction is,

$mac=mv2R$

Substitute $2πRt$ for $v$ in the above equation.

$mac=m(2πRt)2Rmac=4π2mR2Rt2mac=4π2mRt2$ . (1)

Here,

$Fc$ is the centripetal acceleration.

$R$ is the radius of the carousel.

$m$ is the mass of the luggage.

$t$ is the time.

Substitute $30.0 kg$ for $m$ , $7.46 m$ for $R$ , and $38.0 s$ for $t$ in the above equation.

$mac=4π2(30.0 kg)(7.46 m)(38.0 s)2mac=6.118 Nmac≈6.12 N$

Draw the free body diagram for the luggage.

Physics for Scientists and Engineers, Volume 1, Chapters 1-22, Chapter 6, Problem 6.47AP

From the Figure (1), the component of force in $x$ direction is,

$∑Fx=macFscosθ−Nsinθ=macN=Fscosθ−macsinθ$ . (2)

From the Figure (1), the component of force in $y$ direction is,

$∑Fy=mayFssinθ+Ncosθ−mg=may$

Substitute $0 m/s2$ for $ay$ , $Fscosθ−macsinθ$ for $N$ in the above equation.

$Fssinθ+(Fscosθ−macsinθ)cosθ−mg=m(0 m/s2)Fssinθ+(Fscos2θsinθ)−mac(cosθsinθ)−mg=0Fs[sinθ+(cos2θsinθ)]=mac(cotθ)+mgFs[(1sinθ)]=mac(cotθ)+mg$

Simplify the above equation.

$Fs=[mac(cotθ)−mg]sinθ$ (3)

Substitute $30.0 kg$ for $m$ , $6.12 N$ for $mac$ , and $9.8 m/s2$ for $g$ in the equation (3).

$Fs=[(6.12 N)cot20°+(30.0 kg)(9.8 m/s2)]sin20°=105.73 N≈106 N$

Conclusion:

Therefore, the force of static friction exerted by the carousel on the bag is $106 N$ up the incline.

(b)

Expert Solution

To determine

The coefficient of friction between the bag and the carousel.

Answer to Problem 6.47AP

The coefficient of friction between the bag and the carousel is $0.396$ .

Explanation of Solution

Given info: The metallic surface of a luggage carousel slopes downward toward outside at an angle $20.0°$ with the horizontal. The mass of luggage is $30.0 kg$ . The luggage is placed at the position $7.94 m$ measured horizontally from the axis of rotation. The travel bag goes around once in $34.0 s$ .

Substitute $30.0 kg$ for $m$ , $7.94 m$ for $R$ , and $34.0 s$ for $t$ in the equation (1).

$mac=4π2(30.0 kg)7.94 m(34.0 s)2=8.13 N$

Calculate the centripetal force for the bag.

Substitute $30.0 kg$ for $m$ , $8.13 N$ for $mac$ , and $9.8 m/s2$ for $g$ in the equation (3).

$Fs=[(8.13 N)cot20°+(30.0 kg)(9.8 m/s2)]sin20°=107.352 N≈108 N$

Calculate the normal force for the bag when it is on the carousel.

Substitute $108 N$ for $Fs$ , $8.13 N$ for $mac$ and $20.0°$ in the equation (2).

$N=(108 N)cos20.0°−8.13 Nsin20.0°=272.95 N≈273 N$

Formula to calculate the coefficient of friction between the bag and the carousel is,

$μs=FsN$

Substitute $108 N$ for $Fs$ , $273 N$ for $N$ in the above equation.

$μs=108 N273 N=0.3956≈0.396$

Conclusion:

Therefore, the coefficient of friction between the bag and the carousel is $0.396$ .

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Answer 2

Textbook Question

Chapter 6, Problem 6.47AP

(a) A luggage carousel at an airport has the form of a section of a large cone, steadily rotating about its vertical axis. Its metallic surface slopes downward toward the outside, making an angle of 20.0° with the horizontal. A piece of luggage having mass 30.0 kg is placed on the carousel at a position 7.46 m measured horizontally from the axis of rotation. The travel bag goes around once in 38.0 s. Calculate the force of staticfriction exerted by the carousel on the bag. (b) The drive motor is shifted to turn the carousel at a higher constant rate of rotation, and the piece of luggage is bumped to another position, 7.94 m from the axis of rotation. Now going around once in every 34.0 s, the bag is on the verge of slipping down the sloped surface. Calculate the coefficient of static friction between the bag and the carousel.

(a)

Expert Solution

To determine

The force of static friction exerted by the carousel on the bag.

Answer to Problem 6.47AP

The force of static friction exerted by the carousel on the bag is $106 N$ up the incline.

Explanation of Solution

Given info: The metallic surface of a luggage carousel slopes downward toward outside at an angle $20.0°$ with the horizontal. The mass of luggage is $30.0 kg$ . The luggage is placed at the position $7.46 m$ measured horizontally from the axis of rotation. The travel bag goes around once in $38.0 s$ .

Formula to calculate the centripetal force is,

$Fc=mv2R$

The horizontal force acting on the bag in horizontal direction is,

$mac=mv2R$

Substitute $2πRt$ for $v$ in the above equation.

$mac=m(2πRt)2Rmac=4π2mR2Rt2mac=4π2mRt2$ . (1)

Here,

$Fc$ is the centripetal acceleration.

$R$ is the radius of the carousel.

$m$ is the mass of the luggage.

$t$ is the time.

Substitute $30.0 kg$ for $m$ , $7.46 m$ for $R$ , and $38.0 s$ for $t$ in the above equation.

$mac=4π2(30.0 kg)(7.46 m)(38.0 s)2mac=6.118 Nmac≈6.12 N$

Draw the free body diagram for the luggage.

Physics for Scientists and Engineers, Volume 1, Chapters 1-22, Chapter 6, Problem 6.47AP

From the Figure (1), the component of force in $x$ direction is,

$∑Fx=macFscosθ−Nsinθ=macN=Fscosθ−macsinθ$ . (2)

From the Figure (1), the component of force in $y$ direction is,

$∑Fy=mayFssinθ+Ncosθ−mg=may$

Substitute $0 m/s2$ for $ay$ , $Fscosθ−macsinθ$ for $N$ in the above equation.

$Fssinθ+(Fscosθ−macsinθ)cosθ−mg=m(0 m/s2)Fssinθ+(Fscos2θsinθ)−mac(cosθsinθ)−mg=0Fs[sinθ+(cos2θsinθ)]=mac(cotθ)+mgFs[(1sinθ)]=mac(cotθ)+mg$

Simplify the above equation.

$Fs=[mac(cotθ)−mg]sinθ$ (3)

Substitute $30.0 kg$ for $m$ , $6.12 N$ for $mac$ , and $9.8 m/s2$ for $g$ in the equation (3).

$Fs=[(6.12 N)cot20°+(30.0 kg)(9.8 m/s2)]sin20°=105.73 N≈106 N$

Conclusion:

Therefore, the force of static friction exerted by the carousel on the bag is $106 N$ up the incline.

(b)

Expert Solution

To determine

The coefficient of friction between the bag and the carousel.

Answer to Problem 6.47AP

The coefficient of friction between the bag and the carousel is $0.396$ .

Explanation of Solution

Given info: The metallic surface of a luggage carousel slopes downward toward outside at an angle $20.0°$ with the horizontal. The mass of luggage is $30.0 kg$ . The luggage is placed at the position $7.94 m$ measured horizontally from the axis of rotation. The travel bag goes around once in $34.0 s$ .

Substitute $30.0 kg$ for $m$ , $7.94 m$ for $R$ , and $34.0 s$ for $t$ in the equation (1).

$mac=4π2(30.0 kg)7.94 m(34.0 s)2=8.13 N$

Calculate the centripetal force for the bag.

Substitute $30.0 kg$ for $m$ , $8.13 N$ for $mac$ , and $9.8 m/s2$ for $g$ in the equation (3).

$Fs=[(8.13 N)cot20°+(30.0 kg)(9.8 m/s2)]sin20°=107.352 N≈108 N$

Calculate the normal force for the bag when it is on the carousel.

Substitute $108 N$ for $Fs$ , $8.13 N$ for $mac$ and $20.0°$ in the equation (2).

$N=(108 N)cos20.0°−8.13 Nsin20.0°=272.95 N≈273 N$

Formula to calculate the coefficient of friction between the bag and the carousel is,

$μs=FsN$

Substitute $108 N$ for $Fs$ , $273 N$ for $N$ in the above equation.

$μs=108 N273 N=0.3956≈0.396$

Conclusion:

Therefore, the coefficient of friction between the bag and the carousel is $0.396$ .

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Answer 3

Textbook Question

Chapter 6, Problem 6.47AP

(a) A luggage carousel at an airport has the form of a section of a large cone, steadily rotating about its vertical axis. Its metallic surface slopes downward toward the outside, making an angle of 20.0° with the horizontal. A piece of luggage having mass 30.0 kg is placed on the carousel at a position 7.46 m measured horizontally from the axis of rotation. The travel bag goes around once in 38.0 s. Calculate the force of staticfriction exerted by the carousel on the bag. (b) The drive motor is shifted to turn the carousel at a higher constant rate of rotation, and the piece of luggage is bumped to another position, 7.94 m from the axis of rotation. Now going around once in every 34.0 s, the bag is on the verge of slipping down the sloped surface. Calculate the coefficient of static friction between the bag and the carousel.

(a)

Expert Solution

To determine

The force of static friction exerted by the carousel on the bag.

Answer to Problem 6.47AP

The force of static friction exerted by the carousel on the bag is $106 N$ up the incline.

Explanation of Solution

Given info: The metallic surface of a luggage carousel slopes downward toward outside at an angle $20.0°$ with the horizontal. The mass of luggage is $30.0 kg$ . The luggage is placed at the position $7.46 m$ measured horizontally from the axis of rotation. The travel bag goes around once in $38.0 s$ .

Formula to calculate the centripetal force is,

$Fc=mv2R$

The horizontal force acting on the bag in horizontal direction is,

$mac=mv2R$

Substitute $2πRt$ for $v$ in the above equation.

$mac=m(2πRt)2Rmac=4π2mR2Rt2mac=4π2mRt2$ . (1)

Here,

$Fc$ is the centripetal acceleration.

$R$ is the radius of the carousel.

$m$ is the mass of the luggage.

$t$ is the time.

Substitute $30.0 kg$ for $m$ , $7.46 m$ for $R$ , and $38.0 s$ for $t$ in the above equation.

$mac=4π2(30.0 kg)(7.46 m)(38.0 s)2mac=6.118 Nmac≈6.12 N$

Draw the free body diagram for the luggage.

Physics for Scientists and Engineers, Volume 1, Chapters 1-22, Chapter 6, Problem 6.47AP

From the Figure (1), the component of force in $x$ direction is,

$∑Fx=macFscosθ−Nsinθ=macN=Fscosθ−macsinθ$ . (2)

From the Figure (1), the component of force in $y$ direction is,

$∑Fy=mayFssinθ+Ncosθ−mg=may$

Substitute $0 m/s2$ for $ay$ , $Fscosθ−macsinθ$ for $N$ in the above equation.

$Fssinθ+(Fscosθ−macsinθ)cosθ−mg=m(0 m/s2)Fssinθ+(Fscos2θsinθ)−mac(cosθsinθ)−mg=0Fs[sinθ+(cos2θsinθ)]=mac(cotθ)+mgFs[(1sinθ)]=mac(cotθ)+mg$

Simplify the above equation.

$Fs=[mac(cotθ)−mg]sinθ$ (3)

Substitute $30.0 kg$ for $m$ , $6.12 N$ for $mac$ , and $9.8 m/s2$ for $g$ in the equation (3).

$Fs=[(6.12 N)cot20°+(30.0 kg)(9.8 m/s2)]sin20°=105.73 N≈106 N$

Conclusion:

Therefore, the force of static friction exerted by the carousel on the bag is $106 N$ up the incline.

(b)

Expert Solution

To determine

The coefficient of friction between the bag and the carousel.

Answer to Problem 6.47AP

The coefficient of friction between the bag and the carousel is $0.396$ .

Explanation of Solution

Given info: The metallic surface of a luggage carousel slopes downward toward outside at an angle $20.0°$ with the horizontal. The mass of luggage is $30.0 kg$ . The luggage is placed at the position $7.94 m$ measured horizontally from the axis of rotation. The travel bag goes around once in $34.0 s$ .

Substitute $30.0 kg$ for $m$ , $7.94 m$ for $R$ , and $34.0 s$ for $t$ in the equation (1).

$mac=4π2(30.0 kg)7.94 m(34.0 s)2=8.13 N$

Calculate the centripetal force for the bag.

Substitute $30.0 kg$ for $m$ , $8.13 N$ for $mac$ , and $9.8 m/s2$ for $g$ in the equation (3).

$Fs=[(8.13 N)cot20°+(30.0 kg)(9.8 m/s2)]sin20°=107.352 N≈108 N$

Calculate the normal force for the bag when it is on the carousel.

Substitute $108 N$ for $Fs$ , $8.13 N$ for $mac$ and $20.0°$ in the equation (2).

$N=(108 N)cos20.0°−8.13 Nsin20.0°=272.95 N≈273 N$

Formula to calculate the coefficient of friction between the bag and the carousel is,

$μs=FsN$

Substitute $108 N$ for $Fs$ , $273 N$ for $N$ in the above equation.

$μs=108 N273 N=0.3956≈0.396$

Conclusion:

Therefore, the coefficient of friction between the bag and the carousel is $0.396$ .

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Answer 4

Textbook Question

Chapter 6, Problem 6.47AP

(a) A luggage carousel at an airport has the form of a section of a large cone, steadily rotating about its vertical axis. Its metallic surface slopes downward toward the outside, making an angle of 20.0° with the horizontal. A piece of luggage having mass 30.0 kg is placed on the carousel at a position 7.46 m measured horizontally from the axis of rotation. The travel bag goes around once in 38.0 s. Calculate the force of staticfriction exerted by the carousel on the bag. (b) The drive motor is shifted to turn the carousel at a higher constant rate of rotation, and the piece of luggage is bumped to another position, 7.94 m from the axis of rotation. Now going around once in every 34.0 s, the bag is on the verge of slipping down the sloped surface. Calculate the coefficient of static friction between the bag and the carousel.

(a)

Expert Solution

To determine

The force of static friction exerted by the carousel on the bag.

Answer to Problem 6.47AP

The force of static friction exerted by the carousel on the bag is $106 N$ up the incline.

Explanation of Solution

Given info: The metallic surface of a luggage carousel slopes downward toward outside at an angle $20.0°$ with the horizontal. The mass of luggage is $30.0 kg$ . The luggage is placed at the position $7.46 m$ measured horizontally from the axis of rotation. The travel bag goes around once in $38.0 s$ .

Formula to calculate the centripetal force is,

$Fc=mv2R$

The horizontal force acting on the bag in horizontal direction is,

$mac=mv2R$

Substitute $2πRt$ for $v$ in the above equation.

$mac=m(2πRt)2Rmac=4π2mR2Rt2mac=4π2mRt2$ . (1)

Here,

$Fc$ is the centripetal acceleration.

$R$ is the radius of the carousel.

$m$ is the mass of the luggage.

$t$ is the time.

Substitute $30.0 kg$ for $m$ , $7.46 m$ for $R$ , and $38.0 s$ for $t$ in the above equation.

$mac=4π2(30.0 kg)(7.46 m)(38.0 s)2mac=6.118 Nmac≈6.12 N$

Draw the free body diagram for the luggage.

Physics for Scientists and Engineers, Volume 1, Chapters 1-22, Chapter 6, Problem 6.47AP

From the Figure (1), the component of force in $x$ direction is,

$∑Fx=macFscosθ−Nsinθ=macN=Fscosθ−macsinθ$ . (2)

From the Figure (1), the component of force in $y$ direction is,

$∑Fy=mayFssinθ+Ncosθ−mg=may$

Substitute $0 m/s2$ for $ay$ , $Fscosθ−macsinθ$ for $N$ in the above equation.

$Fssinθ+(Fscosθ−macsinθ)cosθ−mg=m(0 m/s2)Fssinθ+(Fscos2θsinθ)−mac(cosθsinθ)−mg=0Fs[sinθ+(cos2θsinθ)]=mac(cotθ)+mgFs[(1sinθ)]=mac(cotθ)+mg$

Simplify the above equation.

$Fs=[mac(cotθ)−mg]sinθ$ (3)

Substitute $30.0 kg$ for $m$ , $6.12 N$ for $mac$ , and $9.8 m/s2$ for $g$ in the equation (3).

$Fs=[(6.12 N)cot20°+(30.0 kg)(9.8 m/s2)]sin20°=105.73 N≈106 N$

Conclusion:

Therefore, the force of static friction exerted by the carousel on the bag is $106 N$ up the incline.

(b)

Expert Solution

To determine

The coefficient of friction between the bag and the carousel.

Answer to Problem 6.47AP

The coefficient of friction between the bag and the carousel is $0.396$ .

Explanation of Solution

Given info: The metallic surface of a luggage carousel slopes downward toward outside at an angle $20.0°$ with the horizontal. The mass of luggage is $30.0 kg$ . The luggage is placed at the position $7.94 m$ measured horizontally from the axis of rotation. The travel bag goes around once in $34.0 s$ .

Substitute $30.0 kg$ for $m$ , $7.94 m$ for $R$ , and $34.0 s$ for $t$ in the equation (1).

$mac=4π2(30.0 kg)7.94 m(34.0 s)2=8.13 N$

Calculate the centripetal force for the bag.

Substitute $30.0 kg$ for $m$ , $8.13 N$ for $mac$ , and $9.8 m/s2$ for $g$ in the equation (3).

$Fs=[(8.13 N)cot20°+(30.0 kg)(9.8 m/s2)]sin20°=107.352 N≈108 N$

Calculate the normal force for the bag when it is on the carousel.

Substitute $108 N$ for $Fs$ , $8.13 N$ for $mac$ and $20.0°$ in the equation (2).

$N=(108 N)cos20.0°−8.13 Nsin20.0°=272.95 N≈273 N$

Formula to calculate the coefficient of friction between the bag and the carousel is,

$μs=FsN$

Substitute $108 N$ for $Fs$ , $273 N$ for $N$ in the above equation.

$μs=108 N273 N=0.3956≈0.396$

Conclusion:

Therefore, the coefficient of friction between the bag and the carousel is $0.396$ .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

The force of static friction exerted by the carousel on the bag.

Answer to Problem 6.47AP

The force of static friction exerted by the carousel on the bag is $106 N$ up the incline.

Explanation of Solution

Given info: The metallic surface of a luggage carousel slopes downward toward outside at an angle $20.0°$ with the horizontal. The mass of luggage is $30.0 kg$ . The luggage is placed at the position $7.46 m$ measured horizontally from the axis of rotation. The travel bag goes around once in $38.0 s$ .

Formula to calculate the centripetal force is,

$Fc=mv2R$

The horizontal force acting on the bag in horizontal direction is,

$mac=mv2R$

Substitute $2πRt$ for $v$ in the above equation.

$mac=m(2πRt)2Rmac=4π2mR2Rt2mac=4π2mRt2$ . (1)

Here,

$Fc$ is the centripetal acceleration.

$R$ is the radius of the carousel.

$m$ is the mass of the luggage.

$t$ is the time.

Substitute $30.0 kg$ for $m$ , $7.46 m$ for $R$ , and $38.0 s$ for $t$ in the above equation.

$mac=4π2(30.0 kg)(7.46 m)(38.0 s)2mac=6.118 Nmac≈6.12 N$

Draw the free body diagram for the luggage.

Physics for Scientists and Engineers, Volume 1, Chapters 1-22, Chapter 6, Problem 6.47AP

From the Figure (1), the component of force in $x$ direction is,

$∑Fx=macFscosθ−Nsinθ=macN=Fscosθ−macsinθ$ . (2)

From the Figure (1), the component of force in $y$ direction is,

$∑Fy=mayFssinθ+Ncosθ−mg=may$

Substitute $0 m/s2$ for $ay$ , $Fscosθ−macsinθ$ for $N$ in the above equation.

$Fssinθ+(Fscosθ−macsinθ)cosθ−mg=m(0 m/s2)Fssinθ+(Fscos2θsinθ)−mac(cosθsinθ)−mg=0Fs[sinθ+(cos2θsinθ)]=mac(cotθ)+mgFs[(1sinθ)]=mac(cotθ)+mg$

Simplify the above equation.

$Fs=[mac(cotθ)−mg]sinθ$ (3)

Substitute $30.0 kg$ for $m$ , $6.12 N$ for $mac$ , and $9.8 m/s2$ for $g$ in the equation (3).

$Fs=[(6.12 N)cot20°+(30.0 kg)(9.8 m/s2)]sin20°=105.73 N≈106 N$

Conclusion:

Therefore, the force of static friction exerted by the carousel on the bag is $106 N$ up the incline.

(b)

Expert Solution

To determine

The coefficient of friction between the bag and the carousel.

Answer to Problem 6.47AP

The coefficient of friction between the bag and the carousel is $0.396$ .

Explanation of Solution

Given info: The metallic surface of a luggage carousel slopes downward toward outside at an angle $20.0°$ with the horizontal. The mass of luggage is $30.0 kg$ . The luggage is placed at the position $7.94 m$ measured horizontally from the axis of rotation. The travel bag goes around once in $34.0 s$ .

Substitute $30.0 kg$ for $m$ , $7.94 m$ for $R$ , and $34.0 s$ for $t$ in the equation (1).

$mac=4π2(30.0 kg)7.94 m(34.0 s)2=8.13 N$

Calculate the centripetal force for the bag.

Substitute $30.0 kg$ for $m$ , $8.13 N$ for $mac$ , and $9.8 m/s2$ for $g$ in the equation (3).

$Fs=[(8.13 N)cot20°+(30.0 kg)(9.8 m/s2)]sin20°=107.352 N≈108 N$

Calculate the normal force for the bag when it is on the carousel.

Substitute $108 N$ for $Fs$ , $8.13 N$ for $mac$ and $20.0°$ in the equation (2).

$N=(108 N)cos20.0°−8.13 Nsin20.0°=272.95 N≈273 N$

Formula to calculate the coefficient of friction between the bag and the carousel is,

$μs=FsN$

Substitute $108 N$ for $Fs$ , $273 N$ for $N$ in the above equation.

$μs=108 N273 N=0.3956≈0.396$

Conclusion:

Therefore, the coefficient of friction between the bag and the carousel is $0.396$ .

Answer 8

(a)

Expert Solution

To determine

The force of static friction exerted by the carousel on the bag.

Answer to Problem 6.47AP

The force of static friction exerted by the carousel on the bag is $106 N$ up the incline.

Explanation of Solution

Given info: The metallic surface of a luggage carousel slopes downward toward outside at an angle $20.0°$ with the horizontal. The mass of luggage is $30.0 kg$ . The luggage is placed at the position $7.46 m$ measured horizontally from the axis of rotation. The travel bag goes around once in $38.0 s$ .

Formula to calculate the centripetal force is,

$Fc=mv2R$

The horizontal force acting on the bag in horizontal direction is,

$mac=mv2R$

Substitute $2πRt$ for $v$ in the above equation.

$mac=m(2πRt)2Rmac=4π2mR2Rt2mac=4π2mRt2$ . (1)

Here,

$Fc$ is the centripetal acceleration.

$R$ is the radius of the carousel.

$m$ is the mass of the luggage.

$t$ is the time.

Substitute $30.0 kg$ for $m$ , $7.46 m$ for $R$ , and $38.0 s$ for $t$ in the above equation.

$mac=4π2(30.0 kg)(7.46 m)(38.0 s)2mac=6.118 Nmac≈6.12 N$

Draw the free body diagram for the luggage.

Physics for Scientists and Engineers, Volume 1, Chapters 1-22, Chapter 6, Problem 6.47AP

From the Figure (1), the component of force in $x$ direction is,

$∑Fx=macFscosθ−Nsinθ=macN=Fscosθ−macsinθ$ . (2)

From the Figure (1), the component of force in $y$ direction is,

$∑Fy=mayFssinθ+Ncosθ−mg=may$

Substitute $0 m/s2$ for $ay$ , $Fscosθ−macsinθ$ for $N$ in the above equation.

$Fssinθ+(Fscosθ−macsinθ)cosθ−mg=m(0 m/s2)Fssinθ+(Fscos2θsinθ)−mac(cosθsinθ)−mg=0Fs[sinθ+(cos2θsinθ)]=mac(cotθ)+mgFs[(1sinθ)]=mac(cotθ)+mg$

Simplify the above equation.

$Fs=[mac(cotθ)−mg]sinθ$ (3)

Substitute $30.0 kg$ for $m$ , $6.12 N$ for $mac$ , and $9.8 m/s2$ for $g$ in the equation (3).

$Fs=[(6.12 N)cot20°+(30.0 kg)(9.8 m/s2)]sin20°=105.73 N≈106 N$

Conclusion:

Therefore, the force of static friction exerted by the carousel on the bag is $106 N$ up the incline.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

The force of static friction exerted by the carousel on the bag.

Answer 13

Answer to Problem 6.47AP

The force of static friction exerted by the carousel on the bag is $106 N$ up the incline.

Answer 14

Explanation of Solution

Given info: The metallic surface of a luggage carousel slopes downward toward outside at an angle $20.0°$ with the horizontal. The mass of luggage is $30.0 kg$ . The luggage is placed at the position $7.46 m$ measured horizontally from the axis of rotation. The travel bag goes around once in $38.0 s$ .

Formula to calculate the centripetal force is,

$Fc=mv2R$

The horizontal force acting on the bag in horizontal direction is,

$mac=mv2R$

Substitute $2πRt$ for $v$ in the above equation.

$mac=m(2πRt)2Rmac=4π2mR2Rt2mac=4π2mRt2$ . (1)

Here,

$Fc$ is the centripetal acceleration.

$R$ is the radius of the carousel.

$m$ is the mass of the luggage.

$t$ is the time.

Substitute $30.0 kg$ for $m$ , $7.46 m$ for $R$ , and $38.0 s$ for $t$ in the above equation.

$mac=4π2(30.0 kg)(7.46 m)(38.0 s)2mac=6.118 Nmac≈6.12 N$

Draw the free body diagram for the luggage.

Physics for Scientists and Engineers, Volume 1, Chapters 1-22, Chapter 6, Problem 6.47AP

From the Figure (1), the component of force in $x$ direction is,

$∑Fx=macFscosθ−Nsinθ=macN=Fscosθ−macsinθ$ . (2)

From the Figure (1), the component of force in $y$ direction is,

$∑Fy=mayFssinθ+Ncosθ−mg=may$

Substitute $0 m/s2$ for $ay$ , $Fscosθ−macsinθ$ for $N$ in the above equation.

$Fssinθ+(Fscosθ−macsinθ)cosθ−mg=m(0 m/s2)Fssinθ+(Fscos2θsinθ)−mac(cosθsinθ)−mg=0Fs[sinθ+(cos2θsinθ)]=mac(cotθ)+mgFs[(1sinθ)]=mac(cotθ)+mg$

Simplify the above equation.

$Fs=[mac(cotθ)−mg]sinθ$ (3)

Substitute $30.0 kg$ for $m$ , $6.12 N$ for $mac$ , and $9.8 m/s2$ for $g$ in the equation (3).

$Fs=[(6.12 N)cot20°+(30.0 kg)(9.8 m/s2)]sin20°=105.73 N≈106 N$

Conclusion:

Therefore, the force of static friction exerted by the carousel on the bag is $106 N$ up the incline.

Answer 15

(b)

Expert Solution

To determine

The coefficient of friction between the bag and the carousel.

Answer to Problem 6.47AP

The coefficient of friction between the bag and the carousel is $0.396$ .

Explanation of Solution

Given info: The metallic surface of a luggage carousel slopes downward toward outside at an angle $20.0°$ with the horizontal. The mass of luggage is $30.0 kg$ . The luggage is placed at the position $7.94 m$ measured horizontally from the axis of rotation. The travel bag goes around once in $34.0 s$ .

Substitute $30.0 kg$ for $m$ , $7.94 m$ for $R$ , and $34.0 s$ for $t$ in the equation (1).

$mac=4π2(30.0 kg)7.94 m(34.0 s)2=8.13 N$

Calculate the centripetal force for the bag.

Substitute $30.0 kg$ for $m$ , $8.13 N$ for $mac$ , and $9.8 m/s2$ for $g$ in the equation (3).

$Fs=[(8.13 N)cot20°+(30.0 kg)(9.8 m/s2)]sin20°=107.352 N≈108 N$

Calculate the normal force for the bag when it is on the carousel.

Substitute $108 N$ for $Fs$ , $8.13 N$ for $mac$ and $20.0°$ in the equation (2).

$N=(108 N)cos20.0°−8.13 Nsin20.0°=272.95 N≈273 N$

Formula to calculate the coefficient of friction between the bag and the carousel is,

$μs=FsN$

Substitute $108 N$ for $Fs$ , $273 N$ for $N$ in the above equation.

$μs=108 N273 N=0.3956≈0.396$

Conclusion:

Therefore, the coefficient of friction between the bag and the carousel is $0.396$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

The coefficient of friction between the bag and the carousel.

Answer 20

Answer to Problem 6.47AP

The coefficient of friction between the bag and the carousel is $0.396$ .

Answer 21

Explanation of Solution

Given info: The metallic surface of a luggage carousel slopes downward toward outside at an angle $20.0°$ with the horizontal. The mass of luggage is $30.0 kg$ . The luggage is placed at the position $7.94 m$ measured horizontally from the axis of rotation. The travel bag goes around once in $34.0 s$ .