Physics for Scientists and Engineers, Technology Update (No access codes included)
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN: 9781305116399
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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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
Check Mark
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 106N 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 30.0kg . The luggage is placed at the position 7.46m measured horizontally from the axis of rotation. The travel bag goes around once in 38.0s .

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

mac=4π2(30.0kg)(7.46m)(38.0s)2mac=6.118Nmac6.12N

Draw the free body diagram for the luggage.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 6, Problem 6.47AP

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

Fx=macFscosθNsinθ=macN=Fscosθmacsinθ . (2)

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

Fy=mayFssinθ+Ncosθmg=may

Substitute 0m/s2 for ay , Fscosθmacsinθ for N in the above equation.

Fssinθ+(Fscosθmacsinθ)cosθmg=m(0m/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 30.0kg for m , 6.12N for mac , and 9.8m/s2 for g in the equation (3).

Fs=[(6.12N)cot20°+(30.0kg)(9.8m/s2)]sin20°=105.73N106N

Conclusion:

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

(b)

Expert Solution
Check Mark
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 30.0kg . The luggage is placed at the position 7.94m measured horizontally from the axis of rotation. The travel bag goes around once in 34.0s .

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

mac=4π2(30.0kg)7.94m(34.0s)2=8.13N

Calculate the centripetal force for the bag.

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

Fs=[(8.13N)cot20°+(30.0kg)(9.8m/s2)]sin20°=107.352N108N

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

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

N=(108N)cos20.0°8.13Nsin20.0°=272.95N273N

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

μs=FsN

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

μs=108N273N=0.39560.396

Conclusion:

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

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Chapter 6 Solutions

Physics for Scientists and Engineers, Technology Update (No access codes included)

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