Physics for Scientists and Engineers with Modern Physics, Technology Update
Physics for Scientists and Engineers with Modern Physics, Technology Update
9th Edition
ISBN: 9781305401969
Author: SERWAY, Raymond A.; Jewett, John W.
Publisher: Cengage Learning
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Chapter 6, Problem 47AP

(a)

To determine

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

(a)

Expert Solution
Check Mark

Answer to Problem 47AP

The static friction force exerted by the carousel on the bag is 106N_.

Explanation of Solution

Write the expression for the angular speed of the luggage as.

  ω=2πT                                                                                                         (I)

Here, ω is the angular speed of the luggage and T is time period to complete one rotation by the luggage.

Write the expression for the angular speed of the luggage in terms on linear speed as.

  ω=vr                                                                                                          (II)

Here, v is the linear speed of the luggage and r is the radius of rotation.

Substitute vr for ω in equation (I)

  vr=2πT

Simplify the above expression for v as.

  v=2πrT                                                                                                       (III)

Write the expression for centripetal acceleration for luggage as.

  ac=v2r                                                                                                     (IV)

Here, ac is centripetal acceleration for luggage.

If the luggage carousel is steadily rotating about its vertical axis then three forces act on the luggage. The static friction force acts along the inclined metallic surface in an outward direction, normal reaction force exerts by the metallic surface on the luggage in normal to the metallic plane and force due to weight acts downward.

As the luggage is rotating in a horizontal plane so centripetal acceleration will acts in x-direction.

The forces act on the luggage on luggage carousel as shown below.

Physics for Scientists and Engineers with Modern Physics, Technology Update, Chapter 6, Problem 47AP

Write the expression corresponding to Newton’s second law of motion in x-direction as.

  fcos20°nsin20°=mac                                                                          (V)

Substitute v2r for ac in equation (V).

  fcos20°nsin20°=mv2r                                                                        (VI)

Multiply cos20° on both sides in equation (VI)

  fcos220°ncos20°sin20°=mv2rcos20°                                            (VII)

Simplify the equation (VII).

  ncos20°sin20°=fcos220°mv2rcos20°                                            (VIII)

Here, f is static friction force, n is normal reaction force exerted by the metallic surface on the luggage and m is mass of the luggage.

Write the expression corresponding to the condition of equilibrium in y-direction as.

  fsin20°+ncos20°mg=0

Re-arrange the terms.

  fsin20°+ncos20°=mg                                                                          (IX)

Multiply sin20° on both sides in the equation (IX).

  fsin220°+nsin20°cos20°=mgsin20°                                                   (X)

Substitute fcos220°mv2rcos20° for ncos20°sin20° in equation (X).

  fsin220°+fcos220°mv2rcos20°=mgsin20°f(sin220°+cos220°)=mv2rcos20°+mgsin20°

Simplify the above expression for f as.

  f=mv2rcos20°+mgsin20°                                                                     (XI)

Conclusion:

Substitute 7.46m for r and 38.0s for T in equation (III).

  v=2π(7.46m)(38.0s)=1.233m/s1.23m/s

Substitute 30kg for m, 7.46m for r, 1.23m/s for v and 9.80m/s2 for g in equation (XI).

  f=(30kg)(1.23m/s)2(7.46m)cos20°+(30kg)(9.80m/s2)sin20°=5.71N+100.55N=106.271N106N

Thus, the static friction force exerted by the carousel on the bag is 106N_.

(b)

To determine

The coefficient of static friction between the bag and the carousel.

(b)

Expert Solution
Check Mark

Answer to Problem 47AP

The coefficient of static friction between the bag and the carousel is 0.39_.

Explanation of Solution

Multiply equation (VI) by sin20° on the both sides.

  fcos20°sin20°+nsin220°=mv2rsin20°

Simplify the above expression for fcos20°sin20°

  fcos20°sin20°=mv2rsin20°+nsin220°                                              (XII)

Multiply cos20° on both sides in the equation (IX).

  fcos20°sin20°+ncos220°=mgcos20°                                              (XIII)

Substitute mv2rsin20°+nsin220° for fcos20°sin20° in equation (XIII).

  mv2rsin20°+nsin220°+ncos220°=mgcos20°n(sin220°+cos220°)=mgcos20°mv2rsin20°

Simplify the above expression for n as.

  n=mgcos20°mv2rsin20°                                                                 (XIV).

Write the expression for the coefficient of static friction as.

  μs=fn                                                                                                     (XV)

Here, μs is static friction coefficient.

Conclusion:

Substitute 7.94m for r and 34.0s for T in equation (III).

  v=2π(7.94m)(34.0s)=1.4673m/s1.47m/s

Substitute 30kg for m, 7.94m for r, 1.47m/s for v and 9.80m/s2 for g in equation (XIV).

  n=(30kg)(9.80m/s2)cos20°(30kg)(1.47m/s)2(7.94m)sin20°=276.2696N2.792N=273.4N273N

Substitute 273N for n and 106N for n  in equation (XV).

  μs=(106N)273N=0.38820.39

Thus, the coefficient of static friction between the bag and the carousel is 0.39_.

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Chapter 6 Solutions

Physics for Scientists and Engineers with Modern Physics, Technology Update

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