What is the coefficient of static friction between the bag and the carousel?

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SUMMARY

The discussion focuses on calculating the coefficient of static friction between a piece of luggage and a rotating luggage carousel, which is inclined at an angle of 18.0°. The mass of the luggage is 30.0 kg, and it is positioned 7.94 m from the axis of rotation, completing a rotation every 30.8 seconds. The calculations involve applying Newton's laws and circular motion equations, specifically F = (m(v^2))/r. The coefficient of static friction is determined to be approximately 0.3304, with a requirement for four decimal places.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with circular motion equations
  • Basic knowledge of trigonometry, particularly sine functions
  • Ability to perform calculations involving force and mass
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  • Study the derivation of the centripetal force equation F = (mv^2)/r
  • Learn how to draw and analyze free body diagrams in physics
  • Explore the concept of static friction and its calculation methods
  • Practice problems involving inclined planes and circular motion
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SpecialOps0
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Homework Statement


(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 18.0° 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 41.5 s. Calculate the force of static friction 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 30.8 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.



Homework Equations


F = (m(v^2))/r
F=force
m=mass
v=velocity
r=radius

F=(v^2)/r


The Attempt at a Solution



A) mgsin(theta) = 294sin(18.5) + 5 = 96 (not correct but close enough to be considered correct)

B)
Value is within 10% of .39 but wants 4 decimal places.

However I know the following that I've done.

Using laws of circular motion, a=v^2 / r

r is given at 7.94

you can find v:

v = distance/time = 2pi r / 38

So it should be .3304?
 
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We could more easily spot your errors if you showed your intermediate work. For part A, you have mgsintheta =mgsintheta + 5, which doesn't make sense. Show how you are applying Newton's laws.
 
[F – mg sin 18]cos 18 = (mv^2)/r = 30x[(2xpix7.94)/38]^2/7.94
 
SpecialOps0 said:
[F – mg sin 18]cos 18 = (mv^2)/r = 30x[(2xpix7.94)/38]^2/7.94
You are tossing out numbers without showing how you arrived at them, making it difficult to see what you are doing. Go step by step. Look in the y (vertical) direction and sum all forces in the y direction = 0, per Newton 1. Then look in the x direction (the horizontal direction) and sum all forces in the x direction = mv^2/r, per Newton 2. Draw a free body diagram.
 
Got it, thanks and I did have an error in my work. I wrote it down to make sure instead of doing it all on the calculator and forgot to change a degree.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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