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

AI Thread Summary
The discussion focuses on calculating the force of static friction and the coefficient of static friction for a luggage bag on a rotating carousel. The carousel's slope is 18.0°, and the bag's mass is 30.0 kg, with initial and adjusted positions affecting the calculations. The first part involves determining the static friction force using circular motion equations, while the second part requires finding the coefficient of static friction when the bag is on the verge of slipping. Participants emphasize the importance of showing intermediate steps in calculations and applying Newton's laws correctly. The final coefficient of static friction is estimated around 0.3304, with a request for more precise decimal places.
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.
 
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