The drawing shows a baggage carousel at an airport. Your suitcase has not slid all the way down the slope and is going around at a constant speed on a circle ((r = 12.0 m) as the carousel turns. The coefficient of static friction between the suitcase and the carousel is 0.840, and the angle in the drawing is 12.9°. How much time is required for your suitcase to go around once? Assumme that the static friction between the suitcase and the carousel is at its maximum. Number i Units

Principles of Physics: A Calculus-Based Text
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Author:Raymond A. Serway, John W. Jewett
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Chapter5: More Applications Of Newton’s Laws
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The drawing shows a baggage carousel at an airport. Your suitcase has not slid all the way down the slope and is going around at a
constant speed on a circle ((r = 12.0 m) as the carousel turns. The coefficient of static friction between the suitcase and the carousel is
0.840, and the angle 0 in the drawing is 12.9°. How much time is required for your suitcase to go around once? Assumme that the static
friction between the suitcase and the carousel is at its maximum.
Number
MO
Units
Transcribed Image Text:The drawing shows a baggage carousel at an airport. Your suitcase has not slid all the way down the slope and is going around at a constant speed on a circle ((r = 12.0 m) as the carousel turns. The coefficient of static friction between the suitcase and the carousel is 0.840, and the angle 0 in the drawing is 12.9°. How much time is required for your suitcase to go around once? Assumme that the static friction between the suitcase and the carousel is at its maximum. Number MO Units
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Follow-up Question

Considering that now I have learned to solve for V & T. My next question is how would I solve for the magnitude & direction of the maximum static frictional force acting on the suitcase? I just began my Physics I course, in your opinion would you consider this word problem to be an intro to physics?

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Follow-up Question

I don't doubt that the answer is correct but I'm having trouble following some of these steps. suppose you could somehow elaborate a little more on step two. a video would be awesome if possible...

Step 2
The centripetal force acting on the suitcase will be equal to the force acting on the suitcase along
the horizontal direction,
mv2r-f cose-N sine
The amount of force acting on the suitcase along the vertical direction,
N cose+f sine-mg
...2
Divide the both equation, we will get
mv2rmg-f cose-N sineN cos+f sinev=rgf cose-N sineN cos+f sine
Friction force can be written as,
f=μN
From the above equation of velocity,
v=rgμN cose-N sineN cose+μN sinv=rgμ cose-sin cose+μ sin
Transcribed Image Text:Step 2 The centripetal force acting on the suitcase will be equal to the force acting on the suitcase along the horizontal direction, mv2r-f cose-N sine The amount of force acting on the suitcase along the vertical direction, N cose+f sine-mg ...2 Divide the both equation, we will get mv2rmg-f cose-N sineN cos+f sinev=rgf cose-N sineN cos+f sine Friction force can be written as, f=μN From the above equation of velocity, v=rgμN cose-N sineN cose+μN sinv=rgμ cose-sin cose+μ sin
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Follow-up Question

It seem that in order to solve the above we have to know what is the value of N. What is the value of N??

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