Friction between table and pool ball

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Homework Help Overview

The discussion revolves around calculating the force of friction and the coefficient of friction between a pool ball and a table, given the weight of the ball and its travel distance and time.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss calculating the normal force and acceleration, with one participant suggesting the use of Newton's second law. There are questions about the validity of calculations and assumptions regarding the normal force.

Discussion Status

Some participants have provided calculations and are seeking verification of their work. There is an ongoing exploration of different interpretations of the problem, particularly regarding the normal force and its implications for the coefficient of friction.

Contextual Notes

There are indications of potential misunderstandings about the relationship between the normal force and the coefficient of friction, as well as the need for clarity on the assumptions made in the calculations.

Drey0287
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Assume the weight of a pool ball is .17kg. If the ball traveled 1.2 m in 5.3 seconds, what is the force of Friction and coefficient of friction between the pool ball and the table?

Now, All I know how to do it calculate the normal force which is (9.8)(.17) = .1666 but where do i get from here?
 
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Drey0287 said:
Assume the weight of a pool ball is .17kg. If the ball traveled 1.2 m in 5.3 seconds, what is the force of Friction and coefficient of friction between the pool ball and the table?

Now, All I know how to do it calculate the normal force which is (9.8)(.17) = .1666 but where do i get from here?

Hints:Use Newton's second law for the rotational motion.(torque is the sum of momenta of all forces).From there u should find µ.The normal force u calculated should be ten times bigger.
[tex]I=\frac{3mr^{2}}{5}[/tex]

Daniel.
 
I find the acceleration to be 0.04272 m/s^2

Therefore, can you not find the force of friction through F=ma?

F(subF)=(.17kg)(0.04272 m/s^2)

mew=F(subF)/F(subN)

mew=0.007262/0.1666

mew=0.0436

check my work, i may be wrong
 
soccerjayl said:
I find the acceleration to be 0.04272 m/s^2

Therefore, can you not find the force of friction through F=ma?

F(subF)=(.17kg)(0.04272 m/s^2)

mew=F(subF)/F(subN)

mew=0.007262/0.1666

mew=0.0436

check my work, i may be wrong

My µ approximately equal to 0,003.

Show your reasoning to depict you mistakes.

Daniel.
 

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