Physics Chapter 2 - Friction

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SungHoonPark97
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Physics Chapter 2 - Friction

Post by SungHoonPark97 » Fri May 25, 2018 10:37 pm

In the textbook, it divides friction force into two components: static and kinetic.

The textbook states that the equation for static friction = (coefficient of static friction)*N(aka applied force)
The textbook states that the equation for kinetic friction = (coefficient of kinetic friction)*N(aka Normal force)

Is this correct? I thought N was normal force for both cases.

In question 3 of chapter 2's independent questions, the explanation states that N is normal force and NOT applied force. So, I was wondering which one is the typo?
NS_Tutor_Katelyn
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Re: Physics Chapter 2 - Friction

Post by NS_Tutor_Katelyn » Thu Jun 07, 2018 2:36 pm

SungHoonPark97 wrote:In the textbook, it divides friction force into two components: static and kinetic.

The textbook states that the equation for static friction = (coefficient of static friction)*N(aka applied force)
The textbook states that the equation for kinetic friction = (coefficient of kinetic friction)*N(aka Normal force)

Is this correct? I thought N was normal force for both cases.

In question 3 of chapter 2's independent questions, the explanation states that N is normal force and NOT applied force. So, I was wondering which one is the typo?
Kinetic friction is the simpler of the two forces: it’s always equal to the coefficient times the normal force.

Static friction is a bit more complicated. The coefficient multiplied by the normal force will give you the maximum possible value of static friction, but the actual value of static friction is always equal to the force that it opposes. So in a scenario where I exert 10 N force on a 10 kg box resting on a flat surface with a coefficient of static friction equal to 0.5, the force of static friction will be equal to 10 N, even though the maximum possible value of static friction would be 50 N.

Let me know if that helps!

-Kate
Katelyn Sawyer
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SungHoonPark97
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Re: Physics Chapter 2 - Friction

Post by SungHoonPark97 » Thu Jun 14, 2018 4:03 pm

Thank you! That helps a lot!
ESTEBAN227
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Re: Physics Chapter 2 - Friction

Post by ESTEBAN227 » Sat Aug 25, 2018 11:43 pm

In your response you stated that for static friction, you multiply the coefficient of static friction with the normal force to find the maximum possible value of static friction. But the actual value of static friction is always equal to the force that it opposes. So in the example you provided, a 10 N force is applied on a 10 kg box resting on a flat surface with a coefficient of static friction equal to 0.5, the force of static friction will be equal to 10 N, does the equation for Fstatic not multiply the coefficient of static friction with the applied force (as stated in the book)? Because then Fstatic would = 5N? or do you just disregard the equation and Fstatic is just equal to the force applied to it?
NS_Tutor_Katelyn
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Re: Physics Chapter 2 - Friction

Post by NS_Tutor_Katelyn » Sun Aug 26, 2018 1:41 pm

ESTEBAN227 wrote:In your response you stated that for static friction, you multiply the coefficient of static friction with the normal force to find the maximum possible value of static friction. But the actual value of static friction is always equal to the force that it opposes. So in the example you provided, a 10 N force is applied on a 10 kg box resting on a flat surface with a coefficient of static friction equal to 0.5, the force of static friction will be equal to 10 N, does the equation for Fstatic not multiply the coefficient of static friction with the applied force (as stated in the book)? Because then Fstatic would = 5N? or do you just disregard the equation and Fstatic is just equal to the force applied to it?
You disregard the equation. Static friction is always equal to the force that it opposes, up to a maximum possible value given by the equation.

-Kate
Katelyn Sawyer
Senior Tutor
katelyn@nextsteptestprep.com
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