Friction and Newton's Third Law

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Discussion Overview

The discussion revolves around the conceptual understanding of friction in the context of Newton's Third Law, specifically in a scenario involving a car accelerating on a level road. Participants explore the nature of forces acting on the car and the road, including the roles of static friction, normal force, and gravitational force.

Discussion Character

  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions the source of friction when the tires exert a force on the road and the road exerts an equal and opposite force on the tires.
  • Another participant suggests that the friction force is the pair of opposite forces described by Newton's Third Law.
  • A clarification is made that friction consists of two equal and opposite forces acting on the car and the road, consistent with the third law.
  • Participants discuss the normal force as being equal and opposite to weight and perpendicular to the contact force, while also addressing the nature of friction as parallel to the contact surface.

Areas of Agreement / Disagreement

Participants express some agreement on the nature of friction as a pair of forces, but there remains some uncertainty regarding the terminology and conceptual clarity around the forces involved.

Contextual Notes

There are unresolved nuances in the definitions and interpretations of friction, normal force, and their interactions, which may depend on the specific context of the problem.

Zoolog
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Hello,

I have a question from a conceptual problem in my textbook as follows:

A car is moving north and speeding up to pass a truck on a level road. The combined contact force exerted on the road by all four tires has vertical component 11.0 kN downward and horizontal component 3.3 kN southward. The drag force exerted on the car by the air is 1.2 kN southward. What is the net force acting on the car?

The answer given in the textbook has the normal and gravitational forces cancelling out; and the net force being 2.1 kN northward on the car; with static friction exceeding the drag force by that much. My confusion is with the source of the friction. If the tires exert 3.3 kN south on the road, and the road exerts a force of 3.3 kN north on the tires where does the friction come in? Is the friction a force in addition to the equal and opposite force? Does the static friction on the car come from moving with the tires to resist this force? Or is the static friction on the car from the road resisting the tires motion?

Thank you
 
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Zoolog said:
If the tires exert 3.3 kN south on the road, and the road exerts a force of 3.3 kN north on the tires where does the friction come in? Is the friction a force in addition to the equal and opposite force? Does the static friction on the car come from moving with the tires to resist this force? Or is the static friction on the car from the road resisting the tires motion?

The "friction" is the pair of opposite 3.3 kN forces. As you said, one force acts on the road and the other force acts on the tires, i.e. on the car.

The question asks you about fhe net force acting on the car.

The force acting on the road does actually change the motion of the Earth by a tiny amount, but it's so small that you can ignore it.
 
O.K, so if I understand you correctly the friction IS the equal and opposite force from the third law?
 
I'm not sure what you mean by "the equal and opposite force" (singular, not plural).

Suppose you hold a block of wood in each hand and rub them together. You can't really say that the "friction force" acts on one block and some other force acts on the other block. I would say there are two equal and opposite friction forces (consistent with the third law), with one force acting on each block.
 
I think I understand now. So the normal force is the force equal and opposite to weight and perpendicular to the contact force. "Friction" is the pair of forces parallel to the contact surface?
 
Zoolog said:
So the normal force is the force equal and opposite to weight and perpendicular to the contact force.
No, the normal force is a pair of equal and opposite forces on the car and ground, and so is gravity. They are the same magnitude here.
"Friction" is the pair of forces parallel to the contact surface?
Yes.
 
Alright, thank you
 

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