Help with a car under braking and friction force

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Discussion Overview

The discussion revolves around understanding the forces acting on a car during braking, specifically focusing on the direction of the frictional force when the car is decelerating. Participants explore the relationships between various forces, including the driving force, braking force, friction, and weight, in the context of a force diagram.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions the direction of the frictional force when a car is decelerating, suggesting it acts opposite to the braking force.
  • Another participant emphasizes that friction acts against the motion of the vehicle during deceleration, while the braking force acts against the road.
  • A later reply corrects an earlier statement about weight, clarifying that weight and upthrust are in equilibrium.
  • Another participant notes that inertia, represented as ##ma##, acts opposite to the direction of acceleration during deceleration, and that the sum of forces must equal this inertia.
  • There is a suggestion that the driving force from the engine is irrelevant to the specific problem being discussed.

Areas of Agreement / Disagreement

Participants express differing views on the roles and directions of forces involved in braking. While there is some agreement on the general principles of force interaction, the specifics regarding the relationships and implications of these forces remain contested.

Contextual Notes

Participants acknowledge the need to consider various forces acting on the vehicle, but there are unresolved aspects regarding the definitions and interactions of these forces, particularly in the context of the force diagram.

eazyduzit
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Got a question for you mechanics enthusiasts out there. Here https://i.imgur.com/Px0Hakg.jpg from what I gather is a basic force diagram depicting a car accelerating along a straight road, Dr represents the driving force of the engine, Fr the friction acting on the tyres against the direction of motion, Nr the normal reaction of the car to the road and W the weight of the car. I have to find the coefficient of friction of the tyres to the road and I'm guessing if I find the value for friction I can then use the formula to solve the coefficient value.My question is in what direction would the frictional force be acting on the car if it were decelerating under a braking force, my common sense would tell me it would be opposite to the braking force but I can't be sure.

I'm talking about something that would look like this https://i.imgur.com/DU42tTp.jpg with Br representing the braking force.

Here https://i.imgur.com/nOoNHWf.jpg is a link to the original question text for anyone wondering.
 
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We need to consider all forces as those acting on the vehicle, not the ground. If we neglect aerodynamic forces we have just the upthrust from the ground and friction.
When the car is decelerating, friction applies a force on the vehicle which acts against the motion.
Braking Force is something that is acting, not against the vehicle but against the road, and is equal and opposite to Friction.
Weight is something that is acting, not against the vehicle, but against the road, and is equal and opposite to upthrust.
 
Please disregard my words regarding weight, which is incorrect. The vehicle experiences the attraction of gravity in equilibrium with the upthrust from the ground. Apologies.
 
You are right, the direction of acceleration ##a## is opposite to the direction of motion when in deceleration.

What is missing from your force diagram is the inertia, i.e. ##ma##. The inertia is always opposite the direction of the acceleration ##a##.

Since the sum of the forces in the x-direction (horizontal) must be equal to ##ma## and the only force acting on the car is the braking (friction) force, then it must be equal and opposite to the inertia ##ma##.

The question doesn't mention the driving force coming from the engine and is irrelevant to find the answer of this particular problem.
 

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