Help with a car under braking and friction force

Click For Summary
SUMMARY

The discussion focuses on understanding the forces acting on a car during deceleration, specifically the relationship between braking force and friction. The key takeaway is that when a car is decelerating, the frictional force acts opposite to the direction of motion, countering the braking force. The participants clarify that the inertia (ma) is also directed opposite to the acceleration, reinforcing the equilibrium of forces. The driving force from the engine is deemed irrelevant for solving the coefficient of friction in this context.

PREREQUISITES
  • Basic understanding of Newton's laws of motion
  • Familiarity with force diagrams and free body diagrams
  • Knowledge of the coefficient of friction and its calculation
  • Concept of inertia and its role in motion
NEXT STEPS
  • Study the calculation of the coefficient of friction using the formula: Friction = Coefficient of Friction x Normal Force
  • Explore the dynamics of deceleration and the role of friction in vehicle handling
  • Learn about free body diagrams and how to represent forces acting on a vehicle
  • Investigate the effects of different surfaces on the coefficient of friction in automotive contexts
USEFUL FOR

Automotive engineers, physics students, mechanics, and anyone interested in vehicle dynamics and braking systems.

eazyduzit
Messages
1
Reaction score
0
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.
 
Engineering news on Phys.org
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.
 

Similar threads

Replies
3
Views
2K
  • · Replies 42 ·
2
Replies
42
Views
4K
  • · Replies 14 ·
Replies
14
Views
2K
Replies
10
Views
3K
  • · Replies 20 ·
Replies
20
Views
14K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 15 ·
Replies
15
Views
7K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
5K