Calculating/Simulating Vehicle Braking Force

Click For Summary
SUMMARY

This discussion focuses on calculating the braking force of a vehicle traveling uphill, specifically using the equation F = mg sin(ang) to determine the gravitational force acting on a 3000 kg vehicle on a 10-degree incline. The calculated tangential gravitational force is 5110.47 N, which translates to a braking torque of 1022.09 Nm when considering a wheel radius of 0.2 m. Participants confirm that static friction must be accounted for, as the tires do not slide, and emphasize that momentum does not need to be considered in this scenario.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Basic knowledge of torque calculations
  • Familiarity with static friction principles
  • Ability to perform trigonometric calculations
NEXT STEPS
  • Research the effects of static friction on vehicle dynamics
  • Explore advanced braking system designs and their torque outputs
  • Learn about the role of momentum in vehicle physics
  • Investigate simulation tools for vehicle dynamics analysis
USEFUL FOR

Automotive engineers, physics students, and anyone involved in vehicle dynamics or braking system design will benefit from this discussion.

Gazzoo
Messages
5
Reaction score
0
Hi All,

I know this type of question has been asked many times but I need help to clarify for myself if I'm doing the right thing. So here goes...

I am trying to simulate/understand the forces on a vehicle as it is traveling on a road. In particular as it is traveling up hills. So if I have my equations correct the primary negative or braking force would be gravity and would be tangential to the inclined plane. This would be calculated using

F = mg sin(ang)

So for an inclined plane/road at 10 deg, and a 3000kg vehicle the tangential gravitational force would be

F = 3000 x 9.81 x sin(10) = 5110.47 N

So is it correct to assume that this force is applied at the wheel contact surface which may have a radius of say 0.2m producing a torque of

T = F x R = 5110.47 * 0.20 = 1022.09 Nm

So in order to mimic this same force/torque using the car brakes they would need to generate a total negative braking torque of the same amount right?

Do I need to bother with speed/momentum in any manner when trying to calculate this braking force as well as the vehicle is traveling up the incline?

Thanks for your help!
 
Physics news on Phys.org
first part is correct, it is mg*sin(ang)

you are talking about a car, a car has tires and tires cause friction
if the car is moving without sliding (I assume this is the case here) you need to consider static friction too, it is static because the contact point of the tire does not move relative to the inclined plane.

the breaks produce a torque that compensates for both of those
and let life stay as simple as it is do not consider any momentum in any of these cases :smile:
 

Similar threads

  • · Replies 19 ·
Replies
19
Views
3K
Replies
4
Views
2K
Replies
3
Views
3K
Replies
3
Views
2K
  • · Replies 12 ·
Replies
12
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 102 ·
4
Replies
102
Views
8K
Replies
14
Views
2K