Calculating/Simulating Vehicle Braking Force

Click For Summary
The discussion focuses on simulating the forces acting on a vehicle traveling uphill, specifically the braking force due to gravity. The primary negative force is calculated using the equation F = mg sin(ang), with a 3000kg vehicle on a 10-degree incline resulting in a tangential gravitational force of approximately 5110.47 N. This force is applied at the wheel contact surface, leading to a torque calculation of 1022.09 Nm. It is emphasized that static friction must be considered since the tires do not slide, and the braking torque must match the gravitational torque to maintain control. Momentum does not need to be factored into the braking force calculations in this scenario.
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:
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 19 ·
Replies
19
Views
3K
Replies
4
Views
2K
Replies
3
Views
3K
Replies
3
Views
1K
  • · Replies 12 ·
Replies
12
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 102 ·
4
Replies
102
Views
7K
Replies
14
Views
1K