Net Force Dispute with Teacher

Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving a 1000 kg automobile that skids to a stop under different pavement conditions. Participants are exploring the calculation of applied forces and coefficients of friction for both dry and wet surfaces, while addressing the implications of frictional forces in the context of net force.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to understand the relationship between applied forces and friction, questioning whether both forces can be considered in calculating coefficients of friction. There is confusion regarding the direction of forces and the definitions of applied and frictional forces.

Discussion Status

The discussion is ongoing, with participants providing insights into the definitions of forces involved. Some guidance has been offered regarding the calculation of acceleration and normal force, but multiple interpretations of the problem are still being explored.

Contextual Notes

Participants are navigating assumptions about the forces acting on the automobile, particularly the roles of friction and applied forces, while also considering the implications of internal versus external forces in the system.

Ninjamonkey474
Messages
2
Reaction score
0
1. A 1000 kg automobile initially moving 72 km/h jams the brakes and skids to a stop in a distance of 24 m on a dry pavement but stops in 81.6 m for a wet pavement. (Neglect reaction delay.)
A. Calculate the applied forces for both weather conditions.
B. What is the coefficient of friction for dry weather?
C. What is the coefficient in wet weather conditions

2. Vfinal2=V02-2ad,
μ=Ffriction/Fnormal,
F=ma

3. I said this was impossible since both friction and applied are pointing in the same direction, you can only calculate the net force and not the applied, without which you cannot find either of the coefficients. She said it was possible but I didn't understand what she meant, can somebody explain to me why I'm wrong or why I'm right?
 
Physics news on Phys.org
Hi Ninjamonkey474! Welcome to PF! :smile:
Ninjamonkey474 said:
A 1000 kg automobile initially moving 72 km/h jams the brakes and skids to a stop in a distance of 24 m on a dry pavement but stops in 81.6 m for a wet pavement.

… both friction and applied are pointing in the same direction …

The friction (from the road) is the applied force. :wink:

(there are no external forces on the car except the weight and the friction from the road)
 
So do I also use the combined brake-road force to calculate the coefficients of friction?
 
Ninjamonkey474 said:
So do I also use the combined brake-road force to calculate the coefficients of friction?
What combined force? The only force you need to worry about is friction from the road.

Or are you thinking that she meant the force that the driver "applied" to the brake? That's not what she meant. And if she did, you would not be able to calculate that. (Note that any force the driver applies to the brake pedal is internal to the system and has no direct affect on the slowing of the car.)
 
When your teacher is talking about the applied forces, what she is talking about is Fnormal and Ffriction. Why don't you start out by first calculating the acceleration rates (or in this example, the deceleration rates) under the two different kinds of road conditions? The next step after that will be to calculate the normal force. The only other force acting on the car besides that is the frictional force exerted by the road on the car, in the direction opposite to the car's motion.

Chet
 

Similar threads

Replies
6
Views
4K
  • · Replies 3 ·
Replies
3
Views
24K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
3
Views
4K
Replies
1
Views
4K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K