Comparing Sideward Static Friction Forces on Cars of Different Masses and Speeds

Click For Summary

Homework Help Overview

The discussion revolves around comparing the sideward static friction forces acting on two cars of different masses and speeds as they round curves of varying radii. The problem involves concepts from dynamics and centripetal force.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between mass, speed, and radius in determining the static friction forces. There are attempts to derive expressions for the centripetal forces acting on each car and compare them.

Discussion Status

Participants are actively engaging with the problem, offering insights and corrections to each other's reasoning. Some have suggested comparing centripetal forces, while others are clarifying the calculations involved. The discussion is evolving with various interpretations being explored.

Contextual Notes

There is a mention of the need for participants to show their efforts in the problem-solving process. Some participants express confusion regarding the calculations and the relationships between the variables involved.

courtney1121
Messages
68
Reaction score
0
Car 1 with mass m rounds a curve of radius R traveling at a constant speed v. Car 2 with a mass 2m rounds a curve of radius 2R at a constant speed 2v. How does the magnitude F2 of the sideward static friction force acting on car 2 compare with the magnitude F1 of the sideward static friction force acting on car 1?

a. F1 = 4F2
b. F1 = 2F2
c. F1 = F2
d. F2 = 2F1
e. F2 = 4F1
 
Physics news on Phys.org
In order to get some help, you should show some efforts first.
 
sorry, I had listed my knowns but it didn't show

Anyways

I know that car 1 has half of everything car 2 has, so wouldn't it make sense that F2 = 2F1?
 
Hint: compare the centripetal forces.
 
Ok the equation of centripetal forces is mv^2/r. The force of car one is simply mv^2/r and the centripteal force of car 2 would be 4mv^2/2r which equals 2mv^2/r. So Force of car 2 is twice that of force of car 1 so F1=2F2?
 
courtney1121 said:
Ok the equation of centripetal forces is mv^2/r. The force of car one is simply mv^2/r and the centripteal force of car 2 would be 4mv^2/2r which equals 2mv^2/r. So Force of car 2 is twice that of force of car 1 so F1=2F2?

The centripetal force exerted on the second car is 2m (2v)^2/(2R) = 4m v^2/R.
 
what happened to 2R?
 
courtney1121 said:
what happened to 2R?

[tex]2m \frac{(2v)^2}{2R}= 2m \frac{4v^2}{2R}= 2m \frac{2v^2}{R}= 4m \frac{v^2}{R}[/tex]. I hope you agree on that. :smile:
 
ok ok i see what you did now
 
  • #10
so then that means F2 = 4F1
 
  • #11
Yes, since the sideward frictional force should balance the centripetal force.
 

Similar threads

Replies
1
Views
3K
Replies
57
Views
3K
  • · Replies 42 ·
2
Replies
42
Views
4K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 5 ·
Replies
5
Views
4K
Replies
15
Views
2K
Replies
17
Views
3K
  • · Replies 3 ·
Replies
3
Views
7K
Replies
9
Views
4K
  • · Replies 1 ·
Replies
1
Views
4K