Just need to clarify something (FRICTION)

  • Thread starter Thread starter asteg123
  • Start date Start date
  • Tags Tags
    Friction
Click For Summary
SUMMARY

The discussion clarifies the calculations for the forces required to move a 60kg package on a level floor with static and kinetic friction coefficients of 0.56 and 0.37, respectively. The maximum static friction force is calculated as 329.28N, indicating that any force greater than this value will initiate motion. To maintain constant velocity, a force of 217.56N is required, derived from the kinetic friction formula. The participant confirms the accuracy of these calculations and seeks validation on the relationship between static and kinetic friction forces.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with friction coefficients
  • Knowledge of basic physics formulas for force calculations
  • Ability to perform calculations involving mass, gravity, and normal force
NEXT STEPS
  • Study the implications of static vs. kinetic friction in real-world applications
  • Learn about the effects of surface materials on friction coefficients
  • Explore advanced topics in dynamics, such as acceleration and deceleration forces
  • Investigate the role of friction in mechanical systems and engineering designs
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the principles of friction and force dynamics in practical scenarios.

asteg123
Messages
12
Reaction score
0
QUESTION: If a 60kg package rests on a level floor with coefficients of static and kinetic frictions 0.56 and 0.37 respectively, what horizontal force is required to start the package's motion? how much force is needed to slide the package across the floor at constant velocity?

using the formulas f(static*max*)=mu * N
where mu(static) = 0.56 and N=m*g = 60*9.8 = 588N

so the maximum static friction would be 329.28N... so the force needed to start the package's motion would be any value GREATER than 329.28N

and that means that the kinetic friction is
f(kinetic) = mu * N
where mu(kinetic) = 0.37
so f(kinetic) = 217.56N
and it takes 217.56N of force to slide the package across the floor at constant velocity..

__________________________________

I was just wondering if what i did was correct... and if the force needed to move an object at a constant velocity less than the force needed to move it from rest...

THX>>>>
 
Physics news on Phys.org
Your answers look good to me :smile: and welcome to PF.
 
They are? CooL... heheheh

Thanks for the welcome... ^^
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
Replies
17
Views
3K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 15 ·
Replies
15
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
18
Views
3K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 42 ·
2
Replies
42
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K