Friction problem Does this make sense?

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

The discussion centers on determining the maximum amplitude of oscillation for a system of two stacked masses, M and m, where M is attached to a spring with a force constant k. The coefficient of static friction between the masses is denoted as w. The correct formula for maximum amplitude x is derived as x = (w(m + M)g) / k, which accounts for the total mass being accelerated and the frictional force preventing slipping. The reasoning provided confirms that the acceleration of the top mass must match the acceleration of the bottom mass at maximum oscillation.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with spring mechanics and Hooke's Law
  • Knowledge of static friction and its coefficient
  • Basic concepts of oscillatory motion and amplitude
NEXT STEPS
  • Study the derivation of Hooke's Law and its applications in oscillatory systems
  • Explore the dynamics of friction in multi-body systems
  • Learn about the effects of mass distribution on oscillation amplitude
  • Investigate the role of acceleration in preventing slipping in stacked masses
USEFUL FOR

Students and educators in physics, mechanical engineers, and anyone studying dynamics and oscillatory motion in multi-body systems.

armolinasf
Messages
195
Reaction score
0

Homework Statement



There are two masses stacked on top of each other. The bottom one, M, is attached to a spring with force constant k. The coefficient of static friction between the M and the top mass, m, is w. What is the maximum amplitude of oscillation such that the top box will not slip on the bottom.



The Attempt at a Solution



Initially my reasoning was that the force of friction, wn, where n is the normal force acting on the top box equal to mg, must equal the pulling force, f=kx

wmg=kx then x=(wmg)/k. The answer in my book gives x= (w(m+M)g)/k

I'm not sure if the way I arrived at that answer is correct:

F=kx=(m+M)a ==> a1=kx/(m+M)

wmg=kx ==>wg=kx/m=a2

But a2 is really equal to a1 since that's the box that's what's being accelerated.

wg=kx/(m+M) ==>w(m+M)g/k = x

I just want to be sure that makes sense. Thanks for the help
 
Physics news on Phys.org
Looks ok to me. You need the mass on the top not to slip at the point of maximum acceleration, which is when x is at its maximum value of oscillation.
 

Similar threads

  • · Replies 13 ·
Replies
13
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 12 ·
Replies
12
Views
1K
Replies
2
Views
2K
Replies
20
Views
3K
Replies
16
Views
3K
  • · Replies 18 ·
Replies
18
Views
2K
Replies
6
Views
2K