Interpret results of suspended mass / tension problem please

Click For Summary
SUMMARY

The discussion revolves around a physics problem involving two masses, m1 (45 kg) and m2 (12 kg), connected by a massless wire over a frictionless pulley. The static friction coefficient for m1 is 0.45, and the kinetic friction coefficient is 0.35. The user seeks clarification on interpreting a negative acceleration value, concluding that it indicates the system is at rest, as the gravitational force on m2 does not exceed the maximum static friction force acting on m1. The key takeaway is that the actual force of static friction is less than or equal to the maximum static friction force, which is critical for determining system behavior.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of static and kinetic friction coefficients
  • Ability to construct free body diagrams (FBDs)
  • Familiarity with basic concepts of equilibrium in physics
NEXT STEPS
  • Study the principles of static and kinetic friction in detail
  • Learn how to construct and analyze free body diagrams (FBDs)
  • Explore Newton's second law and its application in equilibrium problems
  • Investigate the effects of different friction coefficients on system dynamics
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of systems involving friction and pulleys, particularly in the context of static and kinetic friction analysis.

Tyrannosaurus_
Messages
41
Reaction score
2

Homework Statement


Two masses: m1 = 45 kg, m2 = 12kg.
m1 (45kg) sits at rest on a table top with static friction co. 0.45, kinetic friction co. 0.35.
m2 is suspended in the air. m2 is attached to m1 by a massless wire hung over a frictionless pulley

Homework Equations


FBDs, net force equations

The Attempt at a Solution


m1 is at rest, so I'll use the μs = 0.45.
I've attached an image of my solution.

My question is how do I interpret the negative acceleration value? I know it's a vector, but it's not as though the simple mass that is sitting on the table can accelerate, and even pull the 12kg mass even further higher into the air. So, does this mean that the system is at rest and there would be no acceleration because Fg is not greater than the Fk?

Many thanks!
Scan_20150122.png
 
Physics news on Phys.org
Tyrannosaurus_ said:
m1 is at rest, so I'll use the μs = 0.45.
That only makes sense as far as determining whether the system will remain at rest. If not, you need to switch to kinetic friction to discover the acceleration.
Your working is hard to read. You should only post images for diagrams and textbook extracts. Please take the trouble to type other working in directly. Not only is it easier to read, it's easier to identify sections of working when posting replies.
Tyrannosaurus_ said:
how do I interpret the negative acceleration value?
That probably arises from assuming the frictional force equals μsN. μsN is merely the maximum possible frictional force. In equilibrium, it will usually be less than that.
 
  • Like
Likes   Reactions: Tyrannosaurus_
It's too difficult to follow your work, but I think I can deduce the error you made from your answer. You treated m1s as if it were the actual force of friction acting. This is not correct; m1s represents the maximum force that static friction is capable of exerting. The actual force of static friction will only be as much it needs to be to keep the block at rest (unless of course m1s is not large enough, in which case the system will accelerate and only µk will be relevant).
 
Thanks for the help! Much appreciated!
 

Similar threads

  • · Replies 24 ·
Replies
24
Views
9K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 33 ·
2
Replies
33
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
6
Views
2K