How Does Friction Affect the Acceleration in a Pulley System?

  • Thread starter Thread starter algar32
  • Start date Start date
  • Tags Tags
    Friction Pulley
AI Thread Summary
Friction significantly impacts the acceleration in a pulley system, particularly when considering the rough surface of the 12-kg block with a coefficient of kinetic friction of 0.30. The 5.0-kg block accelerates downward when released, and the acceleration must be calculated by accounting for the forces of friction and tension. One user initially calculated the acceleration as 1.6 m/s² but expressed uncertainty about the method used, which involved adding the force of friction to the tension and dividing by the total weight. Other participants confirmed that the approach was generally correct but suggested reviewing previous discussions for clarity. Understanding the role of friction is crucial for accurately determining the system's acceleration.
algar32
Messages
88
Reaction score
1

Homework Statement



A system comprised blocks, a light frictionless pulley, and connecting ropes is shown in the figure. The 9.0-kg block is on a perfectly smooth horizontal table. The surfaces of the 12-kg block are rough, with \mu_k = 0.30 between the block and the table. If the 5.0-kg block accelerates downward when it is released, find its acceleration.

the figure looks like this picture (except M= 5kg):
http://www.cramster.com/answers-feb-11/physics/system-comprising-blocks-system-comprising-blocks-light-frictionless-pulle_1126069.aspx?rec=0

14 minutes ago
- 4 days left to answer.



Homework Equations





The Attempt at a Solution


I want to say it is 1.6 m/s^s but I am not sure If I did it right.
I just added the force friction to the force tension and then divided by the weights (9+5).
Not sure if this is correct or not.
 
Physics news on Phys.org
Looks right to me.
 
This is very similar to the earlier problem you asked.

Use those equations to find the answer :-)
 
tal444 said:
Looks right to me.

EDIT: I was doing this question wrong earlier, refer to emailanmol's posts in your other thread.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top