Dynamics question -- 3 masses on a pulley-rope system on an inclined plane

Click For Summary
SUMMARY

The discussion centers on calculating the acceleration and tensions in a pulley-rope system involving three masses on an inclined plane. User A reported an acceleration of 4.8 m/s² with tensions T1 at 24.5 N and T2 at 34.3 N, while User B obtained a lower acceleration of 4.1 m/s². The conversation highlights the importance of considering mass accelerations and friction in tension calculations, as well as the need for clear equations to facilitate understanding and verification of results.

PREREQUISITES
  • Understanding of Newton's Second Law of Motion
  • Basic principles of static equilibrium
  • Knowledge of inclined plane dynamics
  • Familiarity with tension in pulley systems
NEXT STEPS
  • Study the derivation of tension equations in pulley systems
  • Learn about the effects of friction on inclined planes
  • Explore static equilibrium conditions for multiple masses
  • Practice solving dynamics problems involving multiple masses and pulleys
USEFUL FOR

Physics students, educators, and anyone interested in mastering dynamics problems involving pulley systems and inclined planes.

rabsta00
Messages
3
Reaction score
0
Homework Statement
A mass m is 5kg and another mass, M=6kg. Find the acceleration of this system if the kinetic friction is 0.1 and theta = 30 degrees. Find all tensions of connecting ropes. For which values of M will the system stay in an equilibrium position?Assume that the static friction coefficient is 0.15. Disregard the mass of the pulley and ropes.
Relevant Equations
F=ma
Screen Shot 2021-03-29 at 12.31.31 pm.png

This image was provided, I've completed the first part of the question and got a = 4.8m/s^2 as well as T1= 24.5N and T2=34.3N. not sure about my answers though. also I don't understand the mass in static equilibrium part, can anyone explain how to solve that? Thanks.
 
Physics news on Phys.org
rabsta00 said:
got a = 4.8m/s^2
I get a far smaller value. Please post your working.
 
  • Like
Likes   Reactions: Orodruin
haruspex said:
I get a far smaller value. Please post your working.

IMG_2060.jpg

I tried a different method but ended up getting a = 4.1 which isn't much smaller.
 
You omitted the effect of the mass accelerations and friction on the tensions.
 
rabsta00 said:
View attachment 280501
I tried a different method but ended up getting a = 4.1 which isn't much smaller.
Please take the trouble to type equations in. It makes it much easier to quote lines to comment on.
You have T1=2mg sin(θ)-mg sin(θ). How do you arrive at that?
Your calculation of the acceleration of 2m ignores the string. You just have it sliding down the slope unrestrained.
 

Similar threads

Replies
2
Views
2K
Replies
19
Views
2K
  • · Replies 4 ·
Replies
4
Views
7K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 68 ·
3
Replies
68
Views
13K
  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 3 ·
Replies
3
Views
6K
  • · Replies 4 ·
Replies
4
Views
609
  • · Replies 3 ·
Replies
3
Views
2K
Replies
8
Views
2K