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

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The discussion centers on solving a dynamics problem involving three masses on a pulley-rope system on an inclined plane. One participant calculated an acceleration of 4.8 m/s² and tensions T1 and T2 as 24.5N and 34.3N, respectively, but expressed uncertainty about the accuracy of these values. Another participant obtained a lower acceleration of 4.1 m/s² and pointed out that the original calculations neglected the effects of mass accelerations and friction on the tensions. There is a request for clearer equations to facilitate better understanding and commentary on the calculations. The conversation highlights the complexity of the problem and the need for detailed breakdowns of the equations used.
rabsta00
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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.
 
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rabsta00 said:
got a = 4.8m/s^2
I get a far smaller value. Please post your working.
 
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.
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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