Newtonian Mechanics and Forces Problem

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SUMMARY

The discussion centers on solving a Newtonian mechanics problem involving massless ropes and frictionless pulleys. The user calculated the system's acceleration and tensions, arriving at values of a = 2.613 m/s², T1 = 12.413 N, T2 = 27.438 N, T3 = 5.878 N, and T4 = 35.93 N, based on given masses and a friction coefficient (μ) of 0.5. The user confirmed the correctness of their equations and values through multiple checks, including numerical substitutions. The community provided validation for the calculations, ensuring the user met their deadline.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with tension in strings and pulleys
  • Knowledge of friction coefficients in physics
  • Ability to solve equations of motion
NEXT STEPS
  • Study the principles of static and dynamic equilibrium in mechanics
  • Learn about the effects of friction on tension in pulley systems
  • Explore advanced topics in Newtonian mechanics, such as systems with multiple pulleys
  • Practice solving similar problems using different mass and friction values
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Students studying physics, particularly those focusing on mechanics, as well as educators and tutors looking for practical examples of tension and acceleration calculations in pulley systems.

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Problem, Work, and Solution: http://img375.imageshack.us/img375/9987/physicsfinal.png (You can zoom in if needed)

Sorry for not using the template, this was done in MS Paint and I didn't want to have to type it all out again... Basically, the problem is to find all tensions (ropes/strings are massless) and the acceleration of the system, all in terms of given masses and frictions. The large rectangles are stationary and the circles are massless, frictionless pulleys.

I'd really appreciate if someone could check over this, I did myself more than once and also with numbers, assuming m1 = 1kg, m2 = 2kg, etc and μ = 0.5, it came out to be...
a = 2.613 m/s^2
T1 = 12.413 N
T2 = 27.438 N
T3 = 5.878 N
T4 = 35.93 N
...which seems right because m3 will actually lessen the tension on T3 but T4 will still have to pull all 5 masses so it's higher. Based on the reasonable masses I used and pretty average friction value, I think my answers are right but I still want to be sure.

Thanks!
 
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Anyone? This is due Monday...
 
Equation for acceleration looks right to me.

For T1 however it should be this in my opinion

m1 a = t1 - m1 g

EDIT: Oh, it's the same as yours. Sorry!
T2 also is fine and so should be the rest.
 
Last edited:

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