Modified Atwood Machine Problem

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SUMMARY

The modified Atwood machine problem involves a 20 kg mass on a tabletop and a 5 kg mass hanging off the edge, connected by a massless string over a frictionless pulley. The coefficient of kinetic friction between the tabletop and the 20 kg mass is 0.100. The calculated acceleration of the system is 1.2 m/s², derived from the net force of 30 N divided by the total mass of 25 kg. The tension in the string is determined to be 44 N, with the friction force not affecting the tension calculation for the hanging mass.

PREREQUISITES
  • Understanding of Newton's second law (Fnet = ma)
  • Knowledge of kinetic friction and its calculation (f = μN)
  • Familiarity with basic mechanics involving pulleys
  • Ability to perform calculations involving mass and acceleration
NEXT STEPS
  • Study the effects of friction on different mechanical systems
  • Learn about tension in systems with multiple masses and pulleys
  • Explore advanced problems involving Atwood machines
  • Investigate the role of coefficients of friction in real-world applications
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Students in physics, particularly those studying mechanics, as well as educators looking for examples of Atwood machine problems and their solutions.

plazprestige
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Homework Statement


Consider a modified Atwood machine, where one mass of 20 kg lies on a flat tabletop, and another mass of 5 kg hangs off the edge of the table, where the two masses are connected to each other by a massless string and a frictionless pulley on the edge of the table. The coefficient of kinetic friction between the tabletop and the 20 kg mass is .100. Find a) the acceleration of the system, and b) the tension in the string connecting the masses.

Homework Equations


Fnet = ma
f = (.1)N, where f is the force of kinetic friction and N is the normal force
g = -10 m/s/s (for easy calculations)

The Attempt at a Solution


a) The acceleration of the object (or system) is the net force divided by the total mass of the system. The calculation for net force I executed: (5 kg)*(10 m/s/s) - (.1)(200 N) = 50 - 20 = 30 N. 30 N/ 25 kg = 1.2 m/s/s = a.

b) I'm more unsure about this part. The net force on the 5 kg block is (5 kg)*(10 m/s/s) - T = (5 kg) * a. a must be the same for both masses, so 5 kg * 1.2 m/s/s = 6 N. 6 = 50 - T. - T = -44. T = 44 N.

The part I'm unsure about is whether the fact that the tabletop and the 20 kg block having friction associated with them should factor into the calculation of the tension.

I'd like to know if my two answers are correct. If they are, why is it that part b doesn't have the friction force factored in. If something is wrong, please explain the correct answer.

Thanks.
 
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After some fiddling, I figured out why my answers are correct.
 

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