What would friction be in this pulley system

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SUMMARY

The discussion focuses on calculating the coefficient of kinetic friction (uk) in a pulley system involving two masses, one hanging and one on a table. The equations of motion are derived using Newton's second law (F = ma) and the tension in the rope is analyzed. The final calculation yields a coefficient of kinetic friction of 0.438, confirming the relationship between the forces acting on both masses. The analysis emphasizes the importance of correctly applying the signs for acceleration and gravitational force in the equations.

PREREQUISITES
  • Understanding of Newton's second law (F = ma)
  • Knowledge of tension in a pulley system
  • Familiarity with the concept of kinetic friction
  • Basic algebra for solving equations
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  • Study the principles of tension in pulley systems
  • Learn about the derivation of the coefficient of kinetic friction
  • Explore advanced applications of Newton's laws in multi-body systems
  • Investigate the effects of varying mass and friction on acceleration
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Physics students, engineering students, and anyone interested in mechanics and dynamics of pulley systems.

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1. You have a mass hanging off of a frictionless pulley (massless string). The other mass is on a table. The weight of the mass causes the block on the table to slide at an acceleration. There is friction between the block and the table. What would the coefficient of kinetic friction be for this set up?



2. F = ma
w = mg




3. Well Since the system is accelerating, I thought it was

T(rope) - fk = ma
Acceleration and the masses are known. T isn't.

But you can set this up for both masses

T(rope) - ukm1g = m1a
T(rope) - m2g = m2a
T(rope) = m2g + m2a
m2g + m2a - ukm1g = m1a
-ukm1g = m1a -m2a - m2g
-uk(0.483)(10) = 0.483(1.66) - 0.25(1.66) - 0.25(10)
uk = 0.438

Does this make sense?
 
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In the second equation g and a are in the same direction. Hence they must have the same sign.
 

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