Statics Double-Mass Pulley Incline Problem

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SUMMARY

The discussion centers on solving a static equilibrium problem involving a double-mass pulley system on an incline. The key equation derived is that the tension in the rope, denoted as T, must be analyzed in terms of its components along the incline. The participant, andrewkirk, successfully determined the mass m2 by applying the principle of static equilibrium, specifically by setting the sum of forces in the rotated x-direction to zero. This approach clarifies the relationship between the masses and the incline angle, theta.

PREREQUISITES
  • Understanding of static equilibrium principles
  • Knowledge of vector decomposition in physics
  • Familiarity with pulley systems and tension forces
  • Basic trigonometry, particularly involving angles and sine functions
NEXT STEPS
  • Study static equilibrium problems involving multiple pulleys
  • Learn about vector decomposition techniques in physics
  • Explore the dynamics of inclined planes and forces
  • Review tension force calculations in static systems
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and static equilibrium, as well as educators seeking to enhance their teaching of pulley systems and force analysis.

physicsman32
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Homework Statement


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Homework Equations

The Attempt at a Solution


I'm having trouble coming up with the equations for the system. At first, I thought that m2 = m1sin(theta), but that's not correct. How can I set this problem up?
 
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Let the tension in the rope be T. What must its value be, if the system is static?

To work out the force on the trolley, imagine that, the pulley is locked so that it can't rotate and that instead of going around the pulley, the rope is tied to cleats on the pulley at the two points where it goes from touching to not-touching the pulley. If the tension on both parts of the rope is still T, what will be the net force vector by which the two parts of the rope act on the pulley, and hence on the trolley?
 
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Thank you, andrewkirk. After looking at the problem more closely, I saw that the rope tension in the downward direction could be broken down into components along the rotated axes of the incline. Since the tension in the rope is the same everywhere, I was able to solve for m2 by setting the sum of all forces in the rotated x-direction equal to zero.
 

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