Acceleration on a Pulley: Equal or Different for Connected Masses?

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SUMMARY

The acceleration of two connected masses, one on a frictionless table and the other hanging over a pulley, is equal when released from rest. This conclusion is based on Newton's second law, represented by the equation ∑F=ma. The mass on the table and the hanging mass experience the same magnitude of acceleration due to their connection via a string, ensuring synchronized movement. This principle applies regardless of the individual masses, as long as they are connected in this manner.

PREREQUISITES
  • Understanding of Newton's second law (∑F=ma)
  • Basic knowledge of pulley systems
  • Familiarity with concepts of mass and acceleration
  • Experience with frictionless surfaces in physics
NEXT STEPS
  • Study the effects of friction on pulley systems
  • Explore advanced topics in dynamics, such as tension in strings
  • Learn about the implications of varying mass ratios on acceleration
  • Investigate real-world applications of pulley systems in engineering
USEFUL FOR

Students studying physics, educators teaching mechanics, and anyone interested in understanding the principles of motion and acceleration in connected systems.

Tim Wellens
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Homework Statement


We are dealing with a frictionless table with one mass on the table, while the other one hangs over the edge on a pulley connected by a string... and the mass on the table is greater than the mass hanging from the edge...

When the blocks are released from their resting position, would the magnitude of the acceleration of the mass on the table be more than, less than, or equivalent to the acceleration in the mass hanging over the edge?

Homework Equations


∑F=ma

The Attempt at a Solution


I think when the blocks are released, the magnitude of the acceleration of the mass on the table will be equal to the acceleration of the mass hanging over the edge because they are connected by a string, which would make them move at the same accelerations?

[moderator note: Removed boldface from user text.]
 
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