How Does Friction Affect the Acceleration of Stacked Blocks?

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SUMMARY

The discussion centers on the physics of stacked blocks experiencing friction and acceleration. When a force F is applied to the lower block (mass m2), the acceleration of the lower block can be calculated using the equation a = (F - μ(m1 + m2)g) / m2, where μ is the coefficient of kinetic friction and g is the acceleration due to gravity. The correct formula for the acceleration of the lower block, accounting for the frictional force from the top block, is a2 = (F - μg(2m1 + m2)) / m2. This highlights the importance of considering all frictional forces in multi-block systems.

PREREQUISITES
  • Understanding of Newton's Second Law (F = ma)
  • Knowledge of frictional forces and coefficients (e.g., coefficient of kinetic friction μ)
  • Basic concepts of normal force and gravitational force
  • Ability to manipulate algebraic equations for solving physics problems
NEXT STEPS
  • Study the effects of varying coefficients of friction on acceleration in stacked systems
  • Learn about the dynamics of multi-body systems in classical mechanics
  • Explore the implications of friction in real-world applications, such as vehicle dynamics
  • Investigate the role of static versus kinetic friction in different scenarios
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators and anyone interested in understanding the dynamics of friction in multi-block systems.

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Homework Statement


A block of mass m1 is on top of a block of mass m2. The lower block is on a horizontal surface, and a rope can pull horizontally on the lower block. The coefficient of kinetic friction for all surfaces is μ. What is the resulting acceleration of the lower block if a force F is applied to the rope? Assume that F is sufficiently large so that the top block slips on the lower block.

Homework Equations



F = ma
Frictional Force = Normal Force * μ

The Attempt at a Solution



Normal Force (for lower block) = (m1 + m2) g
Frictional Force (for lower block) = μ * (m1 + m2) g
F - (μ * (m1 + m2) g) = m2 * a
a = ((F - (μ * (m1 + m2) g)) / m2

Correct Answer should be: a2 = (F - μg(2m1 + m2)) / m2
 
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welcome to pf!

hi vjnmath! welcome to pf! :smile:
vjnmath said:
… Assume that F is sufficiently large so that the top block slips on the lower block.

… sooo, you'll need to include the friction force from the top block :wink:
 

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