Finding normal force of moving crates

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SUMMARY

The discussion focuses on calculating the normal force between two moving crates with masses of 49 kg and 29 kg, subjected to a coefficient of kinetic friction of 0.30. The crates are pushed along the floor at a constant velocity, which implies that the net force acting on them is zero. The equations used include the frictional force formula Ff = μkFN, leading to the conclusion that the normal forces acting on each crate (FN1 and FN2) are both 11.7 kg.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with the concept of friction and coefficients of friction
  • Basic algebra for solving equations
  • Knowledge of forces acting on objects in motion
NEXT STEPS
  • Study the effects of varying coefficients of friction on normal force calculations
  • Learn about static vs. kinetic friction and their implications in real-world scenarios
  • Explore advanced topics in dynamics, such as acceleration and net force analysis
  • Investigate the role of normal force in different contexts, such as inclined planes and circular motion
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Physics students, educators, and professionals involved in mechanics, particularly those focusing on forces and motion in practical applications.

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Homework Statement


Your moving company runs out of rope and hand trucks, so you are forced to push two crates along the floor as shown in the figure below. The crates are moving at constant velocity, their masses are m1 = 49 kg and m2 = 29 kg, and the coefficients of kinetic friction between both crates and the floor are 0.30. Find the normal force between the two crates.

Homework Equations


F = ma

The Attempt at a Solution



Ff = μkFN
 
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Ff1 = μk1FN1Ff2 = μk2FN2Ff1 + Ff2 = FN1 + FN2 (μk1m1) + (μk2m2) = FN1 + FN2 (0.30)(49 kg) + (0.30)(29 kg) = FN1 + FN214.7 kg + 8.7 kg = FN1 + FN2 23.4 kg = FN1 + FN2FN1 = 11.7 kgFN2 = 11.7 kg
 

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