Calculating Magnitude of Contact Force Between Boxes 1 & 2

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SUMMARY

The magnitude of the contact force between boxes 1 and 2, when a force of 9.00 N is applied to three boxes with masses m1 = 1.30 kg, m2 = 2.80 kg, and m3 = 4.90 kg, is not equal to 9.00 N. The correct calculation requires considering the total mass being accelerated, which includes boxes 2 and 3. The net force acting on box 1 is determined by the mass of box 1 and the acceleration due to the applied force, leading to a different value for the contact force.

PREREQUISITES
  • Understanding Newton's Second Law (F = ma)
  • Basic knowledge of force and mass interactions
  • Familiarity with calculating net forces in a system
  • Concept of contact forces in a multi-body system
NEXT STEPS
  • Calculate the acceleration of the entire system using F = ma with the total mass.
  • Determine the individual forces acting on each box using free-body diagrams.
  • Explore the concept of contact forces in physics to understand interactions between objects.
  • Review examples of similar problems involving multiple objects and applied forces.
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Students studying physics, particularly those focusing on mechanics and force interactions, as well as educators looking for examples of contact force calculations in multi-body systems.

BugsSport
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Homework Statement


A force of magnitude 9.00 N pushes three boxes with masses m1 = 1.30 kg, m2 = 2.80 kg, and m3 = 4.90 kg
05-20alt.gif

Find the magnitude of the contact force between boxes 1 and 2.

Homework Equations


Fnet,x = ma?


The Attempt at a Solution


I thought that it would be 9.00 N, since the boxes are back-to-back-to-back, but that answer is wrong.
 
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BugsSport said:

Homework Statement


A force of magnitude 9.00 N pushes three boxes with masses m1 = 1.30 kg, m2 = 2.80 kg, and m3 = 4.90 kg
05-20alt.gif

Find the magnitude of the contact force between boxes 1 and 2.

Homework Equations


Fnet,x = ma?

The Attempt at a Solution


I thought that it would be 9.00 N, since the boxes are back-to-back-to-back, but that answer is wrong.

Welcome to PF.

At the interface between 1 and 2, the mass of 2 and 3 has already been accounted for by the force, so the force required to accelerate the remaining mass is given by the force that it needs to still push.

Otherwise, if that force was still the whole 9N, then applying that to the total mass left would result in an acceleration to the left that just wouldn't be there.
 
okay got it thank you
 

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