Contact Forces Question(s) part 2

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SUMMARY

The discussion focuses on solving contact force problems involving three blocks on a rough surface, where m1=2m2=5m3. In Part 2, the equations for applied force and contact forces are modified to include friction, represented by the coefficient μ=0.11. The key equations derived are Fa = 17/2 m3(a+μg), F12 = 7/2 m3(a+μg), and F23 = m3(a+μg). The participant successfully resolves their confusion regarding the cancellation of mass in the equations.

PREREQUISITES
  • Understanding of Newton's second law (F = ma)
  • Knowledge of contact forces and friction
  • Familiarity with mass relationships (m1=2m2=5m3)
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the effects of friction on motion in physics
  • Explore advanced applications of Newton's laws in multi-body systems
  • Learn about the implications of mass ratios in force calculations
  • Investigate real-world examples of contact forces in engineering scenarios
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and forces, as well as educators seeking to explain contact forces and friction in practical applications.

Inertialforce
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Homework Statement


Three blocks are placed onto a frictionless horizontal surface. These blocks are side by side and are in contact with each other. Let m1=2m2=5m3. Answer the following exercises.
(Part 1)
a)If a force is applied to m1, then show that

i)Fa = 17/2m3a

ii)F12 = 7/2m3a where F12 is the contact force of mass 1 on mass 2

iii)F23=m3a

Question: Consider the same three blocks as in the question above. If these blocks rest upon a rough floor (μ=0.11), show that if a force is applied to m1, then:
(Part 2)
i)Fa = 17/2 m3(a+μg)

ii)F12 = 7/2 m3(a+μg)

iii)F23 = m3(a+μg)


Homework Equations


Σ Fx = ma


The Attempt at a Solution


All in all both questions (part 1 and part 2) are more or less the same except this question (part 2) includes the force of friction while the other (part 1) does not. I have already solved all the questions for part 1, and where I am having difficulty is, is in getting rid of the mass (m) in "μmg" for this question so that I could get the values as seen above.
 
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Never mind I figured out where I went wrong and now know what to do for this question so a reply is not needed. But thank you to those who were trying to help by trying to explain the question to me.
 
It would be good to see your attempt at the questions, then maybe I could understand what you are saying a little better. From what I have read you are finding it difficult cancelling the masses in all of part 2.

F = ma
Therefore you have ma2 = 17/2 m3 (a+μg)
Divide by m.
 

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