How Does Newton's Third Law Apply in a Frictionless Two-Block System?

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SUMMARY

The discussion focuses on the application of Newton's Third Law in a frictionless two-block system, where two blocks are accelerated across a horizontal surface. Given a force of 1.2 N and a mass of 1.0 kg, the acceleration of the system is calculated to be 0.4 m/s² to the right. The horizontal force exerted by the bottom block on the top block is also 0.4 N to the right, demonstrating that forces are equal in magnitude and opposite in direction as per Newton's Third Law. The coefficients of friction provided are not relevant to the acceleration since the system is frictionless.

PREREQUISITES
  • Understanding of Newton's Laws of Motion
  • Basic knowledge of force and acceleration (F=ma)
  • Concept of friction coefficients (static and kinetic)
  • Ability to analyze free-body diagrams
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  • Study the implications of Newton's Third Law in multi-body systems
  • Learn about free-body diagrams and their applications in physics problems
  • Explore the effects of friction in different scenarios, including static and kinetic friction
  • Investigate the relationship between force, mass, and acceleration in various contexts
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Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of Newton's Laws in action.

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Homework Statement


Two blocks are accelerated across a horizontal frictionless surface as shown. The coefficient of static friction between the two blocks is 0.7, and the coefficient of kinetic friction is 0.5, use M= 1.0 kg. When F=1.2 N, frictional forces keep the two blocks from sliding relative to each other, and the two move with the same acceleration. In this case, what is the acceleratiob of the two block system? (The picture is of mass 2m on the bottom with mass M on top of it, and F pointing to the right)
2) What is the horizontal component of the force the large block exerts on the small block
3) Suppose that F is increased. What is the maximum acceleration that mass 2M can have, without mass M slipping off?

Homework Equations



F=ma

The Attempt at a Solution


I got the first portion of the problem right. I just solved for f=ma, and plugged in the f=1.2. I figured that the coefficients of friction given were not relevant here because the system is accelerating. For an answer I got a=.4 m/s2 to the right.
2) For the second part of the question: the horizontal component the large block exerts on the small block, I put .4 to the left. I thought that according to Newton's third law the force is equal in magnitude and opposite in direction, but it is saying this answer is wrong. It says the correct answer is .4 to the right. Why is it the same?
 
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I don't understand how you are applying Newton's 3rd law in this case.

Realize that both blocks have the same acceleration, which equals 0.4 m/s^2 to the right. To find the force on the top block (which is exerted by the bottom block), apply Newton's 2nd law.
 

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