Does friction work as an action reaction pair with 2 obects on top of eachother?

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Homework Help Overview

The problem involves two blocks, one resting on top of the other, with a tension force applied to the lower block. The scenario examines the role of friction and action-reaction pairs in determining the tension in a rope holding the upper block in place.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to analyze the forces acting on both blocks, particularly focusing on friction and action-reaction pairs. Some participants question the relevance of certain information and the correctness of the original poster's reasoning regarding the forces on the upper block.

Discussion Status

The discussion is ongoing, with participants expressing differing opinions on the validity of the original poster's approach. Some have offered critiques while others defend the reasoning presented. There is no clear consensus yet, and multiple interpretations of the problem are being explored.

Contextual Notes

Participants are navigating assumptions about the relevance of the lower block's motion and the nature of the forces acting on the upper block. The original poster expresses uncertainty about the action-reaction forces involved.

mrjoe2
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Homework Statement


the 1.0kg block is tied to the wall with a rope. it sits on top of the 2.9kg block. the lower block si pulled to the right with a tension force of 20N. the coefficient of kinetic friction at both the lower and upper surfaces of the 2kg block is mk=.40. what is the tension in the rope holding the 1kg block to the wall?


Homework Equations


Newtons third law, perhaps the second law. ffk=fn *mewk fnet=ma fg=mg


The Attempt at a Solution


when i tried thisi said that the backward friction in the 2kg block was equal to the forward fricttion in the 1kg block (action reaction pair. this also makes sense i think because the 1kg is tending backwards "slipping" as the 2.0kg object accelerates forward). i noticed that the fg of the 2kg box was actually 3*9.8 because it has the 1kg box on top of it. this would also be the Fn of the 2kg box. so ff = 11.76 subbing in fN and mewk. so there is a forward force on box 1kg. then we also have to take into account another action reaction pair. that is, the forward force 20N on box 2kg has a reaction pair on the 1kg block of 20N back. then we have yet another force acting on the 1kg, and that is its own Ffk which is Fk=9.8*1 *.4 = 3.92 Newtons forward. so the tension in the rope would be T = 3.92 + 11.76 - 20 = 4.32 N (back). There are no solutions for these questions and i just want to see if i did it right. i have a feeling its perfection but like in the title i infered that i wasnt sure about the action reaction forces i assumed (through Newtons second law)
 
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Much of the information given here is irrelevant. The friction force is just the coefficient of kinetic friction multiplied by the weight of the top block. As long as the lower block is moving, so that there IS friction, how fast it is moving is irrelevant. Then tension in the line holding the upper block in place is equal to the friction force.
 
i think you're wrong sorry. any other people think they know how to do it?
 
mrjoe2 said:
i think you're wrong sorry. any other people think they know how to do it?

I don't think he's wrong. What horizontal forces are on the 1 kg block?
 
mrjoe2 said:
i think you're wrong sorry. any other people think they know how to do it?
For once, HallsofIvy is not wrong. :smile: (Sorry Halls... couldn't resist.)
 
Well, at least you didn't do what mrjoe2 did- he sent me a pm telling me that if I don't understand a question, I shouldn't respond to it! I was tempted to delete the entire thread.
 

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