How Do You Solve This Newton's Third Law Problem Involving Friction and Tension?

AI Thread Summary
The discussion centers on solving a Newton's Third Law problem involving two blocks with friction and tension. The setup includes a 1.0 kg block tied to a wall resting on a 2.0 kg block, which is subjected to a 20 N tension force. Participants are confused about calculating the frictional forces and the correct application of Newton's second law, particularly regarding the total mass involved in the acceleration. There is a debate over the forces acting on both blocks and the need to consider friction between the blocks and the floor. The conversation highlights the complexities of free body diagrams and the importance of accurately accounting for all forces to determine the correct acceleration.
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Newton Third Law problem. Help pleaseee!

Homework Statement



The 1.0 kg block is tied to the wall with a rope. It sits on top of the 2.0 kg block. The lower block is pulled to the right with a tension force of 20 N. The coefficient of kinetic friction at both the lower and upper surfaces of the 2.0 kg block is uk = 0.370.

What is the acceleration of the 2.0 kg block?


Homework Equations



Manipulations of F=ma.

The Attempt at a Solution


I have given this two tries.

First try,

Fx=Fa2-Fk2-Ft+Fk1=ma
20N-(0.370)(3.0kg)(9.81m/s^2)-3.63N+(0.370)(1.0kg)(9.81)=3a

and I get it wrong..
 
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How did you get 3.63N? Why did you add Fk1? I believe the summation of forces is 2a
 


Hmm but isn't the force being applied to the whole system which is 3kg?

anddd, Fk1 is the kinetic friction acting between block 1 and block 2 and 3.63 is the tension in the rope for block 1.

I was guessing all of the forces on block 1 and 2 should be taking into account...
I guess not?
 


Ops i misread the question didn't notice the 1kg mass is tied to the wall no wonder i don't understand why static coefficient is not given. If we say the force applied is F, is this force F in your free body diagram for the smaller mass? Maybe you can try looking at just the 2kg mass and draw the free body diagram and resolve the forces. Does it give you the correct answer?
 
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I already tried adding up the sum of all the forces only on block 2 and got a wrong answer :(
 


Mind writing out how you do it?
 


20N-0.37*2*9.8=2a
6.4 m/s = a

this is wrong :(
 


0.37*2*9.8 is the frictional force between the smaller mass and the bigger one how about the friction between the bigger mass and the floor?
 
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