Understanding the Ratio of Forces between Blocks in a Frictionless System

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Homework Help Overview

The problem involves understanding the forces between blocks on a frictionless table, specifically the ratio of the force exerted by block 1 on block 2 (F12) to the force exerted by block 7 on block 8 (F78). The context is rooted in Newton's laws of motion and the dynamics of a system of multiple blocks being pushed.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of free body diagrams to visualize forces acting on the blocks. There is uncertainty about how forces interact between the blocks, especially in a frictionless scenario. Questions arise regarding the assumptions about forces exerted by the blocks on each other and the role of Tom's push.

Discussion Status

Some participants have offered hints and guidance on using Newton's laws to set up equations for the system. There is an exploration of different interpretations of how forces are transmitted through the blocks, with no explicit consensus reached yet.

Contextual Notes

Participants are grappling with the implications of a frictionless surface on the forces between the blocks, and there is mention of potential confusion regarding the interactions between the blocks and the external force applied by Tom.

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Homework Statement


Tom now pushes eight identical blocks on the horizontal and frictionless table (he’s compulsive). The force that block 1 exerts on block 2 is F12; the force that block 7 exerts on block 8 is F78. What is the ratio F12/ F78?

Homework Equations


F=ma
Δp = FnetΔt

The Attempt at a Solution


So really my problem is just understanding why the ratio isn't 1. I have no idea how to work it out... this is my attempt and I'm not sure if it is correct.

The force block 1 exerts on block 2 would just be mass*acceleration of the first block. However the force block 7 exerts on block 8 would be mass of one block*7*acceleration as the mass of all the first 7 blocks are being pushed at this acceleration. Therefore the ratio would be 1/7. Is this correct?
 
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Have you tried drawing a freebody diagram? I think that would help in this case. On the first block you obviously have tom pushing up. Is that it? You can repeat this for all the blocks.
 
Yeah I tried that... but I don't think I am doing it correctly haha.
The blocks are lying on the table and he is exerting a horizontal force on block 1 causing all of them to move.

They way I thought of doing a freebody diagram is that the forces on 1 is the force from book 2 and Tom. Then on 2 you have the force from book 1 and the force from book 3. And then for 3 you have the force from book 2 and book 4.

But thinking about that again and wouldn't the boxes not push back on each other because the table is a frictionless surface? The force exerted by book 2 onto 1 is generated by there being friction. So my second guess is the only force on any blocks are the forces inbetween the two blocks. So force exerted on block 1 is just Tom. Force on block 2 is the force exerted by block 1. Force on block 3 is the force exerted by block 2 and so on. But then wouldn't the force just be equal to Tom's force once you iterated through all of that? Or should the force acting on block 2 be Tom and the force exerted by block 1?
 
Hint: Using the Newton's second and third laws you can write 8 separate equations. Add up all of this 8 equations to get the force F exerted to the first block.bI think it is 8ma, where a is the same acceleration for all bodies; since they are moving at the same constant rate of change of speed. Using the first equation you can obtain the force F12 which should be 7ma. Using this and the last equation you can come up with the ratio that should be 7.
 
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OHHH. So the force exerted from block two to block one is equal to the 7 masses of the blocks just sitting there * acceleration. And then the force from the last block only has one block to push against so its just m*a. My logic was just really bad on this problem haha thank you very much.
 

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