For 2) wouldn't there also be an equal and opposite force on block B by block A?

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Homework Help Overview

The discussion revolves around a problem involving three blocks on a frictionless table, where a hand applies a force to the leftmost block, causing all blocks to accelerate. Participants are exploring the forces acting on each block and the relationships between them, particularly focusing on the net forces and action-reaction pairs as described by Newton's laws.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of forces acting on block A and the implications of the system's acceleration on the force applied by the hand. Questions arise regarding the net force on block B and the contributions from adjacent blocks. Some participants question the assumptions about forces between blocks and the interpretation of Newton's third law.

Discussion Status

The discussion is ongoing, with participants providing insights into the relationships between the forces acting on the blocks. Some guidance has been offered regarding the application of Newton's laws, but there is no explicit consensus on the correct approach to determining the net force on block B.

Contextual Notes

Participants are grappling with the implications of action-reaction pairs and the net forces acting on individual blocks, indicating a need for clarity on these foundational concepts. There is also a mention of a potential typo in one of the posts, which may affect the understanding of the discussion.

Gooner5
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Three blocks, each of mass 13 kg are on a frictionless table. A hand pushes on the left most box (A) such that the three boxes accelerate in the positive horizontal direction as shown at a rate of a = 1.4 m/s2.

1)What is the magnitude of the force on block A from the hand?

2)What is the net horizontal force on block B?


For number 1 I was able to get the answer by adding up all the blocks and multiplying that by the acceleration but I am confused by why this is right. Before I did this I thought that the answer would be the mass of A * 1.4. Why does the force by the hand on block A include the other blocks as well? Is it because the whole system is accelerating and for that to happen the force has to take into account the other blocks as well?

For number 2 I found the force done on the system to be (13+13+13)*1.4 = 54.6N. I then found that the force that object C does on block B and the force that object A does on block B would 13*1.4= 18.2N. I then added the force done on the system by the hand with block A and subtracted that by the force by C on B to get 54.6, but this is not the right answer. What should I be looking for in this problem?

Thanks
 
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For 1, the hand doesn't care whether it's three separate blocks that happen to be adjacent, or three blocks glued together, or a single block mass 39kg. Alternatively, consider the force pair (action and reaction) at each interface. Each block accelerates according to the net force on it.

For 2), all block B "knows" is that it experiences two forces, one from block A and one from block C, of opposite sign. The net force on B is simply the sum of those two. Alternatively, consider that for each block separately ΣF=Fnet=ma, where the sum is over the forces acting directly on the block. You tried using the first part of that, ΣF=Fnet, to find Fnet (but summed the wrong forces); the alternative is to use the second part, Fnet=ma.
 
Last edited:
For 2) wouldn't there also be an equal and opposite force on block B by block C?
 
Gooner5 said:
For 2) wouldn't there also be an equal and opposite force on block B by block C?

Do you care about that?

When using Newtons law F=ma what does F stand for? Hint... the answer is NOT just "Force".
 
Gooner5 said:
For 2) wouldn't there also be an equal and opposite force on block B by block C?
I made a typo in my post. Now corrected.
 

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