Magnitude of Force Exerted on Middle Block in a Three-Block System

[ConfusedMonkey]

Homework Statement

Blocks of 1.0, 2.0, and 3.0 kg are lined up on a frictionless table with a 12-N force applied to the leftmost block. What's the magnitude of the force that the rightmost block exerts on the middle one?

The blocks are lined up in such fashion: [1][2][3] and the 12 N force being applied to the leftmost block is obviously to the rightward direction.

Homework Equations

The Attempt at a Solution

The force being applied to the entire system is 12 N and all of the blocks accelerate at the same rate, so letting $m_i$ be the mass of the i'th block, we have $12 = (m_1 + m_2 + m_3)a \implies a = \frac{12}{m_1 + m_2 + m_3}$. Now, let $F_{23}$ denote the force that the second block exerts on the third block. We have $F_{23} = \frac{12m_3}{m_1 + m_2 + m_3}$. Plugging in the values for each $m_i$ gives $a = 6$. I know that $F_{32}$, which is the force that the third block exerts on the second is then $-6$ N by the third law.

I have two questions.

1) The textbook gives an answer of $6$ N to the right. Is this wrong? How could the force that the third block exerts on the second block be to the right?

2) If I try to find $F_{32}$ directly, without using the third law, I get $F_{32} = m_2 \cdot (-a) = m_2 \cdot \frac{-12}{m_1 + m_2 + m_3} \neq -6$. Is this because the acceleration, $a$, that I'm plugging in is the acceleration of the entire system, whereas I would need to plug in the acceleration that block 3 induces on block 2?

 

[haruspex]

to the right. Is this wrong?

Yes, it is clearly wrong.

F32=m2⋅(−a)

You are forgetting the other force on the middle block.

 

[ConfusedMonkey]

The only other horizontal force is $F_{12}$ (the force the first block exerts on the second block), but we are told to only find $F_{32}$.

 

[haruspex]

The only other horizontal force is $F_{12}$ (the force the first block exerts on the second block), but we are told to only find $F_{32}$.

I understand that, but the law you are trying to apply to the middle block is ΣF=ma. ΣF must include all the horizontal forces acting on the middle block.

 

[scottdave]

I tend to agree with you about the direction. Let's look at a slightly simpler problem, to see if it can help with yours.
Take two blocks. The left one (call it A) is 3 kg, and the right one (call it B) is 2 kg. Now push (10 Newtons toward the right) on A. Total mass is 5 kg, so (10 N)/(5 kg) = 2 m/s². Now let's look at B. It is 2 kg, and it is accelerating at 2 m/s² so it must have 4 Newtons pushing on it (toward the right). So it is pushing back 4 Newtons onto A (to the left). Now A has 10 N right and 4 N left for a net force of 6 N (to the right).
Let's see if we did it right. A has 6 Newtons net force, and it's mass is 3 kg, so (6 N)/(3 kg) = 2 m/s², which is what we expect. Can you expand this thought process to your problem of 3 blocks?

 

[ConfusedMonkey]

Thanks for the help guys.

 

[ConfusedMonkey]

I understand that, but the law you are trying to apply to the middle block is ΣF=ma. ΣF must include all the horizontal forces acting on the middle block.

I thought I had solved it but I realized I made some mistakes...

Okay, the only horizontal forces that are on Block 2 are $F_{12}$ and $F_{32}$. Therefore, $F_{12} + F_{32} = m_2a = 2m_2$ (since a = 2). Therefore, $F_{32} = 2m_2 - F_{12} = 2m_2 - 2m_2 = 0$ since $F_{12} = 2m_2$. This is clearly wrong. I feel as if I am missing something obvious...

 

[haruspex]

since F₁₂=2m₂

How so?

 

[ConfusedMonkey]

How so?

$F_{12}$ is the force that the first block exerts on the second. The object the force is being exerted on is block 2, Hence $F_{12} = m_2(a) = 2m_2$

 

[haruspex]

$F_{12}$ is the force that the first block exerts on the second. The object the force is being exerted on is block 2, Hence $F_{12} = m_2(a) = 2m_2$

That's making the same mistake as before, but from the other side. The acceleration of the middle block results from the net force on it.

 

[ConfusedMonkey]

Okay, I see your point. The acceleration I need for $F_{12}$ is the acceleration the middle block gains from its contact with block 1 alone (right?)

But I don't see how to calculate that acceleration.

 

[haruspex]

Okay, I see your point. The acceleration I need for $F_{12}$ is the acceleration the middle block gains from its contact with block 1 alone (right?)

But I don't see how to calculate that acceleration.

You could start with the 12N and subtract the net force needed to accelerate block 1 to find F₁₂.

 

[ConfusedMonkey]

You could start with the 12N and subtract the net force needed to accelerate block 1 to find F₁₂.

That solves the question, but why does subtracting the 2N from 12N give me F₁₂?

 

[haruspex]

That solves the question, but why does subtracting the 2N from 12N give me F₁₂?

Because for block 1, 2N=ma=ΣF=12N+F₂₁= 12N-F₁₂. So F₁₂=12N-2N.

 

[ConfusedMonkey]

Thank you for your help and patience. I have to say, I think the first way I solved the problem was a lot easier!