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

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In summary: I don't remember the formula off the top of my head, but that should get you close.You could start with the 12N and subtract the net force needed to accelerate block...I don't remember the formula off the top of my head, but that should get you close.
  • #1
ConfusedMonkey
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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?
 
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  • #2
ConfusedMonkey said:
to the right. Is this wrong?
Yes, it is clearly wrong.
ConfusedMonkey said:
F32=m2⋅(−a)
You are forgetting the other force on the middle block.
 
  • #3
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}##.
 
  • #4
ConfusedMonkey said:
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.
 
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  • #5
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?
 
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  • #6
Thanks for the help guys.
 
  • #7
haruspex said:
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...
 
  • #8
ConfusedMonkey said:
since F12=2m2
How so?
 
  • #9
haruspex said:
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##
 
  • #10
ConfusedMonkey said:
##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.
 
  • #11
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.
 
  • #12
ConfusedMonkey said:
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.
 
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  • #13
haruspex said:
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?
 
  • #14
ConfusedMonkey said:
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.
 
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  • #15
Thank you for your help and patience. I have to say, I think the first way I solved the problem was a lot easier!
 

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

1. What is the definition of "force exerted on block"?

The force exerted on a block is the external push or pull that is applied to the block, causing it to accelerate or move in a certain direction.

2. How is the force exerted on a block calculated?

The force exerted on a block can be calculated using Newton's second law of motion, which states that force equals mass times acceleration (F=ma).

3. What are the different types of forces that can be exerted on a block?

There are several types of forces that can be exerted on a block, including gravitational force, frictional force, normal force, and applied force.

4. How does the force exerted on a block affect its motion?

The force exerted on a block determines its acceleration and direction of motion. If the force is greater than the opposing forces, the block will accelerate in the direction of the force. If the forces are balanced, the block will remain at rest or continue to move at a constant velocity.

5. Can the force exerted on a block change over time?

Yes, the force exerted on a block can change over time if there are changes in the magnitude or direction of the force, or if other forces are introduced. This can result in changes in the block's motion or acceleration.

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