- #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?