Three blocks on a fricionless horizontal surface

AI Thread Summary
The discussion revolves around the dynamics of three blocks of equal mass on a frictionless surface when a force is applied to the first block. It clarifies that the contact force exerted by block one on block two is derived from the total force applied and the masses of all blocks involved. The key equation shows that this contact force is proportional to the combined mass of blocks two and three, divided by the total mass of all three blocks. Newton's second law is applied to each block to derive the relationships between the forces and accelerations. Understanding these interactions emphasizes the principles of action and reaction forces in a system of connected masses.
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OK, three blocks on a fricionless horizontal surface are in contact on each other, a force is applied to block 1, m1, m2, m3, respectively. And m1=m2=m3.

how come the force of contact that block one exerts on block is F(m2+m3) / (m1+m2+m3) instead of just F*m2/(m1+m2+m3). ( I know that a is F/(m1+m2+m3).)
 
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Think about Action and Reaction Forces [Contact Forces] acting on each block.
 
Right direction : Positive; Left direction: Negative.
<br /> \begin{multline*}<br /> \begin{split}<br /> &amp;Newton&#039;s\ 2nd\ Law:\\<br /> &amp;\sum \vec{F}=m\vec{a}\\<br /> &amp;Consider\ m_1:\\<br /> &amp;F+(-Fm_1m_2)=m_1a\\<br /> &amp;F=m_1a+(Fm_1m_2)\ ...(1)\\<br /> &amp;Consider\ m_2:\\<br /> &amp;Fm_2m_1+(-Fm_2m_3)=m_2a\ ...(2)\\<br /> &amp;Consider\ m_3:\\<br /> &amp;Fm_3m_2=m_3a\ ...(3)\\<br /> &amp;(3)\rightarrow(2):\\<br /> &amp;Fm_2m_1=m_2a+m_3a\ ...(4)\\<br /> &amp;(4)\rightarrow(1):\\<br /> &amp;F=m_1a+m_2a+m_3a\ ...(5)\\<br /> &amp;From\ (4):\\<br /> &amp;Fm_2m_1=(m_2+m_3)a\\<br /> &amp;=\frac{F(m_2+m_3)}{m_1+m_2+m_3}\\<br /> \end{split}<br /> \end{multline*}<br />
 

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