Three Bodies in a Line

The situation

Three bodies of masses $m_1$, $m_2$, $m_3$ are connected in a line by two strings on a smooth horizontal surface. Force $F$ is applied to $m_1$ (the front body). String 1 connects $m_1$ to $m_2$ with tension $T_1$. String 2 connects $m_2$ to $m_3$ with tension $T_2$.

$m_1 \xrightarrow{F} \text{---}T_1\text{---} m_2 \text{---}T_2\text{---} m_3$

Acceleration of the system

Treating all three as one system, only $F$ is the external force:

$F = (m_1 + m_2 + m_3)a$

$\boxed{a = \frac{F}{m_1 + m_2 + m_3}}$

Tension $T_1$ (between $m_1$ and $m_2$)

$T_1$ must accelerate $m_2$ and $m_3$ together:

Apply Newton's Second Law to the system of ($m_2 + m_3$):

$T_1 = (m_2 + m_3)a = \frac{(m_2 + m_3)F}{m_1 + m_2 + m_3}$

Tension $T_2$ (between $m_2$ and $m_3$)

$T_2$ only needs to accelerate $m_3$:

$T_2 = m_3 a = \frac{m_3 F}{m_1 + m_2 + m_3}$

The pattern

Each tension equals the force needed to accelerate the bodies behind that string:

$T_1 = \frac{(m_2+m_3)}{m_1+m_2+m_3} \cdot F, \quad T_2 = \frac{m_3}{m_1+m_2+m_3} \cdot F$

The closer to the rear of the chain, the smaller the tension — each string only pulls what is behind it.

$T_2 < T_1 < F$

Extending to $n$ bodies

For $n$ bodies with masses $m_1, m_2, \ldots, m_n$, force $F$ applied to $m_1$:

$a = \frac{F}{\sum_{i=1}^n m_i}$

Tension in string between $m_k$ and $m_{k+1}$:

$T_k = \frac{\sum_{i=k+1}^n m_i}{\sum_{i=1}^n m_i} \cdot F$

Each tension equals $F$ multiplied by the fraction of total mass that lies behind that string.

Verification

Apply Newton's Second Law to $m_1$ alone:

$F - T_1 = m_1 a$

$T_1 = F - m_1 a = F - \frac{m_1 F}{m_1+m_2+m_3} = \frac{(m_2+m_3)F}{m_1+m_2+m_3} \checkmark$