Two Bodies Connected by String — Acceleration and Tension

The situation

Two bodies of masses $m_1$ and $m_2$ are connected by a light inextensible string on a smooth horizontal surface. A force $F$ is applied to $m_2$ (the rear body). Find the acceleration of the system and the tension $T$ in the string.

Treating the system as a whole

Since the string is inextensible, both bodies have the same acceleration $a$.

Apply Newton's Second Law to the entire system (both bodies together):

$F = (m_1 + m_2)a$

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

The external force accelerates the total mass — exactly as expected.

Finding tension

Apply Newton's Second Law to $m_1$ alone. The only horizontal force on $m_1$ is the tension $T$ pulling it forward:

$T = m_1 a = m_1 \cdot \frac{F}{m_1 + m_2}$

$\boxed{T = \frac{m_1 F}{m_1 + m_2}}$

Why tension is less than F

The force $F$ accelerates both masses. The tension $T$ only needs to accelerate $m_1$. Since $m_1 < m_1 + m_2$, we have $T < F$.

The string transmits only the fraction of the force needed for $m_1$'s acceleration.

If force is applied to $m_1$ instead

$a = \frac{F}{m_1 + m_2} \quad \text{(same)}$

Now tension pulls $m_2$ forward:

$T = m_2 a = \frac{m_2 F}{m_1 + m_2}$

The acceleration is the same regardless of which end the force is applied. The tension changes depending on which mass is being pulled through the string.

With friction

If the surface has friction coefficient $\mu$ and both blocks have the same $\mu$:

$F - \mu(m_1 + m_2)g = (m_1 + m_2)a$

$a = \frac{F - \mu(m_1+m_2)g}{m_1+m_2} = \frac{F}{m_1+m_2} - \mu g$

Tension in string (force is applied to $m_2$, string connects to $m_1$):

$T - \mu m_1 g = m_1 a \implies T = m_1(a + \mu g) = \frac{m_1 F}{m_1+m_2}$

Interestingly, the tension formula is the same as the frictionless case when $\mu$ is the same for both blocks.