Body on Surface Connected to Hanging Body

The situation

Mass $m_1$ sits on a smooth horizontal surface. It is connected by a light inextensible string over a smooth pulley at the edge of the table to mass $m_2$, which hangs freely. The system is released from rest. Find the acceleration and tension.

Setting up

The string is inextensible, so both masses have the same acceleration $a$.

$m_1$ accelerates horizontally (along the surface). $m_2$ accelerates vertically downward.

For $m_1$ (horizontal):

The only horizontal force is tension $T$ (surface is smooth, no friction):

$T = m_1 a \quad \text{...(1)}$

For $m_2$ (vertical, downward positive):

$m_2 g - T = m_2 a \quad \text{...(2)}$

Solving

Add (1) and (2):

$m_2 g = (m_1 + m_2)a$

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

From (1):

$\boxed{T = m_1 a = \frac{m_1 m_2 g}{m_1 + m_2}}$

Checking

$T < m_2 g$: The hanging mass accelerates downward, so net force on it is downward: $m_2 g - T > 0$, confirming $T < m_2 g$. ✓

$a < g$: Since $m_1 + m_2 > m_2$, we have $a = \frac{m_2 g}{m_1+m_2} < g$. The hanging mass does not fall as fast as free fall — $m_1$ resists via the string. ✓

$m_1 \to 0$: $a \to g$, $T \to 0$ — $m_2$ falls freely, no tension needed. ✓

$m_1 \to \infty$: $a \to 0$, $T \to m_2 g$ — huge mass on table cannot be moved, tension equals weight of $m_2$. ✓

With friction on the surface

If the surface has kinetic friction coefficient $\mu$:

For $m_1$:

$T - \mu m_1 g = m_1 a$

For $m_2$:

$m_2 g - T = m_2 a$

Adding:

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

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

For motion to occur: $m_2 > \mu m_1$ (hanging weight must overcome friction).

$T = m_1(a + \mu g) = \frac{m_1 m_2 g(1+\mu) - \mu m_1^2 g}{m_1+m_2}$

Or more cleanly from $m_2$'s equation: $T = m_2(g - a) = \frac{m_1 m_2 g(1 + \mu)}{m_1+m_2}$... let me use the direct route:

From $m_2$: $T = m_2 g - m_2 a = m_2\left(g - \frac{(m_2-\mu m_1)g}{m_1+m_2}\right) = \frac{m_2 g(m_1+m_2 - m_2+\mu m_1)}{m_1+m_2} = \frac{m_1 m_2 g(1+\mu)}{m_1+m_2}$