A body of mass m_1 = 2 kg lies on a body of mass m_2 = 1 kg…

A body of mass $m_1 = 2\ \mathrm{kg}$ lies on a body of mass $m_2 = 1\ \mathrm{kg}$, which rests on a smooth horizontal table. The two are connected by a light rope passing around a pulley fixed to a wall. What force must act on the lower body so that it moves with acceleration $a = g/2$? The coefficient of friction between the bodies is $\mu = 0.5$; the table is frictionless.

Explanation of Method

Two stacked bodies connected over a wall pulley.

Step 1: How the bodies move

When the lower body is pulled away from the wall, the rope drags the upper body toward the wall with the same acceleration $a$. They slide over each other, so the sliding friction $f = \mu m_1 g = 9.81\ \mathrm{N}$ acts on both: backward on each one relative to its motion.

Step 2: Equations

\[ \text{upper: } T - \mu m_1 g = m_1 a, \]

\[ \text{lower: } F - T - \mu m_1 g = m_2 a. \]

\[ T = m_1 a + \mu m_1 g = 9.81 + 9.81 = 19.62\ \mathrm{N}. \]

Step 3: Force

\[ F = (m_1 + m_2)\,a + 2\mu m_1 g = 3 \cdot 4.905 + 19.62. \]

\[F \approx 34.3\ \mathrm{N}\]

$F$ $m_1 = 2$ $m_2 = 1$ $a$ $f$ $f$ $F = (m_1 + m_2)a + 2\mu m_1 g$ $\approx 34.3\ \mathrm{N}$

Assumed arrangement: the rope from the upper body runs round a pulley fixed to the wall and back to the lower body. When the lower body is pulled right, the upper one moves left; friction between them (orange) brakes both of them.

Result

\[\boxed{\,F \approx 34.3\ \mathrm{N}\,}\]

Remark

About the original solution. It included friction on the upper body but forgot it on the lower one: by Newton's third law the upper body rubs back on the lower one with the same 9.81 N. That gave 24.53 N instead of 34.3 N. (The arrangement, with the rope around a wall pulley, is the standard one for this problem; the figure is not available.)