Three bodies of masses m_1 = 4 kg , m_2 = 16 kg and m_3 = 8…
Three bodies of masses $m_1 = 4\ \mathrm{kg}$, $m_2 = 16\ \mathrm{kg}$ and $m_3 = 8\ \mathrm{kg}$ are connected by ropes and arranged as shown in the figure; the coefficient of friction with the surface is $\mu = 0.1$. Find the direction of motion and the acceleration.
Explanation of Method
Three connected bodies: direction and acceleration.
Step 1: Assumed arrangement
The figure is not available. The question about the direction of motion fits the classic layout: $m_2$ on a horizontal table, with $m_1$ and $m_3$ hanging over pulleys on opposite edges. Only $m_2$ touches the surface, so only it feels friction.
Step 2: Direction
The heavier side, $m_3$, pulls harder: $(m_3 - m_1)g = 39.2\ \mathrm{N}$, which exceeds the largest possible friction $\mu m_2 g = 15.7\ \mathrm{N}$. So the system moves toward $m_3$.
Step 3: Acceleration
\[ a = \frac{(m_3 - m_1)g - \mu m_2 g}{m_1 + m_2 + m_3} = \frac{39.24 - 15.70}{28}. \]
\[a \approx 0.84\ \mathrm{m/s^2}, \ \text{toward } m_3\]
Rope tensions: $T_1 = m_1(g + a) \approx 42.6\ \mathrm{N}$, $T_3 = m_3(g - a) \approx 71.8\ \mathrm{N}$.
Result
\[\boxed{\,a \approx 0.84\ \mathrm{m/s^2}, \ \text{toward } m_3\,}\]
Remark
About the original solution. It let the full weight of $m_3$ pull the system (so $m_3$ hangs) but also subtracted friction on $m_3$ and gave $m_1$ no weight term, which does not match any single arrangement; its answer was 1.82 m/s². If $m_1$ and $m_2$ are both on the table and only $m_3$ hangs, the consistent result is $a = (m_3 - \mu(m_1 + m_2))g/28 \approx 2.10\ \mathrm{m/s^2}$. The answer depends on the missing figure.