1. Analyzing the Forces

Let $v_1, v_2, v_3$ be the velocities of the top, middle, and bottom bars respectively. The viscous drag force between surfaces is proportional to relative velocity: $F = -k v_{rel}$.

The equations of motion for the three masses ($m$) are:

• Bar 1: Drag from Bar 2. $$ m\frac{dv_1}{dt} = -k(v_1 – v_2) $$
• Bar 2: Drag from Bar 1 (forward) and Bar 3 (backward). $$ m\frac{dv_2}{dt} = k(v_1 – v_2) – k(v_2 – v_3) $$
• Bar 3: Drag from Bar 2 (forward) and Floor (backward). $$ m\frac{dv_3}{dt} = k(v_2 – v_3) – k(v_3 – 0) $$

2. Integrating to Find Displacement

We can rewrite acceleration as $\frac{dv}{dt}$. Multiplying by $dt$ and integrating over the entire motion (from $t=0$ to $t=\infty$) allows us to relate momentum change to total displacement. Note that $\int v \, dt = x$.

Initial State: $v_1 = u$, $v_2 = 0$, $v_3 = 0$.
Final State: $v_1 = 0$, $v_2 = 0$, $v_3 = 0$.

Integrating the equations:

1. $$ \int m \, dv_1 = -k \int (v_1 – v_2) \, dt $$ $$ m(0 – u) = -k(x_1 – x_2) \implies mu = k(x_1 – x_2) \quad \text{— (i)} $$
2. $$ \int m \, dv_2 = k \int (v_1 – v_2) \, dt – k \int (v_2 – v_3) \, dt $$ $$ m(0 – 0) = k(x_1 – x_2) – k(x_2 – x_3) \implies x_1 – x_2 = x_2 – x_3 \quad \text{— (ii)} $$
3. $$ \int m \, dv_3 = k \int (v_2 – v_3) \, dt – k \int v_3 \, dt $$ $$ m(0 – 0) = k(x_2 – x_3) – k x_3 \implies x_2 – x_3 = x_3 \quad \text{— (iii)} $$

3. Solving the System

From (iii): $$ x_2 = 2x_3 $$

Substitute into (ii): $$ x_1 – 2x_3 = 2x_3 – x_3 \implies x_1 – 2x_3 = x_3 \implies x_1 = 3x_3 $$

Substitute into (i): $$ mu = k(3x_3 – 2x_3) \implies mu = k x_3 $$

Thus, we find the displacements:

$$ x_3 = \frac{mu}{k} $$ $$ x_2 = 2x_3 = \frac{2mu}{k} $$ $$ x_1 = 3x_3 = \frac{3mu}{k} $$

The displacements are:

Top bar: $\frac{3mu}{k}$, Middle bar: $\frac{2mu}{k}$, Bottom bar: $\frac{mu}{k}$