Step 1: System Dynamics

Let $x$ be the length of the bar currently on the smooth (lower) section. The remaining length $(L-x)$ is on the rough (upper) section.

The smooth part wants to accelerate at $g \sin \theta$. The rough part wants to accelerate at $g \sin \theta – \mu g \cos \theta$. Since the front is accelerating faster than the back, the bar is in tension.

Total mass $m$. Linear mass density $\lambda = m/L$.

Total Force along the incline:

$$F_{net} = (\lambda x)g \sin \theta + \lambda(L-x)(g \sin \theta – \mu g \cos \theta)$$ $$F_{net} = mg \sin \theta – \mu \frac{m}{L}(L-x)g \cos \theta$$

Acceleration of the whole bar:

$$a = \frac{F_{net}}{m} = g \sin \theta – \mu g \cos \theta \left( \frac{L-x}{L} \right)$$

Step 2: Calculating Tension at the Interface

The tension $T$ is maximum at the interface between the smooth and rough parts of the bar. Consider the equation of motion for the lower segment (length $x$, mass $m_x = \lambda x$) which is on the smooth surface.

Forces on this segment: Gravity component ($m_x g \sin \theta$) pulling down, Tension ($T$) pulling up (back).

$$m_x a = m_x g \sin \theta – T$$ $$T = m_x (g \sin \theta – a)$$

Substitute $m_x = \frac{m}{L}x$ and the expression for $a$:

$$T = \frac{m}{L}x \left[ g \sin \theta – \left( g \sin \theta – \mu g \cos \theta \frac{L-x}{L} \right) \right]$$ $$T = \frac{m}{L}x \left[ \mu g \cos \theta \frac{L-x}{L} \right]$$ $$T(x) = \frac{\mu m g \cos \theta}{L^2} (Lx – x^2)$$

Step 3: Maximizing Tension

To find the maximum tension, we maximize the function $f(x) = Lx – x^2$ with respect to $x$.

$$\frac{df}{dx} = L – 2x = 0 \implies x = \frac{L}{2}$$

The tension is maximum when the bar is exactly half-way across the boundary.

Substitute $x = L/2$ into the tension equation:

$$T_{max} = \frac{\mu m g \cos \theta}{L^2} \left( L\left(\frac{L}{2}\right) – \left(\frac{L}{2}\right)^2 \right)$$ $$T_{max} = \frac{\mu m g \cos \theta}{L^2} \left( \frac{L^2}{4} \right)$$ $$T_{max} = 0.25 \mu m g \cos \theta$$