Consider a uniform metal rod of length $L = 1.0\,\text{m}$ placed on a rough horizontal surface. When the temperature of different portions of the rod is changed according to the profile $\Delta \theta(x)$, the rod undergoes thermal expansion (or contraction).

Let $\alpha$ be the coefficient of linear expansion. The local strain due to temperature change is given by $\epsilon_{th} = \alpha \Delta \theta(x)$. Assuming the rod expands freely first, the displacement of any point $x$ relative to the left end ($x=0$) would be: $$ Y(x) = \int_0^x \alpha \Delta \theta(x’) \, dx’ $$ The total change in length of the rod is: $$ \Delta l = Y(L) = \int_0^L \alpha \Delta \theta(x) \, dx $$

Since the tabletop is rough, friction acts on the rod. In the steady state, the rod is in equilibrium, meaning the net friction force on the rod must be zero.

Let $y(x)$ be the actual displacement of a point at position $x$. The direction of the friction force at any point opposes the direction of displacement.

• If a portion moves to the right ($y(x) > 0$), friction acts to the left.
• If a portion moves to the left ($y(x) < 0$), friction acts to the right.

For a uniform rod, the friction force per unit length $f$ is constant magnitude. The condition for zero net force is: $$ \sum F_{ext} = \int_0^L f \cdot (-\text{sgn}(y(x))) \, dx = 0 $$ This implies that the total length of the rod moving to the left must equal the total length moving to the right. $$ L_{left} = L_{right} = \frac{L}{2} = 0.5\,\text{m} $$

The actual displacement $y(x)$ is related to the free expansion $Y(x)$ by a constant shift $C$, which represents the displacement of the left end ($x=0$). $$ y(x) = Y(x) – C $$ The constant $C$ is determined such that the equilibrium condition ($L_{left} = L_{right}$) is met. For the specific temperature profile shown in the problem graph (which alternates in sign), the function $y(x)$ crosses the zero axis at three distinct points (as shown in the diagram above).

Let $u_L$ be the displacement of the left end and $u_R$ be the displacement of the right end. $$ u_L = y(0) = -C $$ $$ u_R = y(L) = \Delta l – C $$ The magnitudes of displacement must distribute the total extension $\Delta l$ such that the friction balances. Based on the areas defined by the graph geometry which satisfy the $L/2$ condition, the shift is found to be: $$ C = \frac{1}{6} \Delta l $$

Therefore, the displacements are: $$ u_L = -\frac{1}{6} \Delta l \quad (\text{Directed towards Left}) $$ $$ u_R = \Delta l – \frac{1}{6} \Delta l = \frac{5}{6} \Delta l \quad (\text{Directed towards Right}) $$