classical mechanics - The Parallelepiped on a Sloping Plane: A Reconsideration of an Old Friend

Consider the following elementary situation, which all the students of physics meet during their first-year study of classical mechanics.

A homogeneous rigid parallelepiped is at rest on a sloping plane, under the action of the weight force. Let $\alpha$ be the slope of the plane with respect to the horizontal ground.

Is there some elementary way of determining the distribution of the reaction forces acting at each point of the parallelepiped's base? Clearly, if $\alpha=0$, we expect that no friction is present and that the perpendicular reaction force is constant everywhere. But what is the answer for a general $\alpha$?

Mathematically, the problem is indeterminate, since we consider the parallelepiped a rigid body, and there are for sure infinitely many distributions (of reaction forces and internal forces) compatible with this static situation. But I think some reasonable assumption should let us find a unique solution. I am particularly interested in the distribution of the perpendicular reaction forces, that is in the pressure.

Given the symmetry of the problem, we can reduce it to a 2-dimensional one, in which the parallelepiped is replaced by a rectangle with surface density $\sigma$ which has a base along a straight line $x$ which meets the ground at an angle $\alpha$. Let $Oxy$ be a cartesian system, and let $R(\xi)=(R_{x}(\xi),R_{y}(\xi))$ be the reaction force per unit length acting on the rectangle at the point $(\xi,0)$ of its base. The issue is to determine $R(\xi)$. Note that, whatever the distribution of the friction forces $R_{x}$, the total moment of $R_{x}$ with respect to the center of mass of the rectangle is determined (its magnitude is equal to $\frac{h}{2} Mg \sin \alpha$, where $h$ is the height of the rectangle and $M$ its mass). So, the total moment of the forces $R_{y}$ with respect to the center of mass is known too: it must be opposite to that of the $R_{x}$. What assumption can we add to find the distribution of $R_{y}$?

In the special case $\alpha=0$, we can reason as follows. If you consider a vertical little strip of the rectangle, it is reasonable to assume that the neighboring parts of the rectangle exert no force on it, so that $R_{y}(\xi)=\sigma g h$ for all $\xi$ on the base. But how to reason for a general $\alpha$?

Thank you very much in advance.

PS I have never taken a course in continuum mechanics, so I apologize if my question should turn out to be trivial.

Answer

Actually, our problem is indeterminate also in the special case in which $\alpha=0$ and there is no friction, as the following reasoning shows.

Let $L$ be the length of our original rectangle $T$, and imagine to cut $T$ along the segment $S$ of extremes $(\frac{h}{2},0)$ and $(\frac{h}{2}+\epsilon,h)$ for small enough $\epsilon > 0$ so that each one of the two resulting trapezoids $T_1$ and $T_2$ is in equilibrium on the plane (with no contact forces between $T_1$ and $T_2$). This corresponds mathematically to assume that there are no internal forces in $T$ along $S$. Call $R(\xi)$ a solution of this configuration. Then $R_{y}(\xi)$ cannot be constant, since we must have \begin{equation} \int_{0}^{\frac{h}{2}} R_{y}(\xi) d \xi = g \sigma h \left( L + \frac{\epsilon}{2} \right) > \int_{\frac{h}{2}}^{h} R_{y}(\xi) d \xi = g \sigma h \left( L - \frac{\epsilon}{2} \right). \end{equation} Since this solution is as "physically reasonable" as the one found above in which $R_{y}(\xi)$ is constant, we conclude that our problem cannot be solved without information about the internal forces inside $T$.