## 2 Problem Description  
In this section, the mathematical description of a rigid block rocking on an elastic beam is presented. A sketch of the problem is given in Figure 1. We start by describing the motion of the beam and block due to arbitrary loading and then the constraints imposed by contact are given. Figure 2 shows a simply supported beam of unit mass per length \(m\), cross-sectional area \(A\) and second moment of area \(I\), [PERSON]'s modulus \(E\), and a damping coefficient per unit of length \(c\), loaded with arbitrary loading in the \(x\) and \(y\) directions \(p_{x}(x,t)\) and \(p_{y}(x,t)\). The displacements of the beam axis in the \(x\) and \(y\) directions are denoted as \(u(x,t)\) and \(v(x,t)\). By applying [PERSON]'s second law to an infinitesimal segment of the beam and assuming [PERSON]'s beam theory, the equations of motion for the beam are obtained as follows [40]:  
\[m\frac{\partial^{2}v(x,t)}{\partial t^{2}}+c\frac{\partial v(x,t)}{\partial t }+EI\frac{\partial^{4}v(x,t)}{\partial x^{4}}=p_{y}(x,t), \tag{1}\]  
\[m\frac{\partial^{2}u(x,t)}{\partial t^{2}}+c\frac{\partial u(x,t)}{\partial t }-EA\frac{\partial^{2}u(x,t)}{\partial x^{2}}=p_{x}(x,t). \tag{2}\]  
Figure 1: Rigid block rocking on an elastic beam. Figure 2: Simply supported beam loaded with arbitrarily distributed loading in the \(x\) and \(y\) directions. Figure 3a shows a rigid block at an arbitrary position in the \(xy\) plane. The block position is determined by the triplet \(X_{k}\), \(Y_{k}\), and \(\varphi_{k}\) that represents, respectively, the \(x\) and \(y\) coordinates of the block's centre of gravity and the rotation angle of the block. Furthermore, \(F_{x}\) and \(F_{y}\) denote, respectively, the \(x\) and \(y\) components of the resultant of all forces acting on the block's centre of gravity, while \(M_{k}\) is the moment with respect to the centre of gravity caused by all loads acting on the block (Figure 3b). We denote the block mass and moment of inertia as \(m_{k}\) and \(I_{k}\). Now, by [PERSON]'s second law, the equations of motion for the block are  
\[m_{k}\ddot{X}_{k}(t) =F_{x}(t), \tag{3}\] \[m_{k}\ddot{Y}_{k}(t) =F_{y}(t),\] (4) \[I_{k}\ddot{\varphi}_{k}(t) =M_{k}(t). \tag{5}\]  
Regarding the contact between the block and the beam, in this work, a simplification is introduced by assuming that contact is happening only between the block vertices and the beam axis. Both normal and frictional contact are considered. Figure 4 shows the forces \(r_{x}\) and \(r_{y}\) developed during contact and acting on each body. We now rewrite Equations (1)-(5) including the contact forces:  
\[m\frac{\partial^{2}v(x,t)}{\partial t^{2}}+c\frac{\partial v(x,t)}{\partial t }+EI\frac{\partial^{4}v(x,t)}{\partial x^{4}}=p_{y}(x,t)-r_{y}(t), \tag{6}\]  
\[m\frac{\partial^{2}u(x,t)}{\partial t^{2}}+c\frac{\partial u(x,t)}{\partial t }-EA\frac{\partial^{2}u(x,t)}{\partial x^{2}}=p_{x}(x,t)-r_{x}(t), \tag{7}\]  
\[m\frac{\partial^{2}v(x,t)}{\partial t^{2}}+c\frac{\partial v(x,t)}{\partial t }+EI\frac{\partial^{4}v(x,t)}{\partial x^{4}}=p_{y}(x,t), \tag{1}\]0 \[m\frac{\partial^{2}v(x,t)}{\partial t^{2}}+c\frac{\partial v(x,t)}{\partial t }+EI\frac{\partial^{4}v(x,t)}{\partial x^{4}}=p_{y}(x,t), \tag{1}\]1 (9) \[m\frac{\partial^{2}v(x,t)}{\partial t^{2}}+c\frac{\partial v(x,t)}{\partial t }+EI\frac{\partial^{4}v(x,t)}{\partial x^{4}}=p_{y}(x,t), \tag{1}\]2  
\[m\frac{\partial^{2}v(x,t)}{\partial t^{2}}+c\frac{\partial v(x,t)}{\partial t }+EI\frac{\partial^{4}v(x,t)}{\partial x^{4}}=p_{y}(x,t), \tag{1}\]3 Contact forces acting on the beam and block at the contact point. \[m\frac{\partial^{2}v(x,t)}{\partial t^{2}}+c\frac{\partial v(x,t)}{\partial t }+EI\frac{\partial^{4}v(x,t)}{\partial x^{4}}=p_{y}(x,t), \tag{1}\]4 (a) Block position coordinates; (b) forces acting on the block’s centre of gravity.