Shear force and moment

Consider a uniform beam of mass $12\text{kg}$ and length $20\text{m}$ . Let $x$ be the horizontal distance, in meters, from the left hand end of the beam. Two supports A and B support the beam at $x=0\text{m}$ and $x=16\text{m}$ respectively. A (rectangular) mass $M_1$ of length $6\text{m}$ is placed on the beam, with its geometric center at $x=5\text{m}$ . Its mass density, in $\text{kg/m}$ , is modeled by $\rho(x)=x$ . Another mass $M_2$ is also placed on the beam with its left edge at $x=19\text{m}$ . If the equations for the shear force and bending moment along the beam are $V(x)$ and $M(x)$ respectively, calculate $\frac{M(4)+V(4)}{g}$ .

Details and Assumptions :

• Positive shear force is downwards; positive moment is anti-clockwise. And vice versa.
• Ignore the width of $M_1$ .
• $M_2$ is a square mass of side length $2\text{m}$ and mass density $5\text{kg/m}^2$ .
• $g$ is the gravity constant, $g=9.81\text{ms}^{-2}$ .