The mass of the beam is\(m = 110\;{\rm{kg}}\).

The mass of the piano is \(M = 320\;{\rm{kg}}\).

In this question, the equilibrium conditions for the beam will be used to calculate the forces that the supports exert on the beam. Initially, calculate the torque about the left side of the beam.

The following is the free body diagram.

The relation to calculate the net torque can be written as:

\(\begin{array}{c}\sum \tau = 0\\\left( {{F_{\rm{R}}} \times l} \right) - \left( {mg \times \frac{l}{2}} \right) - \left( {Mg \times \frac{l}{4}} \right) = 0\\{F_{\rm{R}}} = g\left( {\frac{m}{2} + \frac{M}{4}} \right)\end{array}\)

Here, \(g\) is the gravitational acceleration and \({F_{\rm{R}}}\) is the force on the right end of the beam.

On plugging the values in the above relation, you get:

\(\begin{array}{l}{F_{\rm{R}}} = \left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {\frac{{110\;{\rm{kg}}}}{2} + \frac{{320\;{\rm{kg}}}}{4}} \right)\\{F_{\rm{R}}} = 1323\;{\rm{N}}\end{array}\)

The relation to calculate the force on the left end of the beam can be written as:

\(\begin{array}{c}\sum {F_{\rm{y}}} = 0\\{F_{\rm{L}}} + {F_{\rm{R}}} - mg - Mg = 0\end{array}\)

On plugging the values in the above relation, you get:

\(\begin{array}{c}{F_{\rm{L}}} + \left( {1323\;{\rm{N}}} \right) - \left( {110\;{\rm{kg}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right) - \left( {320\;{\rm{kg}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right) = 0\\{F_{\rm{L}}} = 2891\;{\rm{N}}\end{array}\)

Thus, the magnitudes of the forces on the left and right ends of the beam are \({F_{\rm{L}}} = 2891\;{\rm{N}}\) and \({F_{\rm{R}}} = 1323\;{\rm{N,}}\) respectively.