91Ó°ÊÓ

The force on point A is\({F_{\rm{A}}}\).

The force on point B is\({F_{\rm{B}}}\).

The mass of the cantilever is \(m = 1200\;{\rm{kg}}\).

In order to determine the forces at points A and B, first, find the torque about the left end of the beam and apply the conditions of equilibrium.

The following is the free body diagram.

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

\(\begin{array}{c}\sum \tau = 0\\\left( {{F_{\rm{B}}} \times 20\;{\rm{m}}} \right) - \left( {mg \times 25\;{\rm{m}}} \right) = 0\end{array}\)

Here, \(g\) is the gravitational acceleration and \({F_{\rm{B}}}\) is the force on point B.

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

\(\begin{array}{c}\left( {{F_{\rm{B}}} \times 20\;{\rm{m}}} \right) - \left( {1200\;{\rm{kg}} \times 9.8\;{\rm{m/}}{{\rm{s}}^2} \times 25\;{\rm{m}}} \right) = 0\\{F_{\rm{B}}} = 14700\;{\rm{N}}\end{array}\)

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

\(\begin{array}{c}\sum {F_{\rm{y}}} = 0\\{F_{\rm{A}}} + {F_{\rm{B}}} - mg = 0\end{array}\)

Here, \({F_{\rm{A}}}\) is the force on point A.

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

\(\begin{array}{c}{F_{\rm{A}}} + \left( {14700\;{\rm{N}}} \right) - \left( {1200\;{\rm{kg}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right) = 0\\{F_{\rm{A}}} = - 2928\;{\rm{N}}\end{array}\)

Thus, the forces on points A and B are \({F_{\rm{A}}} = - 2928\;{\rm{N}}\) and \({F_{\rm{B}}} = 14700\;{\rm{N}}\), respectively.