Step 1

We are given the mass of the beam is $m = 120\;{\rm{kg}}$ and the point load acting over the beam is $F = 800\;{\rm{N}}$.

We are asked to determine the horizontal and vertical components of the initial reaction at the pin A, if the cord at B suddenly fails.

 
Step 2

The free body diagram of the system is shown as:

We have the length of the beam is $L = 2\;{\rm{m}} + 2\;{\rm{m}} = 4\;{\rm{m}}$.

We have the distance between points A and B is ${d_2} = 2\;{\rm{m}}$.

We have the distance between points B and free end is ${d_1} = 2\;{\rm{m}}$.

 
Step 3

The formula to calculate the mass moment of inertia of the beam about the point A is,

\[\begin{array}{l} {I_A} = {I_G} + m{\left( {{d_2}} \right)^2}\\ {I_A} = \frac{{m{L^2}}}{{12}} + m{\left( {{d_2}} \right)^2} \end{array}\]
 
Step 4

Substitute the values in the above expression.

\[\begin{array}{l} {I_A} = \frac{{\left( {120\;{\rm{kg}}} \right){{\left( {4\;{\rm{m}}} \right)}^2}}}{{12}} + \left( {120\;{\rm{kg}}} \right){\left( {2\;{\rm{m}}} \right)^2}\\ {I_A} = 640\;{\rm{kg}} \cdot {{\rm{m}}^2} \end{array}\]
 
Step 5

The formula to calculate the normal acceleration of the beam is,

\[{\left( {{a_G}} \right)_n} = {\omega ^2}{d_2}\]

Initially, the beam is in rest condition. i.e., $\omega = 0$.

 
Step 6

Substitute the values in the above expression.

\[\begin{array}{c} {\left( {{a_G}} \right)_n} = {\left( 0 \right)^2}\left( {2\;{\rm{m}}} \right)\\ {\left( {{a_G}} \right)_n} = 0 \end{array}\]
 
Step 7

The formula to calculate the tangential acceleration of the beam is,

\[{\left( {{a_G}} \right)_t} = \alpha {d_2}\]
 
Step 8

The formula to calculate the angular acceleration of the beam is,

\[FL + mg{d_2} = {I_A}\alpha \]

Here, g is the gravitational acceleration having a standard value of $9.81\;{\rm{m/}}{{\rm{s}}^2}$.

 
Step 9

Substitute the values in the above expression.

\[\begin{array}{c} \left[ \begin{array}{c} \left( {800\;{\rm{N}}} \right)\left( {4\;{\rm{m}}} \right) \times \left( {\frac{{1\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^2}}}{{1\;{\rm{N}}}}} \right)\\ + \left( {120\;{\rm{kg}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {2\;{\rm{m}}} \right) \end{array} \right] = \left( {640\;{\rm{kg}} \cdot {{\rm{m}}^2}} \right)\alpha \\ \alpha = 8.68\;{\rm{rad/}}{{\rm{s}}^2} \end{array}\]
 
Step 10

The formula to calculate the normal reaction at point A is,

\[{A_n} = m{\left( {{a_G}} \right)_n}\]
 
Step 11

Substitute the values in the above expression.

\[\begin{array}{c} {A_n} = \left( {120\;{\rm{kg}}} \right)\left( 0 \right)\\ {A_n} = 0 \end{array}\]
 
Step 12

The formula to calculate the tangential reaction at point A is,

\[\begin{array}{c} F + mg + {A_t} = m{\left( {{a_G}} \right)_t}\\ F + mg + {A_t} = m\left( {\alpha {d_2}} \right) \end{array}\]
 
Step 13

Substitute the values in the above expression.

\[\begin{array}{c} \left[ \begin{array}{c} \left( {800\;{\rm{N}}} \right)\left( {\frac{{1\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^2}}}{{1\;{\rm{N}}}}} \right)\\ + \left( {120\;{\rm{kg}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^2}} \right) + {A_t} \end{array} \right] = \left( {120\;{\rm{kg}}} \right)\left( {8.68\;{\rm{rad/}}{{\rm{s}}^2}} \right)\left( {2\;{\rm{m}}} \right)\\ {A_t} = 106\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^2} \times \left( {\frac{{1\;{\rm{N}}}}{{1\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^2}}}} \right)\\ {A_t} = 106\;{\rm{N}} \end{array}\]