Step 1

We are given the triangular force per unit length $F = 400\;{\rm{N/m}}$ at $a = 3\;{\rm{m}}$ from point $A$ and $b = 3\;{\rm{m}}$ from $B$.

We are asked to determine the reactions at the support.

 
Step 2

First draw the free body diagram for the given beam.

Figure-(1)

 
Step 3

Find the angle of slope using the following relation.

\[\tan \phi = \frac{x}{y}\]

Substitute the known values in the above expression.

\[\begin{array}{c} \tan \phi = \frac{3}{4}\\ \phi = {\tan ^{ - 1}}\left( {0.75} \right)\\ \phi = 36.87^\circ \end{array}\]
 
Step 4

Find the force acting on the beam using the following relation.

\[F' = \frac{1}{2}\left( F \right)\left( {a + b} \right)\]

Substitute the known values in the above expression.

\[\begin{array}{c} F' = \frac{1}{2}\left( {400\;{\rm{N/m}}} \right)\left( {3\;{\rm{m}} + 3\;{\rm{m}}} \right)\\ = 1200\;{\rm{N}} \end{array}\]
 
Step 5

Now find vertical reaction force at $B$ taking moment at $A$ equal to zero using the following relation.

\[ - F' \times a + {B_x} \times 0 + {B_y} \times \left( {a + b} \right) = 0\]

Substitute the known values in the above expression.

\[\begin{array}{c} - \left( {1200\;{\rm{N}}} \right) \times \left( {3\;{\rm{m}}} \right) + {B_x} \times 0 + {B_y} \times \left( {3\;{\rm{m}} + 3\;{\rm{m}}} \right) = 0\\ 3600\;{\rm{N}} \cdot {\rm{m}} = {B_y} \times \left( {6\;{\rm{m}}} \right)\\ {B_y} = 600\;{\rm{N}} \end{array}\]
 
Step 6

Now find vertical reaction force at $A$ taking moment at $B$ equal to zero using the following relation.

\[F' \times b + {N_A}\cos \phi \times \left( {a + b} \right) - {N_A}\sin \phi \times \left( 0 \right) = 0\]

Substitute the known values in the above expression.

\[\begin{array}{c} \left( {1200\;{\rm{N}}} \right) \times \left( {3\;{\rm{m}}} \right) - {N_A}\cos 36.87^\circ \times \left( {3\;{\rm{m}} + 3\;{\rm{m}}} \right) + 0 = 0\\ 3600\;{\rm{N}} \cdot {\rm{m}} = \left( {4.8\;{\rm{m}}} \right){N_A}\\ {N_A} = 750\;{\rm{N}} \end{array}\]
 
Step 7

Find sum of all the forces acting in $x$ direction using the following relation.

\[{B_x} - {N_A}\sin \phi = 0\]

Substitute the known values in the above expression.

\[\begin{array}{c} {B_x} - \left( {750\;{\rm{N}}} \right)\left( {\sin 36.87^\circ } \right) = 0\\ {B_x} = 450\;{\rm{N}} \end{array}\]