Step 1

We are given the roller at A can support up to a load is ${R_A} < 4\;{\rm{kN}}$, and the pin at B can support up to a load is ${R_B} < 8\;{\rm{kN}}$.

We are asked the maximum intensity of the distributed load so that failure of the supports does not occur.

 
Step 2

The following is the free body diagram of the system:

We have the normal reaction exerted at point A is ${R_A}$.

We have the horizontal reaction exerted at point B is ${B_x}$.

We have the vertical reaction exerted at point B is ${B_y}$.

We have the length of horizontal member is ${d_1} = 4\;{\rm{m}}$.

We have the length of inclined member is ${d_2} = 3\;{\rm{m}}$.

 
Step 3

The normal reaction exerted at point A by moment of all the forces about point B is,

\[\begin{array}{c} \Sigma {M_B} = 0\\ \left[ \begin{array}{c} w{d_1}\left( {\frac{{{d_1}}}{2}} \right) - {R_A}\sin 30^\circ \left( {{d_2}\sin 30^\circ } \right)\\ - {R_A}\cos 30^\circ \left( {{d_2}\cos 30^\circ + {d_1}} \right) \end{array} \right] = 0 \end{array}\]
 
Step 4

Substitute the values in the above expression.

\[\begin{array}{c} \left[ \begin{array}{c} w\left( {4\;{\rm{m}}} \right)\left( {\frac{{4\;{\rm{m}}}}{2}} \right) - {R_A}\sin 30^\circ \left( {3\sin 30^\circ \;{\rm{m}}} \right)\\ - {R_A}\cos 30^\circ \left( {3\cos 30^\circ \;{\rm{m}} + 4\;{\rm{m}}} \right) \end{array} \right] = 0\\ 8w\;{{\rm{m}}^2} - 0.75{R_A}\;{\rm{m}} - 5.7{R_A}\;{\rm{m}} = 0\\ 6.45{R_A}\;{\rm{m}} = 8w\;{{\rm{m}}^2}\\ {R_A} = 1.24w\;{\rm{m}} \end{array}\]
 
Step 5

The horizontal reaction exerted at point B by equilibrium condition is,

\[\begin{array}{c} \Sigma {F_x} = 0\\ {B_x} = {R_A}\sin 30^\circ \end{array}\]
 
Step 6

Substitute the values in the above expression.

\[\begin{array}{l} {B_x} = \left( {1.24w\;{\rm{m}}} \right)\sin 30^\circ \\ {B_x} = 0.62w\;{\rm{m}} \end{array}\]
 
Step 7

The vertical reaction exerted at point B by equilibrium condition is,

\[\begin{array}{c} \Sigma {F_y} = 0\\ {B_y} + {R_A}\cos 30^\circ - w{d_1} = 0 \end{array}\]
 
Step 8

Substitute the values in the above expression.

\[\begin{array}{c} {B_y} + \left( {1.24w\;{\rm{m}}} \right)\cos 30^\circ - \left( w \right)\left( {4\;{\rm{m}}} \right) = 0\\ {B_y} = 2.93w\;{\rm{m}} \end{array}\]
 
Step 9

The normal reaction exerted at point B is,

\[{R_B} = \sqrt {{{\left( {{B_x}} \right)}^2} + {{\left( {{B_y}} \right)}^2}} \]
 
Step 10

Substitute the values in the above expression.

\[\begin{array}{l} {R_B} = \sqrt {{{\left( {0.62w\;{\rm{m}}} \right)}^2} + {{\left( {2.93w\;{\rm{m}}} \right)}^2}} \\ {R_B} = 3w\;{\rm{m}} \end{array}\]
 
Step 11

The maximum intensity of the distributed load so the roller will not fail is,

\[{R_A} < 4\;{\rm{kN}}\]
 
Step 12

Substitute the values in the above expression.

\[\begin{array}{c} 1.24w\;{\rm{m}} < 4\;{\rm{kN}}\\ w < 3.22\;{\rm{kN/m}} \end{array}\]
 
Step 13

The maximum intensity of the distributed load so the pin will not fail is,

\[{R_B} < 8\;{\rm{kN}}\]
 
Step 14

Substitute the values in the above expression.

\[\begin{array}{c} 3w\;{\rm{m}} < 8\;{\rm{kN}}\\ w < 2.67\;{\rm{kN/m}} \end{array}\]

The maximum intensity of the distributed load so that failure of the supports does not occur is $2.67\;{\rm{kN/m}}$.