Let, frictional force is f and normal force is N.
Balancing horizontal forces :
[tex]50cos \ 30^o - f = 12 \times 4.13\\\\f = 50cos \ 30^o - 49.56\\\\f = -6.26\ N[/tex]
Also, balancing vertical forces, we get :
[tex]N = mg + 50sin \ 30^o\\\\N =( 12\times 9.8 ) + (50\times sin \ 30^o)\\\\N = 142.6\ N[/tex]
Therefore, the frictional force and normal force is -6.26 N and 142.6 N.