Question 3.23: A force P = 5000 N is applied at the center C of the beam AB...

Considering FBD of the beam

ΣX = 0,               R_{A X}=5000 \cos 30^{\circ}

R_{A X}= 4330.12 N

ΣY = 0,              R_{A Y}+R_{B}=5000 \sin 30^{\circ}=2500  N

R_{B} = 1250 N

from equation (1), R_{AY}= 2500 −1250 = 1250 N

R_{A}=\sqrt{R_{A H}^{2}+R_{A V}^{2}}

=\sqrt{(4330.12)^{2}+(1250)^{2}}

R_{A} = 4506.93 N

\tan \alpha=\frac{R_{A Y}}{R_{A X}}=\frac{1250}{4330.12}

\alpha=16.10°