Authors: Russell C. Hibbeler
ISBN-13: 978-0133915426
See our solution for Question 6FP from Chapter 5 from Hibbeler's Engineering Mechanics.
We are given that the force applied by the person is $F = 250\;{\rm{N}}$.
We are asked to calculate the reaction force on points A, B and C.

We have the distance between points A and B is ${d_1} = 0.15\;{\rm{m}}$.
We have the distance between point B and C is ${d_2} = 0.2\;{\rm{m}}$.
We have the distance between point C and arm is ${d_3} = 0.4\;{\rm{m}}$.
The free body diagram of the system is shown as,

Applying the force equilibrium equation along x-axis:
\[\begin{array}{c} \sum {F_x} = 0\\ {F_C}\cos 60^\circ - F\cos 30^\circ = 0 \end{array}\]Substitute the known values in the equation:
\[\begin{array}{c} {F_C}\left( {0.5} \right) - \left( {250\;{\rm{N}}} \right)\left( {0.866} \right) = 0\\ {F_C} = \frac{{216.5\;{\rm{N}}}}{{0.5}}\\ = 433\;{\rm{N}} \end{array}\]Applying the moment of forces at point B:
\[ - {F_A}\sin 30^\circ {d_1} - {F_C}{d_2} + F\cos 30^\circ \left( {{d_2} + {d_3}} \right) = 0\]Substitute the known values in the equation:
\[\begin{array}{c} \left\{ \begin{array}{l} - {F_A}\left( {0.5} \right)\left( {0.15\;{\rm{m}}} \right) - \\ \left( {433\;{\rm{N}}} \right)\left( {0.2\;{\rm{m}}} \right) + \\ \left( {250\;{\rm{N}}} \right)\left( {0.866} \right)\left( {0.2\;{\rm{m}} + 0.4\;{\rm{m}}} \right) \end{array} \right\} = 0\\ {F_A}\left( {0.075\;{\rm{m}}} \right) = 129.9\;{\rm{N}} \cdot {\rm{m}} - 86.6\;{\rm{N}} \cdot {\rm{m}}\\ {F_A} = \frac{{\left( {43.3\;{\rm{N}} \cdot {\rm{m}}} \right)}}{{\left( {0.075\;{\rm{m}}} \right)}}\\ = 577.33\;{\rm{N}} \end{array}\]Applying the force equilibrium equation along y-axis:
\[ - F\sin 30^\circ + {F_C}\sin 60^\circ + {F_B} - {F_A} = 0\]Substitute the known values in the equation:
\[\begin{array}{c} \left\{ \begin{array}{l} - \left( {250\;{\rm{N}}} \right)\left( {0.5} \right) + \\ \left( {433\;{\rm{N}}} \right)\left( {0.866} \right) + \\ {F_B} - 577.33\;{\rm{N}} \end{array} \right\} = 0\\ {F_B} = 125\;{\rm{N}} - 374.98\;{\rm{N}} + 577.33\;{\rm{N}}\\ = 327.35\;{\rm{N}} \end{array}\]