Let the acceleration of the center of gravity of the pendulum be

     a_cg = a_xi + a_yj  

For simplicity, the FBD and MAD are combined into one diagram. The origin of the x-y axis is placed at the pin support, and the reaction forces of the pin support are labeled. Newton's Second Law can be used to sum the forces in both directions, giving

     ΣF_x = A_x = ma_x

     ΣF_y = A_y - mg = ma_y

The angular acceleration, α, is assumed to go in the counterclockwise direction (if wrong it will be negative). The angular acceleration is the same around any point since the α vector actually points directly out of the screen. Summing the moments about A gives

     ΣM_A = I_A α

     -mgd sinθ = I_A α

or

For θ = 0, which is the position requested by the problem, angular acceleration becomes simply

     α(0) = (-1/7.289) (3.17) (32.2) (1.442) sin0 = 0

The two force equation (F_x and F_y) can be used to find the forces A_x and A_y, but first the acceleration of the center of gravity a_xi and a_yj need to be determined.

The acceleration of the center of gravity of the pendulum can be expressed in terms of the acceleration of the pivot point A as

     a_cg = a_A + α × r_cg/A - ω²r

           = a_xi + a_yj  

Because point A is pinned, its acceleration is zero,

     a_A = 0

Recall, α(0) = 0, which reduces the equation to

     a_cg-θ=0 => -ω²r = a_xi + a_yj

To this equation, the angular velocity ω when θ = 0 needs to be determined. Recall that

which can be integrated to find an expression for ω giving

For θ = 0, the angular velocity becomes

     ω(0) = 1.173 rad/s

Substituting gives,

     a_cg-θ=0 = -(1.173)² (-1.442)j

                 = 1.984j ft/s² = a_xi + a_yj

Equate like terms,

      a_x-θ=0 = 0         a_y-θ=0 = 1.984 ft/s²

Substitute back into the initial force equations,

     F_x = A_x = ma_x

     A_x-θ=0 = 0

     A_y-θ=0 = m (a_y + g)

                = 3.17 (1.984 + 32.2) = 108.4 lb