Answer
$C_x=4.97KN$
$D_x=2.23KN$
$D_y=7.2KN$
Work Step by Step
According to the principle of impulse and momentum
$\Sigma M_A=\frac{dm}{dt}(d_{DB}v_B-d_{DA}v_A)$
We plug in the known values to obtain:
$-C_x\times 2=600(0-(1.5-0.12)\times 12)$
$\implies C_x=4968N=4.97KN$
We know that the resultant force in the $x$-direction is given as
$\Sigma F_x=\frac{dm}{dt}(v_B)_x-(v_A)_x$
$\implies D_x+C_x=600(12-0)$
This simplifies to:
$D_x=2232N=2.23KN$
Now the resultant force in the $y$-direction is given as
$\Sigma F_y=\frac{dm}{dt}[(v_B)_y-(v_A)_y]$
We plug in the known values to obtain:
$D_y=600(0-(-12))$
This simplifies to:
$D_y=7200N=7.2KN$
