Answer
$F_{CD}=288.87KN$
Work Step by Step
We can determine the required force as follows:
We know that
$s_B+2s_L=L$
Differentiating both sides, we obtain:
$v_B+2v_L=0$
Again differentiating the above equation, we obtain:
$a_B+2a_L=0$
$\implies a_B=-2a_L$
$\implies 2=-2a_L$
$\implies a_L=-1m/s^2$
The negative sign shows that its direction is upward.
According to the equation of motion in y-direction
$\Sigma F_y=ma_y$
$\implies T+T-W=ma_L$
We plug in the known values to obtain:
$2T-8000(9.81)=8000\times 1$
This simplifies to:
$T=43240N$
The sum of of moments about is given as
$\Sigma M_A=0$
$\implies F_{CD}(4cos60)-W_B(6cos60)-2T(12cos60)=0$
$\implies F_{CD}(2)-(2000\times 9.81)(3)-2(43240)(6)=0$
This simplifies to:
$F_{CD}=288870N=288.87KN$