Answer
$T=25.1KN$
$a=2.74m/s^2$
Work Step by Step
We can determine the required tension and acceleration as follows:
We know that the sum of moments about $A$ is given as
$\Sigma M_A=0$
$\implies F_{CD}(4cos60)-W_B(6cos60)-2T(12cos60)=0$
We plug in the known values to obtain:
$(180000)(2)-(2000\times 9.81)(3)-2T(6)=0$
This simplifies to:
$T=25.1KN$
Now, we apply the equation of motion in y-direction
$\Sigma F_y=ma_y$
$\implies 2T-W=ma$
We plug in the known values to obtain:
$2(25095)-4000(9.81)=4000a$
This simplifies to:
$a=2.74m/s^2$
