The load is lifted at constant velocity (in dynamic equilibrium), thus, the forces involve is similar to forces under static equilibrium.

$\Sigma F_V = 0$

$C_V = 120 + 120(\frac{3}{5})$

$C_V = 192 \, \text{kN}$

$\Sigma F_H = 0$

$C_H = 120(\frac{4}{5})$

$C_H = 96 \, \text{kN}$

By symmetry of vertical forces

$R_A = D_V = \frac{1}{2}(192)$

$R_A = D_V = 96 \, \text{kN}$

$\Sigma F_H = 0$

$D_H = 96 \, \text{kN}$

**At Joint A**

$\Sigma F_V = 0$

$F_{AB}(\frac{3}{5}) = 96$

$F_{AB} = 160 \, \text{kN}$

$\Sigma F_H = 0$

$F_{AC} = \frac{4}{5}F_{AB} = \frac{4}{5}(160)$

$F_{AC} = 128 \, \text{kN}$

**At Joint C**

$\Sigma F_V = 0$

$F_{BC} = 192 \, \text{kN}$

$\Sigma F_H = 0$

$F_{CD} + 96 = 128$

$F_{CD} = 32 \, \text{kN}$

**At Joint B**

$\Sigma F_H = 0$

$F_{BD}(\frac{4}{5}) = 160(\frac{4}{5})$

$F_{BD} = 160 \, \text{kN}$

$\Sigma F_V = 0$

$F_{BD}(\frac{3}{5}) + 160(\frac{3}{5}) = 192$

$160(\frac{3}{5}) + 160(\frac{3}{5}) = 192$

$192 = 192$ *Check!*

**At Joint D**

$\Sigma F_V = 0$

$96 = 160(\frac{3}{5})$

$96 = 96$ *Check!*

$\Sigma F_H = 0$

$32 + 96 = 160(\frac{4}{5})$

$128 = 128$ *Check!*

**Summary**

AB = 160 kN compression

AC = 128 kN tension

BC = 192 kN tension

CD = 32 kN tension

BD = 160 kN compression

**With the roller support at D and the hinge support at A**