Answer:
Tension in cable BE= 196.2 N
Reactions A and D both are 73.575 N
Explanation:
The free body diagram is as attached sketch. At equilibrium, sum of forces along y axis will be 0 hence
[tex]T_{BE}-W=0[/tex] hence
[tex]T_{BE}=W=20*9.81=196.2 N[/tex]
Therefore, tension in the cable, [tex]T_{BE}=196.2 N[/tex]
Taking moments about point A, with clockwise moments as positive while anticlockwise moments as negative then
[tex]196.2\times 0.125- 196.2\times 0.2+ D_x\times 0.2=0[/tex]
[tex]24.525-39.24+0.2D_x=0[/tex]
[tex]D_x=73.575 N[/tex]
Similarly,
[tex]A_x-D_y=0[/tex]
[tex]A_x=73.575 N[/tex]
Therefore, both reactions at A and D are 73.575 N
