Question

The spring has a stiffness of k=800N/m and an unstreches length of 200mm. Determine the force in cables BC and BD when the spring is held in the position shown.

Answer 1

Let F_BC and F_BD are force in the cables BC and BD.

From figure,

F_AB = k*x

= 800*(0.5 - 0.2)

= 240 N

let theta1 and theta2 are the angle made by F_BC and F_BD with +x axis.

theta1 = tan^-1(400/400)

= 45 degrees

theta2 = tan^-1(300/400)

= 36.87 degrees

As the point B is in equilibrium, net force acting on it must be zero.

Apply, Fnety on point B = 0

F_BC*sin(45) - F_BD*sin(36.87) = 0

F_BD = F_BC*sin(45)/sin(36.87)

F_BD = 1.5645*F_BC

now Apply, Fnetx on point B= 0

F_BC*cos(45) + F_BD*cos(36.87) - F_AB = 0

F_BC*cos(45) + 1.5645*F_BC*cos(36.87) = F_AB

1.9587*F_BC = 240

F_BC = 240/1.9587

= 122.5 N <<<<<<<<<------------Answer

F_BD = 1.5645*122.5

= 191.7 N <<<<<<<<<------------Answer