A homogenous square plate with a total weight of W 4 kN is

A homogenous square plate with a total weight of W = 4 kN is supported by four cables as shown in Figure 3. These four cables are connected to a spring FE at point E If the magnitude of the moment created by the tension in cable AE. around point F is 2 Squareroot 2 kN middot m. calculate the upstretched length of the spring FF. O is located at the geometric center of the plate. The coordinates for point E is (0, 0, 1 m) Neglect the thickness of the plate as well as the weights of the cables and spring. Also, neglect the elongation in the cables.

Solution

The length of the cable AE = (2 x 2^2 + 1)^0.5 = 3 m.

Assume tension in the cable AE is T_AE.

Co-ordinate of point A is (2, -2 , 0) and point E is (0, 0, 1).

So, the unit vector along AE direction = {k - (2i -2j) )}/3 =- 2/3 i + 2/3 j +1/3 k

So, cosine along z direction is 1/3 i.e T_AE x 1/3 = W/4

Or, T_AE = 3 kN

Elongation of the spring due to the weight of the plate = 4/10 = 0.4 m.

Assume the unstretched length of the spring = l₀.

Moment in the z-direction, Mz : W/4 x 2^0.5

Moment in the y-direction, My : T_AE x 2/3 x (l₀ + 0.4 + 0.5)

Moment in the x-direction, Mx : T_AE x -2/3 x (l₀ + 0.4 + 0.5)

So, total moment = (Mx^2 + My^2 + Mz^2)^0.5 = 2x1.414

By solving this equation, we can get l₀ .