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## Homework Statement

The two uniform rectangular plates each weighing 800 kg are freely hinged about their common edge and suspended by the central cable and four symmetrical corner cables. Calculate the tension T in each of the corner cables and the tension T

_{0}in the center cable.

All dimensions in the figure are in meters.

2. Homework Equations

The scalar force equilibrium equations in three mutually perpendicular directions x-,y- and z-,i.e,

ΣF

_{x}=0 ΣF

_{y}=0 and ΣF

_{z}=0

The scalar moment equilibrium equations about three mutually perpendicular axes through a point,i.e,

ΣM

_{x}=0 ΣM

_{y}=0 and ΣM

_{z}=0[/B]

## The Attempt at a Solution

The Free Body Diagram of the plate assembly is drawn below along with the choice of the coordinate axes.

The tensions in the corner cables 'T' can be resolved into its horizontal and vertical components as T

_{xy}and T

_{z}respectively.The horizontal component T

_{xy}can then be resolved into components along x- and y- directions as T

_{x}and T

_{y}respectively in the x-y plane.The angles which orient the line of action of a corner cable tension T can be determined as illustrated in the figure below.

The force equilibrium equations in the x- and y- directions are already satisfied since the identical x- and y- components of all the corner cable tensions cancel each other.

The force equilibrium equation in the z-direction yields,

ΣF

_{z}=0

⇒4Tsinα+T

_{0}-2.800(9.81)=0

⇒4T.##\frac {5}{\sqrt {46}}##+T

_{0}=15696

⇒2.95T+T

_{0}

**=15696**

The moment equilibrium equation about the x-axis through O gives,

ΣM

⇒2Tsinα(6)+T

⇒12T.[/B]

The moment equilibrium equation about the x-axis through O gives,

ΣM

_{x}=0⇒2Tsinα(6)+T

_{0}(3)-2.800(9.81)(3)=0⇒12T.

**##\frac {5}{\sqrt {46}}##+3T**

⇒8.85T

This is the same equation as obtained from_{0}=47088⇒8.85T

**+3T**_{0}=47088This is the same equation as obtained from

**ΣF**

_{z}=0.**The moment equilibrium equation about the y-axis through O gives,**

ΣM

⇒ -2Tsinα(2##\sqrt {12}##)+800(9.81)(##\sqrt {12}##+##\frac {\sqrt {12}} {2}##)-T

⇒ΣM

_{y}=0⇒ -2Tsinα(2##\sqrt {12}##)+800(9.81)(##\sqrt {12}##+##\frac {\sqrt {12}} {2}##)-T

_{0}(##\sqrt {12}##)+800(9.81)(##\frac {\sqrt {12}} {2}##)=0⇒

**-2T.##\frac {5} {\sqrt {46}}##.(2##\sqrt {12}##)+800(9.81)(##\sqrt {12}##+##\frac {\sqrt {12}} {2}##+##\frac {\sqrt {12}} {2}##) -T**

⇒ 10.22T+##\sqrt {12}##T

which is again the same equation as obtained before.

The moment equilibrium equation about the z-axis through O is already satisfied.

So as it is evident the equations of equilibrium are resulting into a single equation in T and T

Where is the problem?_{0}(##\sqrt {12}##)=0⇒ 10.22T+##\sqrt {12}##T

_{0}=54372.5which is again the same equation as obtained before.

The moment equilibrium equation about the z-axis through O is already satisfied.

So as it is evident the equations of equilibrium are resulting into a single equation in T and T

_{0}.We need at least two dissimilar equations in T and T_{0}to solve for T and T_{0}.Where is the problem?