Statics: Problem about Equilibrium in 3-dimensions

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The discussion revolves around calculating the tensions in the corner and center cables of a system with two hinged rectangular plates, each weighing 800 kg. The force equilibrium equations in three dimensions are established, but the attempt to solve for the tensions results in a single equation, indicating insufficient information. The need for an additional equation arises from the fact that the plates are hinged, which complicates the analysis compared to a rigid body scenario. The participants question whether the center cable's effect should be considered separately for each plate and explore how the problem would change if the plates were welded into a single rigid body. The conclusion emphasizes the necessity of incorporating the hinge condition to derive the required equations for tension calculations.
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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 T0 in the center cable.
Plate Assembly.jpg

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,

ΣFx=0 ΣFy=0 and ΣFz=0

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

ΣMx=0 ΣMy=0 and ΣMz=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.
FBD of Plate Assembly.jpg

The tensions in the corner cables 'T' can be resolved into its horizontal and vertical components as Txy and Tz respectively.The horizontal component Txy can then be resolved into components along x- and y- directions as Tx and Ty 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.
Force Resolution Geometry.jpg


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,

ΣFz=0

⇒4Tsinα+T0-2.800(9.81)=0

⇒4T.##\frac {5}{\sqrt {46}}##+T0=15696

⇒2.95T+T0=15696

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

ΣMx=0

⇒2Tsinα(6)+T0(3)-2.800(9.81)(3)=0

⇒12T.##\frac {5}{\sqrt {46}}##+3T0=47088

⇒8.85T+3T0=47088

This is the same equation as obtained from ΣFz=0.

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

ΣMy=0

⇒ -2Tsinα(2##\sqrt {12}##)+800(9.81)(##\sqrt {12}##+##\frac {\sqrt {12}} {2}##)-T0(##\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}##) -T0(##\sqrt {12}##)=0

⇒ 10.22T+##\sqrt {12}##T0=54372.5
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 T0.We need at least two dissimilar equations in T and T0 to solve for T and T0.

Where is the problem?
[/B]
 
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Soumalya said:
Where is the problem?
All of your equations would still be true if the pair plates were to form a single rigid body. You need one which depends on their being hinged.
 
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haruspex said:
All of your equations would still be true if the pair plates were to form a single rigid body. You need one which depends on their being hinged.
I had a thought about this approach earlier when a couple of things pushed me into doubts.Firstly,if we treat the two plates hinged together separately as two different bodies should we consider the effect of the centre cable on both the plates?

Secondly,if instead of being hinged together about their common edge the plates were welded together to form the assembly as shown(so that they could be treated as a single rigid body) would it be possible to determine the values of the tensions in that case using the conditions for static equilibrium?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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