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Introductory Physics Homework Help
Statics: Problem about Equilibrium in 3-dimensions
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[QUOTE="Soumalya, post: 5515424, member: 507042"] [h2]Homework Statement[/h2] 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[SUB]0[/SUB] in the center cable. [ATTACH=full]102973[/ATTACH] 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[SUB]x[/SUB]=0 ΣF[SUB]y[/SUB]=0 and ΣF[SUB]z[/SUB]=0 The scalar moment equilibrium equations about three mutually perpendicular axes through a point,i.e, ΣM[SUB]x[/SUB]=0 ΣM[SUB]y[/SUB]=0 and ΣM[SUB]z[/SUB]=0[/B][h2]The Attempt at a Solution[/h2] The Free Body Diagram of the plate assembly is drawn below along with the choice of the coordinate axes. [ATTACH=full]102974[/ATTACH] The tensions in the corner cables 'T' can be resolved into its horizontal and vertical components as T[SUB]xy[/SUB] and T[SUB]z[/SUB] respectively.The horizontal component T[SUB]xy[/SUB] can then be resolved into components along x- and y- directions as T[SUB]x[/SUB] and T[SUB]y[/SUB] 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. [ATTACH=full]102975[/ATTACH] 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[SUB]z[/SUB]=0 ⇒4Tsinα+T[SUB]0[/SUB]-2.800(9.81)=0 ⇒4T.##\frac {5}{\sqrt {46}}##+T[SUB]0[/SUB]=15696 ⇒2.95T+T[SUB]0[/SUB][B]=15696 The moment equilibrium equation about the x-axis through O gives, ΣM[SUB]x[/SUB]=0 ⇒2Tsinα(6)+T[SUB]0[/SUB](3)-2.800(9.81)(3)=0 ⇒12T.[B]##\frac {5}{\sqrt {46}}##+3T[SUB]0[/SUB]=47088 ⇒8.85T[B][B][B]+3T[SUB]0[/SUB]=47088[/B][/B][/B] This is the same equation as obtained from [B]ΣF[SUB]z[/SUB]=0. [B][B]The moment equilibrium equation about the y-axis through O gives, ΣM[SUB]y[/SUB]=0 ⇒ -2Tsinα(2##\sqrt {12}##)+800(9.81)(##\sqrt {12}##+##\frac {\sqrt {12}} {2}##)-T[SUB]0[/SUB](##\sqrt {12}##)+800(9.81)(##\frac {\sqrt {12}} {2}##)=0 ⇒[B][B][B][B][B][B] -2T.##\frac {5} {\sqrt {46}}##.(2##\sqrt {12}##)+800(9.81)(##\sqrt {12}##+##\frac {\sqrt {12}} {2}##+##\frac {\sqrt {12}} {2}##) -T[SUB]0[/SUB](##\sqrt {12}##)=0 ⇒ 10.22T+##\sqrt {12}##T[SUB]0[/SUB]=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 T[SUB]0[/SUB].We need at least two dissimilar equations in T and T[SUB]0[/SUB] to solve for T and T[SUB]0[/SUB]. Where is the problem?[/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B] [/QUOTE]
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Introductory Physics Homework Help
Statics: Problem about Equilibrium in 3-dimensions
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