Solve for equilibrium ? Jun 24, 2016

${T}_{1} \approx 3954.7 k g f \mathmr{and} {T}_{2} \approx 1797.6 k g f$

Explanation: The situation given in the question has been shown in the figure.

• $\text{weight of the shaft} = 5.097 M g f = 5097 k g f$

• $\text{O is the point of suspension, OP and OQ are chains of 4m }$

• $\text{R is CG , OR is the vertical line along which total weight acts}$

• PR =1.25m and RQ = 2.75m

• ${T}_{1} = \text{Tension along PQ" and T_2 =" Tension along OQ}$

• In $\Delta P O Q , O P = O Q = Q P = 4 m \implies \Delta P O Q \text{ equilateral}$

• So In $\Delta P O Q , \text{ Each angle} = {60}^{\circ}$

Let
$\angle P O R = x \text{ then } \angle Q O R = 60 - x$

Now in $\Delta O P R , \frac{P R}{\sin} x = \frac{O R}{\sin} 60. \ldots \ldots \left(1\right)$

And in $\Delta O R Q , \frac{Q R}{\sin} \left(60 - x\right) = \frac{Q R}{\sin} 60. \ldots \left(2\right)$

Coparing (1) and (2) we get

$\frac{P R}{\sin} x = \frac{Q R}{\sin} \left(60 - x\right)$

$\implies \sin \frac{60 - x}{\sin} x = \frac{Q R}{P R} = \frac{2.75}{1.25} = \frac{11}{5} \ldots \ldots \left(3\right)$

Now considering the equilibrium of forces we can say that hrizontal components of ${T}_{1} \mathmr{and} {T}_{2}$ are equal in magnitude.
So
${T}_{1} \sin x = {T}_{2} \sin \left(60 - x\right)$

$\implies {T}_{1} / {T}_{2} = \sin \frac{60 - x}{\sin} x \ldots \ldots \left(4\right)$

Comparing (3) and (4) we can write

${T}_{1} / {T}_{2} = \frac{11}{5} = 2.2 \implies {T}_{1} = 2.2 \cdot {T}_{2}$

Now from equilibrium point of view the magnitude of Resultant of two tensions ${T}_{1} \mathmr{and} {T}_{2}$ acting at angle ${60}^{\circ}$ will be equal to
weight of the shaft i.e. $5097 k g f$

So we can write

${T}_{1}^{2} + {T}_{2}^{2} + 2 {T}_{1} \cdot {T}_{2} \cos {60}^{\circ} = {5097}^{2}$

Inserting ${T}_{1} = 2.2 {T}_{2} \mathmr{and} \cos {60}^{\circ} = \frac{1}{2}$ we get

${2.2}^{2} {T}_{2}^{2} + {T}_{2}^{2} + 2 \times 2.2 \cdot {T}_{2}^{2} \cdot \frac{1}{2} = {5097}^{2}$

$\implies {2.2}^{2} {T}_{2}^{2} + {T}_{2}^{2} + \cancel{2} \times 2.2 \cdot {T}_{2}^{2} \cdot \frac{1}{\cancel{2}} = {5097}^{2}$

$\implies 8.04 {T}_{2}^{2} = {5097}^{2}$

$\implies {T}_{2} = \frac{5097}{\sqrt{8.04}} = 1797.6 k g f$
and
${T}_{1} = 2.2 \times {T}_{2} = 2.2 \times 1797.6 = 3954.7 k g f$

Jun 24, 2016

$\left\mid {t}_{1} \right\mid = 3954.66$ and $\left\mid {t}_{3} \right\mid = 1797.57$

Explanation:

When in equilibrium, resultant weigth force passes across the shaft gravity center. The chain and the bar segment between anchored chains, form a equilateral triangle.

Let ${p}_{1} , {p}_{2} , {p}_{3}$ be the triangle vertices, and ${p}_{g}$ the point where the gravity center.

${p}_{1} = \left\{0 , 0\right\}$
${p}_{2} = \left\{2 , 2 \sqrt{3}\right\}$
${p}_{3} = \left\{4 , 0\right\}$
${p}_{g} = \left\{1.25 , 0\right\}$

The weight passes along the line defined by ${p}_{2} , {p}_{g}$ so if we have

${\vec{e}}_{1} = \frac{{p}_{2} - {p}_{1}}{\left\lVert {p}_{2} - {p}_{1} \right\rVert}$
${\vec{e}}_{3} = \frac{{p}_{2} - {p}_{3}}{\left\lVert {p}_{2} - {p}_{3} \right\rVert}$
$\vec{f} = \frac{{p}_{g} - {p}_{2}}{\left\lVert {p}_{g} - {p}_{2} \right\rVert}$

we can equate

$M g \vec{f} = {t}_{1} {\vec{e}}_{1} + {t}_{3} {\vec{e}}_{3}$

and also

$M g \left\langle\vec{f} , {\vec{e}}_{1}\right\rangle = {t}_{1} + {t}_{3} \left\langle{\vec{e}}_{3} , {\vec{e}}_{1}\right\rangle$
$M g \left\langle\vec{f} , {\vec{e}}_{2}\right\rangle = {t}_{1} \left\langle{\vec{e}}_{1} , {\vec{e}}_{3}\right\rangle + {t}_{3}$

Solving for ${t}_{1} , {t}_{3}$ we obtain

$\left\{{t}_{1} = - \frac{11 M g}{\sqrt{201}} , {t}_{2} = - \frac{5 M g}{\sqrt{201}}\right\}$

but $M g = 5097$ then $\left\mid {t}_{1} \right\mid = 3954.66$ and $\left\mid {t}_{3} \right\mid = 1797.57$

Jun 24, 2016

${T}_{1} = 3965.48$
${T}_{2} = 1802.49$

Explanation: $\text{All forces and their components}$ $\text{torque according to the point A:}$

$m g \cdot \cos \theta \cdot 1.25 = {T}_{2.} \sin \alpha \cdot 4$

$\cos \theta = 0.98$

${T}_{2} = \frac{m \cdot g \cdot \cos \theta \cdot 1.25}{4 \cdot \sin \alpha} = \frac{5097 \cdot 0.98 \cdot 1.25}{4 \cdot 0.866}$

${T}_{2} = \frac{6243.825}{3.464}$

${T}_{2} = 1802.49$ $\text{torque according to the point B:}$

${T}_{1} \cdot \sin \alpha \cdot 4 = m g \cdot \cos \theta \cdot 2.75$

${T}_{1} = \frac{m \cdot g \cdot \cos \theta \cdot 2.75}{4 \cdot \sin \alpha}$

${T}_{1} = \frac{5097 \cdot 0.98 \cdot 2.75}{4 \cdot 0.866}$

${T}_{1} = \frac{13736.415}{3.464}$

${T}_{1} = 3965.48$