Problem

The cables of the system will each safely support a tension of 1500 kN. Based on this crit...

The cables of the system will each safely support a tension of 1500 kN. Based on this criterion, what is the largest safe value of the tension in cable AB?

Step-by-Step Solution

Solution 1

Given data:

The locations of point \(A\) is \((3.4,1,0) \mathrm{m}\)

The locations of point \(B\) is \((2,1,0) \mathrm{m}\)

The locations of point \(C\) is \((2.2,0,1) \mathrm{m}\)

The locations of point \(D\) is \((2.2,0,-1) \mathrm{m}\)

The locations of point \(E\) is \((1,1.2,0) \mathrm{m}\)

The locations of point \(F\) is \((0,1,4,1.2) \mathrm{m}\)

The locations of point \(G\) is \((0,1.4,-1.2) \mathrm{m}\)

Maximum safe tension for cable is \(1500 \mathrm{kN}\)

Position vector of cable \(B A\)

$$ \begin{aligned} &\mathbf{r}_{B A}=(3.4 \mathbf{i}+1 \mathbf{j}+0 \mathbf{k})-(2 \mathbf{i}+1 \mathbf{j}+0 \mathbf{k}) \\ &\mathbf{r}_{B A}=(3.4 \mathbf{i}+0 \mathbf{j}+0 \mathbf{k}) \end{aligned} $$

Position vector of cable \(B C\)

$$ \begin{aligned} &\mathbf{r}_{B C}=(2.2 \mathbf{i}+0 \mathbf{j}+1 \mathbf{k})-(2 \mathbf{i}+1 \mathbf{j}+0 \mathbf{k}) \\ &\mathbf{r}_{n C}=(0.2 \mathbf{i}-1 \mathbf{j}+\mathbf{k}) \end{aligned} $$

Position vector of cable \(B D\)

$$ \begin{aligned} &\mathbf{r}_{B D}=(2.2 \mathbf{i}+0 \mathbf{j}-1 \mathbf{k})-(2 \mathbf{i}+1 \mathbf{j}+0 \mathbf{k}) \\ &\mathbf{r}_{n D}=(0.2 \mathbf{i}-1 \mathbf{j}-1 \mathbf{k}) \end{aligned} $$

Position vector of cable \(B E\)

$$ \begin{aligned} &\mathbf{r}_{B E}=(1 \mathbf{i}+1.2 \mathbf{j}+0 \mathbf{k})-(2 \mathbf{i}+1 \mathbf{j}+0 \mathbf{k}) \\ &\mathbf{r}_{B E}=(-1 \mathbf{i}+0.2 \mathbf{j}+0 \mathbf{k}) \end{aligned} $$

Position vector of cable \(E F\)

$$ \begin{aligned} &\mathbf{r}_{E F}=(0 \mathbf{i}+1.4 \mathbf{j}+1.2 \mathbf{k})-(1 \mathbf{i}+1.2 \mathbf{j}+0 \mathbf{k}) \\ &\mathbf{r}_{E F}=(-1 \mathbf{i}+0.2 \mathbf{j}+1.2 \mathbf{k}) \end{aligned} $$

Position vector of cable EG

$$ \begin{aligned} &\mathbf{r}_{E G}=(0 \mathbf{i}+1.4 \mathbf{j}-1.2 \mathbf{k})-(1 \mathbf{i}+1.2 \mathbf{j}+0 \mathbf{k}) \\ &\mathbf{r}_{E G}=(-1 \mathbf{i}+0.2 \mathbf{j}-1.2 \mathbf{k}) \end{aligned} $$

Free body diagram showing the direction of forces:

Tension in the cable BA

$$ \begin{aligned} \mathbf{T}_{\mathrm{BA}} &=T_{B A} \frac{\mathbf{r}_{B A}}{\left|\mathbf{r}_{B \Lambda}\right|} \\ &=T_{B M} \frac{3.4 \mathbf{i}+0 \mathbf{j}+0 \mathbf{k}}{\sqrt{3.4^{2}+0^{2}+0^{2}}} \\ &=T_{B A}(\mathbf{i}) \mathrm{N} \end{aligned} $$

Tension in the cable \(B C\)

$$ \begin{aligned} \mathbf{T}_{\mathrm{BC}} &=T_{B C} \frac{\mathbf{r}_{B C}}{\left|\mathbf{r}_{B C}\right|} \\ &=T_{B C} \frac{0.2 \mathbf{i}-1 \mathbf{j}+\mathbf{l k}}{\sqrt{0.2^{2}+1^{2}+1^{2}}} \\ &=T_{B C}(0.14 \mathbf{i}-0.7 \mathbf{j}+0.7 \mathbf{k}) \mathrm{N} \end{aligned} $$

Tension in the cable BD

$$ \begin{aligned} \mathbf{T}_{\mathrm{BD}} &=T_{B D} \frac{\mathbf{r}_{B D}}{\left|\mathbf{r}_{B D}\right|} \\ &=T_{B D} \frac{0.2 \mathbf{i}-1 \mathbf{j}-1 \mathbf{k}}{\sqrt{0.2^{2}+1^{2}+1^{2}}} \\ &=T_{B D}(0.14 \mathbf{i}-0.7 \mathbf{j}-0.7 \mathbf{k}) \mathrm{N} \end{aligned} $$

Tension in the cable BE

$$ \begin{aligned} \mathbf{T}_{B E} &=T_{B E} \frac{\mathbf{r}_{B E}}{\left|\mathbf{r}_{B E}\right|} \\ &=T_{B E} \frac{-\mathbf{l}+0.2 \mathbf{j}+0 \mathbf{k}}{\sqrt{1^{2}+0.2^{2}+0^{2}}} \\ &=T_{B E}(-0.98 \mathbf{i}+0.196 \mathbf{j}) \mathrm{N} \end{aligned} $$

Tension in the cable EF

$$ \begin{aligned} \mathbf{T}_{E F} &=T_{E F} \frac{\mathbf{r}_{E F}}{\left|\mathbf{r}_{E F}\right|} \\ &=T_{E F} \frac{-1 \mathbf{i}+0.2 \mathbf{j}+1.2 \mathbf{k}}{\sqrt{1^{2}+0.2^{2}+1.2^{2}}} \\ &=T_{E F}(-0.635 \mathbf{i}+0.127 \mathbf{j}+0.762 \mathbf{k}) \mathrm{N} \end{aligned} $$

Tension in the cable EG

$$ \begin{aligned} \mathbf{T}_{\mathrm{EG}} &=T_{E G} \frac{\mathbf{r}_{E G}}{\left|\mathbf{r}_{E G}\right|} \\ &=T_{E G} \frac{-\mathrm{li}+0.2 \mathbf{j}-1.2 \mathbf{k}}{\sqrt{1^{2}+0.2^{2}+1.2^{2}}} \\ &=T_{E G}(-0.635 \mathbf{i}+0.127 \mathbf{j}-0.762 \mathbf{k}) \mathrm{N} \end{aligned} $$

From the equilibrium condition at \(\mathrm{B}\).

From equations (1), (2), (3) and (4)

$$ \begin{aligned} &\sum F=0 \\ &\mathbf{T}_{\mathrm{BA}}+\mathbf{T}_{\mathrm{BD}}+\mathbf{T}_{\mathrm{BC}}+\mathbf{T}_{\mathrm{BE}}=\mathbf{0} \ldots \ldots .(7) \\ &T_{B M}(\mathrm{i})+T_{B C}(0.14 \mathbf{i}-0.7 \mathbf{j}+0.7 \mathbf{k})+T_{B D}(0.14 \mathbf{i}-0.7 \mathbf{j}-0.7 \mathbf{k})+T_{B D}(-0.98 \mathbf{i}+0.196 \mathbf{j})=0 \\ &\left\{\begin{array}{l} \left(900 \mathrm{kN} \times 1+0.14 T_{B C}+0.14 T_{B D}-0.98 T_{B E}\right) \mathbf{i}+\left(0-0.7 T_{B C}-0.7 T_{s D}+0.196 T_{\text {sE }}\right) \mathbf{j} \\ +\left(0+0.7 T_{B C}-0.7 T_{B D}\right) \mathbf{k} \end{array}\right\}=0 \end{aligned} $$

Equating the \(\mathbf{i}\) coefficients

$$ \left(900 \mathrm{kN} \times 1+0.14 T_{B C}+0.14 T_{B D}-0.98 T_{B E}\right)=0 \ldots \ldots $$

Equating the \(\mathrm{j}\) coefficients

$$ \left(0-0.7 T_{B C}-0.7 T_{B D}+0.196 T_{B E}\right)=0 $$

Equating the \(k\) coefficients

$$ \sum F_{z}=\left(0+0.7 T_{B C}-0.7 T_{B D}\right) \mathbf{k}=0 \ldots \ldots $$

From equations (8), (9) and (10)

Therefore, the tensions in the cable are \(T_{B C}=133.928 \mathrm{~N}\), \(T_{B D}=133.928 \mathrm{~N}\), and \(T_{B E}=956.632 \mathrm{~N}\)

From the equilibrium condition at \(E\).

From equations (4), (5) and (6)

$$ \begin{aligned} &\sum F=0 \\ &\mathbf{T}_{E F}+\mathbf{T}_{\mathrm{EG}}+\mathbf{T}_{\mathrm{nE}}=0 \ldots \ldots \text { (11) } \\ &\left\{\begin{array}{l} T_{E F}(-0.635 \mathbf{i}+0.127 \mathbf{j}+0.762 \mathbf{k})+T_{E G}(-0.635 \mathbf{i}+0.127 \mathbf{j}-0.762 \mathbf{k}) \\ +T_{B E}(-0.98 \mathbf{i}+0.196 \mathbf{j}) \end{array}\right\}=0 \\ &\left\{\begin{array}{l} \left(+0.98 \times T_{B E}-0.635 T_{E F}-0.635 T_{E G}\right) \mathbf{i}+\left(-0.196 \times T_{B E}+0.127 T_{E F}+0.127 T_{E G}\right) \mathbf{j} \\ +\left(+0.762 T_{E F}-0.762 T_{E G}\right) \mathbf{k} \end{array}\right\}=0 \end{aligned} $$

Equating the \(\mathbf{i}\) coefficients

$$ \left(900 \mathrm{kN} \times 1+0.14 T_{B C}+0.14 T_{B D}-0.98 T_{B E}\right)=0 \ldots \ldots $$

Equating the jcoefficients

$$ \left(0-0.7 T_{B C}-0.7 T_{B D}+0.196 T_{B E}\right)=0 \ldots $$

Equating the \(\mathrm{k}\) coefficients

$$ \sum F_{z}=\left(0+0.7 T_{B C}-0.7 T_{B D}\right) \mathbf{k}=0 \ldots \ldots $$

From equations \((7),(8)\) and (9)

Therefore, the tensions in the cable are \(T_{E F}=738.18 \mathrm{kN}\) and \(T_{E G}=738.18 \mathrm{kN}\).

For tension of \(\mathrm{BA}\) is \(T_{\mathrm{B}}=900 \mathrm{kN}\) the tension in the cable \(\mathrm{BE}\) is \(T_{\mathrm{BE}}=956.632 \mathrm{~N}\) For given tension \(\mathrm{BE}\) is \(T_{\mathrm{RE}}=1500 \mathrm{~N}\) the tension in the cable \(\mathrm{AB}\) is

$$ \begin{aligned} T_{A B} &=\frac{1500}{956.632} \times 900 \\ &=1411.2 \mathrm{~N} \end{aligned} $$

Therefore, the maximum permissible tension in \(\mathrm{AB}\) is \(T_{A B}=1411.2 \mathrm{~N}\)

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