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 $$T_{A B}=1411.2 \mathrm{~N}$$