Chapter 2.4, Problem 2.96P

To determine

The tensions in cables $AB$ and $AD$.

Answer to Problem 2.96P

The tensions in cables $AB$ and $AD$ are $\underset{\_}{490\text{N}}$ and $\underset{\_}{515\text{N}}$ respectively.

Explanation of Solution

Write the equation to find the magnitude of the vector $AB$.

$\left|AB\right|=\sqrt{A{B}_x^2+A{B}_y^2+A{B}_z^2}$ (I)

Write the equation to find the unit vector along $\overrightarrow{AB}$.

$\lambda =\frac{\overrightarrow{AB}}{\left|AB\right|}$ (II)

Write the equation of $T_{AB}$ in terms of its rectangular components.

$T_{AB}=T_{AB}\lambda$ (III)

Here, $T_{AB}$ is the tension in the cable $AB$, and $\lambda$ is the unit vector along $\overrightarrow{AB}$.

Write the equation to find the magnitude of the vector $AC$.

$\left|AC\right|=\sqrt{A{C}_x^2+A{C}_y^2+A{C}_z^2}$ (IV)

Write the equation to find the unit vector along $\overrightarrow{AC}$.

$\lambda =\frac{\overrightarrow{AC}}{\left|AC\right|}$ (V)

Write the equation of $T_{AC}$ in terms of its rectangular components.

$T_{AC}=T_{AC}\lambda$ (VI)

Here, $T_{AC}$ is the tension in the cable $AC$, and $\lambda$ is the unit vector along $\overrightarrow{AC}$.

Write the equation to find the magnitude of the vector $AD$.

$\left|AD\right|=\sqrt{A{D}_x^2+A{D}_y^2+A{D}_z^2}$ (VII)

Write the equation to find the unit vector along $\overrightarrow{AD}$.

$\lambda =\frac{\overrightarrow{AD}}{\left|AD\right|}$ (VIII)

Write the equation of $T_{AD}$ in terms of its rectangular components.

$T_{AD}=T_{AD}\lambda$ (IX)

Here, $T_{AD}$ is the tension in the cable $AD$, and $\lambda$ is the unit vector along $\overrightarrow{AD}$.

Write the equation to find the resultant of the forces exerted by the three cables at $A$.

$R=T_{AB}+T_{AC}+T_{AD}$ (X)

Conclusion:

Refer Fig.P2.89 and calculate the vector coordinates of the vector $AB$.

$\overrightarrow{AB}=-\left(320\text{mm}\right)i-\left(480\text{mm}\right)j+\left(360\text{mm}\right)k$

Substitute $-320\text{mm}$ for $A{B}_x$, $-480\text{mm}$ for $A{B}_y$, and $360\text{mm}$ for $A{B}_z$ in the equation (I).

$\begin{array}{c}AB=\sqrt{{\left(-320\text{mm}\right)}^2+{\left(-480\text{mm}\right)}^2+{\left(360\text{mm}\right)}^2}\\ =680\text{mm}\end{array}$

Substitute $-\left(320\text{mm}\right)i-\left(480\text{mm}\right)j+\left(360\text{mm}\right)k$ for $\overrightarrow{AB}$, and $680\text{mm}$ for $AB$ in equation (II).

$\begin{array}{c}\lambda =\frac{-\left(320\text{mm}\right)i-\left(480\text{mm}\right)j+\left(360\text{mm}\right)k}{680\text{mm}}\\ =-\frac{320}{680}i-\frac{480}{680}j+\frac{360}{680}k\end{array}$

Substitute $408\text{N}$ for $T_{AB}$, and $-\frac{320}{680}i-\frac{480}{680}j+\frac{360}{680}k$ for $\lambda$ in equation (III).

$T_{AB}=T_{AB}\left(-\frac{320}{680}i-\frac{480}{680}j+\frac{360}{680}k\right)$

Refer Fig.P2.89 and calculate the vector coordinates of the vector $AC$.

$\overrightarrow{AC}=+\left(450\text{mm}\right)i-\left(480\text{mm}\right)j+\left(360\text{mm}\right)k$

Substitute $450\text{mm}$ for $A{C}_x$, $-480\text{mm}$ for $A{C}_y$, and $360\text{mm}$ for $A{C}_z$ in the equation (IV).

$\begin{array}{c}AC=\sqrt{{\left(450\text{mm}\right)}^2+{\left(-480\text{mm}\right)}^2+{\left(360\text{mm}\right)}^2}\\ =750\text{mm}\end{array}$

Substitute $+\left(450\text{mm}\right)i-\left(480\text{mm}\right)j+\left(360\text{mm}\right)k$ for $\overrightarrow{AC}$, and $750\text{mm}$ for $AC$ in equation (V).

$\begin{array}{c}\lambda =\frac{+\left(450\text{mm}\right)i-\left(480\text{mm}\right)j+\left(360\text{mm}\right)k}{750\text{mm}}\\ =\frac{450}{750}i-\frac{480}{750}j+\frac{360}{750}k\end{array}$

Substitute $54\text{N}$ for $T_{AC}$ and $\frac{450}{750}i-\frac{480}{750}j+\frac{360}{750}k$ for $\lambda$ in the above equation (VI).

$\begin{array}{c}{T}_{AC}=\left(54\text{N}\right)\left(\frac{450}{750}i-\frac{480}{750}j+\frac{360}{750}k\right)\\ =\left(32.4\text{N}\right)i-\left(34.560\right)j+\left(25.920\right)k\end{array}$

Refer Fig.P2.89 and calculate the vector coordinates of the vector $AD$.

$\overrightarrow{AD}=\left(250\text{mm}\right)i-\left(480\text{mm}\right)j-\left(360\text{mm}\right)k$

Substitute $250\text{mm}$ for $A{D}_x$, $-480\text{mm}$ for $A{D}_y$, and $-360\text{mm}$ for $A{D}_z$ in the equation (VII).

$\begin{array}{c}AD=\sqrt{{\left(250\text{mm}\right)}^2+{\left(-480\text{mm}\right)}^2+{\left(-360\text{mm}\right)}^2}\\ =650\text{mm}\end{array}$

Substitute $\left(250\text{mm}\right)i-\left(480\text{mm}\right)j-\left(360\text{mm}\right)k$ for $\overrightarrow{AD}$, and $650\text{mm}$ for $AD$ in equation (VIII).

$\begin{array}{c}\lambda =\frac{\left(250\text{mm}\right)i-\left(480\text{mm}\right)j-\left(360\text{mm}\right)k}{650\text{mm}}\\ =\frac{250}{650}i-\frac{480}{650}j-\frac{360}{650}k\end{array}$

Substitute $\frac{250}{650}i-\frac{480}{650}j-\frac{360}{650}k$ for $\lambda$ in the above equation (IX).

$T_{AD}=T_{AD}\left(\frac{250}{650}i-\frac{480}{650}j-\frac{360}{650}k\right)$

Substitute $T_{AB}\left(-\frac{320}{680}i-\frac{480}{680}j+\frac{360}{680}k\right)$ for $T_{AB}$, $\left(32.4\text{N}\right)i-\left(34.560\right)j+\left(25.920\right)k$ for $T_{AC}$ and $T_{AD}\left(\frac{250}{650}i-\frac{480}{650}j-\frac{360}{650}k\right)$ for $T_{AD}$ in equation (X).

$\begin{array}{l}R=T_{AB}\left(-\frac{320}{680}i-\frac{480}{680}j+\frac{360}{680}k\right)+\left(32.4\right)i-\left(34.56\right)j+\left(25.920\right)k+T_{AD}\left(\frac{250}{650}i-\frac{480}{650}j-\frac{360}{650}k\right)\\ =\left(-\frac{320}{680}{T}_{AB}+32.4+\frac{250}{650}{T}_{AD}\right)i+\left(-\frac{480}{680}{T}_{AB}-\left(34.56\right)-\frac{480}{650}{T}_{AD}\right)j+\left(\frac{360}{680}{T}_{AB}+\left(25.920\right)-\frac{360}{650}{T}_{AD}\right)k\end{array}$

Substitute $0$ for the coefficients of $i$ and $k$ from the above equation.

$-\frac{320}{680}{T}_{AB}+32.4+\frac{250}{650}{T}_{AD}=0$ (XI)

$\frac{360}{680}{T}_{AB}+\left(25.920\right)-\frac{360}{650}{T}_{AD}=0$ (XII)

Multiply the equation (XI) by $3.6$.

$-\frac{144}{85}{T}_{AB}+116.64+\frac{18}{13}{T}_{AD}=0$ (XIII)

Multiply the equation (XI1) by $2.5$.

$\frac{45}{34}{T}_{AB}+64.8-\frac{18}{13}{T}_{AD}=0$ (IX)

Add the equations (XIII) and (IX).

$\begin{array}{c}-\frac{144}{85}{T}_{AB}+116.64+\frac{18}{13}{T}_{AD}+\frac{45}{34}{T}_{AB}+64.8-\frac{18}{13}{T}_{AD}=0\\ 181.44-\frac{63}{170}{T}_{AB}=0\\ T_{AB}=\left(181.44\right)\left(\frac{170}{63}\right)\\ =489.60\text{N}\end{array}$ (X)

Substitute $489.60\text{N}$ for $T_{AB}$ in equation (X) and solve for $T_{AD}$.

$\begin{array}{c}\frac{360}{680}\left(489.60\text{N}\right)+\left(25.920\right)-\frac{360}{650}{T}_{AD}=0\\ T_{AD}=\left(285.12\right)\frac{650}{360}\\ =514.8\text{N}\end{array}$

Therefore, the tensions in cables $AB$ and $AD$ are $\underset{\_}{490\text{N}}$ and $\underset{\_}{515\text{N}}$ respectively.