Chapter 2.5, Problem 2.112P

A transmission tower is held by three guy wires attached to a pin at A and anchored by bolts at B, C, and D. If the tension in wire AC is 590 lb, determine the vertical force P exerted by the tower on the pin at

To determine

The vertical force $P$ exerted by the tower on the pin at $A$ , if the tension in the wire $AC$ is $590\text{lb}$.

Answer to Problem 2.112P

The vertical force $P$ exerted by the tower on the pin at $A$ is $\underset{\_}{2000\text{lb}}$.

Explanation of Solution

Free body diagram at $\text{A}$ is shown in figure1.

Here, $T_{AB}$ is the magnitude of tension in cable $AB$, $T_{AC}$ is the magnitude of tension in cable $AC$, $T_{AD}$ is the magnitude of tension in the cable $AD$ , $p$ is the magnitude of force $P$ exerted by the tower on the pin at $A$.

The tension in the cable $AB$ is $840\text{lb}$.

The sketch of plate supported by three cables is shown in figure2.

Let $T_{AB}$, $T_{AC}$,$T_{AD}$ and $P$ are the tension vector in cable $AB$, $AC$, $AD$ and force exerted by the tower at $A$ in the upward direction

Let $i$ , $j$ and $k$ are the unit vectors along the of $x,y\text{and}z$ direction.

Write the equation of vector distance $AB$.

$\overrightarrow{AB}=\left(x_2-x_1\right)i+(y_2-y_1)j+(z_2-z_1)k$ (I)

Here, $\overrightarrow{AB}$ is the vector distance of the cable $AB$, the variables $x_{1,}{y}_1{\text{and z}}_1$ are the coordinates of point $A$ and $x_2,y_2{\text{and z}}_2$ are the coordinates of point $B$.

Write the vector distance of the cable $AC$.

$\overrightarrow{AC}=\left(x_3-x_1\right)i+(y_3-y_1)j+(z_3-z_1)k$ (II)

Here, $\overrightarrow{AC}$ is the vector distance of the cable $AC$, the variables $x_{1,}{y}_1{\text{and z}}_1$ are the coordinates of point $A$ and $x_3,y_3{\text{and z}}_3$ are the coordinates of point $C$.

Write the vector distance of the cable $AD$.

$\overrightarrow{AD}=\left(x_4-x_1\right)i+(y_4-y_1)j+(z_4-z_1)k$ (III)

Here, $\overrightarrow{AD}$ is the vector distance of the cable $AD$, the variables $x_{1,}{y}_1{\text{and z}}_1$ are the coordinates of point $A$ and $x_4,y_4{\text{and z}}_4$ are the coordinates of point $D$.

Write the equation of tension in the cable $AB$.

$T_{AB}={\lambda }_{AB}{T}_{AB}$ (IV)

Here, $T_{AB}$ is the tension in the cable $AB$ , $T_{AB}$ is the magnitude of the tension in the cable $AB$ and ${\lambda }_{AB}$ is the unit vector in the direction of $AB$.

Write the equation of ${\lambda }_{AB}$.

${\lambda }_{AB}=\frac{\overrightarrow{AB}}{AB}$ (V)

Write the equation of tension in the cable $AC$.

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

Here, $T_{AC}$ is the tension in the cable $AC$ , $T_{AC}$ is the magnitude of the tension in the cable $AC$ and ${\lambda }_{AC}$ is the unit vector in the direction of $AC$.

Write the equation of ${\lambda }_{AC}$.

${\lambda }_{AC}=\frac{\overrightarrow{AC}}{AC}$ (VII)

Write the equation of tension in the cable $AD$.

$T_{AD}={\lambda }_{AD}{T}_{AD}$ (VIII)

Here, $T_{AD}$ is the tension in the cable $AD$ , $T_{AD}$ is the magnitude of the tension in the cable $AD$ and ${\lambda }_{AD}$ is the unit vector in the direction of $AD$.

Write the equation of ${\lambda }_{AD}$.

${\lambda }_{AD}=\frac{\overrightarrow{AD}}{AD}$ (IX)

Write the equation of force exerting at point $A$ along $y$ direction.

$P=Pj$ (X)

Here, $P$ is the force exerted at by the tower at pin at point $A$.

Write the equilibrium condition for the forces at $A$.

${\displaystyle \sum F}=0$

Here, F is the force

The above equation implies that at equilibrium, total force acting on the cable at $A$ is zero.

Refer figure 2 and write the equation of equilibrium of forces at $A$.

$T_{AB}+T_{AC}+T_{AD}+P=0$

Conclusion:

Substitute $0\text{ft}$ for $x_1$ , $100\text{ft}$ for $y_1$ , $0\text{ft}$ for $z_1$ , $-20\text{ft}$ for $x_2$ , $0\text{ft}$ for $y_2$ and $25\text{ft}$ for $z_2$ in equation (I) to get $\overrightarrow{AB}$.

$\overrightarrow{AB}=-\left(20\text{ft}\right)i-(100\text{ft})j+(25\text{ft})k$

Calculate the magnitude of $\overrightarrow{AB}$.

$\begin{array}{c}AB=\sqrt{{\left(-20\text{ft}\right)}^2+{(-100\text{ft})}^2+{(25\text{ft})}^2}\\ =\text{105}\text{ft}\end{array}$

Substitute $0\text{ft}$ for $x_1$ , $100\text{ft}$ for $y_1$ , $0\text{ft}$ for $z_1$ , $60\text{ft}$ for $x_3$ , $0\text{ft}$ for $y_3$ and $25\text{ft}$ for $z_3$ in equation (II) to get $\overrightarrow{AC}$.

$\overrightarrow{AC}=\left(60\text{ft}\right)i-(100\text{ft})j+(18\text{ft})k$

Calculate the magnitude of $\overrightarrow{AC}$.

$\begin{array}{c}AC=\sqrt{{\left(60\text{ft}\right)}^2+{(-100\text{ft})}^2+{(18\text{ft})}^2}\\ =118\text{ft}\end{array}$

Substitute $0\text{ft}$ for $x_1$ , $100\text{ft}$ for $y_1$ , $0\text{ft}$ for $z_1$ , $-20\text{ft}$ for $x_4$ , $0\text{ft}$ for $y_4$ and $-74\text{ft}$ for $z_4$ in equation (III) to get $\overrightarrow{AD}$.

$\overrightarrow{AD}=\left(-20\text{ft}\right)i-(100\text{ft})j-(74\text{ft})k$

Calculate the magnitude of $\overrightarrow{AD}$.

$\begin{array}{c}AD=\sqrt{{\left(-20\text{ft}\right)}^2+{(-100\text{ft})}^2+{(-74\text{ft})}^2}\\ =126\text{ft}\end{array}$

Substitute $-\left(20\text{ft}\right)i-(100\text{ft})j+(25\text{ft})k$ for $\overrightarrow{AB}$ and $105\text{ft}$ for $AB$ in equation (V) to get ${\lambda }_{AB}$.

$\begin{array}{c}{\lambda }_{AB}=\frac{-\left(20\text{ft}\right)i-(100\text{ft})j+(25\text{ft})k}{105\text{ft}}\\ =-\frac{4}{21}i-\frac{20}{21}j+\frac{5}{21}k\end{array}$

Substitute $-\frac{4}{21}i-\frac{20}{21}j+\frac{5}{21}k$ for ${\lambda }_{AB}$ in equation (IV) to get $T_{AB}$.

$T_{AB}=\left(-\frac{4}{21}i-\frac{20}{21}j+\frac{5}{21}k\right)T_{AB}$

Substitute $\left(60\text{ft}\right)i-(100\text{ft})j+(18\text{ft})k$ for $\overrightarrow{AC}$ and $118\text{ft}$ for $AC$ in equation
(VII) to get ${\lambda }_{AC}$.

$\begin{array}{c}{\lambda }_{AC}=\frac{\left(60\text{ft}\right)i-(100\text{ft})j+(18\text{ft})k}{118\text{ft}}\\ =\frac{30}{59}i-\frac{50}{59}j+\frac{9}{59}k\end{array}$

Substitute $\frac{30}{59}i-\frac{50}{59}j+\frac{9}{59}k$ for ${\lambda }_{AC}$ in equation (VI) to get $T_{AC}$.

$T_{AC}=\left(\frac{30}{59}i-\frac{50}{59}j+\frac{9}{59}k\right)T_{AC}$

Substitute $\left(-20\text{ft}\right)i-(100\text{ft})j-(74\text{ft})k$ for $\overrightarrow{AD}$ and $\text{126}\text{ft}$ for $AD$ in equation (IX) to get ${\lambda }_{AD}$.

$\begin{array}{c}{\lambda }_{AD}=\frac{\left(-20\text{ft}\right)i-(100\text{ft})j-(74\text{ft})k}{\text{126}\text{ft}}\\ =-\frac{10}{63}i-\frac{50}{63}j-\frac{37}{63}k\end{array}$

Substitute $-\frac{10}{63}i-\frac{50}{63}j-\frac{37}{63}k$ for ${\lambda }_{AD}$ in equation (VIII) to get $T_{AD}$.

$T_{AD}=\left(-\frac{10}{63}i-\frac{50}{63}j-\frac{37}{63}k\right)T_{AD}$

Substitute $\left(-\frac{4}{21}i-\frac{20}{21}j+\frac{5}{21}k\right)T_{AB}$ for $T_{AB}$ , $\left(\frac{30}{59}i-\frac{50}{59}j+\frac{9}{59}k\right)T_{AC}$ for $T_{AC}$ , $\left(-\frac{10}{63}i-\frac{50}{63}j-\frac{37}{63}k\right)T_{AD}$, $Pj$ for $P$ in the in equation (XI) to get force $P$ exerted at point $A$.

$\begin{array}{l}\left(-\frac{4}{21}i-\frac{20}{21}j+\frac{5}{21}k\right)T_{AB}+\left(\frac{30}{59}i-\frac{50}{59}j+\frac{9}{59}k\right)T_{AC}+\left(-\frac{10}{63}i-\frac{50}{63}j-\frac{37}{63}k\right)T_{AD}+Pj=0\\ \left(-\frac{4}{21}{T}_{AB}+\frac{30}{59}{T}_{AC}-\frac{10}{63}{T}_{AD}\right)i+\left(-\frac{20}{21}{T}_{AB}-\frac{50}{59}{T}_{AC}-\frac{50}{63}{T}_{AD}+P\right)j+\left(+\frac{5}{21}{T}_{AB}+\frac{9}{59}{T}_{AC}-\frac{37}{63}{T}_{AD}\right)k=0\end{array}$

Since total force is zero. Equate force along each direction as zero.

$-\frac{4}{21}{T}_{AB}+\frac{30}{59}{T}_{AC}-\frac{10}{63}{T}_{AD}=0$ (XII)

$-\frac{20}{21}{T}_{AB}-\frac{50}{59}{T}_{AC}-\frac{50}{63}{T}_{AD}+P=0$ (XIII)

$+\frac{5}{21}{T}_{AB}+\frac{9}{59}{T}_{AC}-\frac{37}{63}{T}_{AD}\text{=}\text{0}$ (XIV)

Substitute $590\text{lb}$ for $T_{AC}$ in equation (XII) and (XIV) to modify the equation.

$-\frac{4}{21}{T}_{AB}+\frac{30}{59}\left(590\text{lb}\right)-\frac{10}{63}{T}_{AD}=0$

$-\frac{4}{21}{T}_{AB}+300\text{lb}-\frac{10}{63}{T}_{AD}=0$ (XV)

$+\frac{5}{21}{T}_{AB}+\frac{9}{59}\left(590\text{lb}\right)-\frac{37}{63}{T}_{AD}\text{=}\text{0}$

$+\frac{5}{21}{T}_{AB}+90\text{lb}-\frac{37}{63}{T}_{AD}\text{=}\text{0}$ (XVI)

Multiply equation (XV) by $5$ and XVI by $-10$ and add to get $T_{AC}$.

$\begin{array}{c}-\frac{20}{21}{T}_{AB}+1500\text{lb}-\frac{50}{63}{T}_{AD}+\frac{20}{21}{T}_{AB}+360\text{lb}-\frac{148}{63}{T}_{AD}\text{=}\text{0}\\ 1860\text{lb}-\frac{198}{63}{T}_{AD}\text{=}\text{0}\\ {\text{T}}_{AD}=591.82\text{lb}\end{array}$

Substitute $591.82\text{lb}$ for $T_{AD}$ in equation (XVI) to get $T_{AB}$.

$\begin{array}{c}+\frac{5}{21}{T}_{AB}+90\text{lb}-\frac{37}{63}\left(591.82\text{lb}\right)\text{=}\text{0}\\ {\text{T}}_{AB}=1081.82\text{lb}\end{array}$

Substitute $1081.82\text{lb}$ for $T_{AB}$, $590\text{lb}$ for $T_{AC}$ and $591.82\text{lb}$ for $T_{AD}$ in equation (XIII) to get $P$.

$\begin{array}{c}-\frac{20}{21}\left(1081.82\text{lb}\right)-\frac{50}{59}\left(590\text{lb}\right)-\frac{50}{63}\left(591.82\text{lb}\right)+P=0\\ P=2000\text{lb}\end{array}$

Therefore, the force $P$ exerted by the tower at the pin positioned at $A$ is equal to $\underset{\_}{2000\text{lb}}$.