Chapter 4.3, Problem 4.110P

To determine

Find the tension in each brace and the reaction at C.

Answer to Problem 4.110P

The tension in brace BD is $\underset{\_}{T_{BD}=176.8\text{lb}}$.

The tension in brace BE is $\underset{\_}{T_{BE}=176.8\text{lb}}$.

The reaction at C is $\underset{\_}{C=-\left(50\text{lb}\right)\text{j}+\left(216.5\text{lb}\right)\text{k}}$.

Explanation of Solution

Given information:

The length of the flag pole AC is $10\text{ft}$.

The inclination of the flag pole AC is $30°$.

The distance BC is $3\text{ft}$.

Calculation:

Assumption:

Apply the sign convention for calculating the equations of equilibrium as shown below.

• For the horizontal forces equilibrium condition, take the force acting towards right side as positive $\left(\to +\right)$ and the force acting towards left side as negative $\left(\leftarrow -\right)$.
• For the vertical forces equilibrium condition, take the upward force as positive $\left(\uparrow +\right)$ and the downward force as negative $\left(\downarrow -\right)$.
• For moment equilibrium condition, take the clockwise moment as negative and counterclockwise moment as positive.

Sketch the Free Body Diagram as shown in Figure 1.

Calculate the position vector $\left(\text{r}\right)$ as shown below.

The position of A is;

$\begin{array}{c}{\text{r}}_A=\left(10\sin 30°\right)\text{j}-\left(10\cos 30°\right)\text{k}\\ =\left(5\text{ft}\right)\text{j}-\left(8.66\text{ft}\right)\text{k}\end{array}$

The position of B is;

$\begin{array}{c}{\text{r}}_B=\left(3\sin 30°\right)\text{j}+\left(3\cos 30°\right)\text{k}\\ =\left(1.5\text{ft}\right)\text{j}+\left(2.598\text{ft}\right)\text{k}\end{array}$

The position of D is;

${\text{r}}_D=-\left(3\text{ft}\right)\text{i}+\left(3\text{ft}\right)\text{j}$

The position of E is;

${\text{r}}_E=\left(3\text{ft}\right)\text{i}+\left(3\text{ft}\right)\text{j}$

Calculate the position of brace BD as shown below.

$\overrightarrow{BD}={\text{r}}_D-{\text{r}}_B$

Substitute $-\left(3\text{ft}\right)\text{i}+\left(3\text{ft}\right)\text{j}$ for ${\text{r}}_D$ and $\left(1.5\text{ft}\right)\text{j}+\left(2.598\text{ft}\right)\text{k}$ for ${\text{r}}_B$.

$\begin{array}{c}\overrightarrow{BD}=\left[-\left(3\text{ft}\right)\text{i}+\left(3\text{ft}\right)\text{j}\right]-\left[\left(1.5\text{ft}\right)\text{j}+\left(2.598\text{ft}\right)\text{k}\right]\\ =-\left(3\text{ft}\right)\text{i}+\left(1.5\text{ft}\right)\text{j}-\left(2.598\text{ft}\right)\text{k}\end{array}$

Calculate the length of brace BD as shown below.

$\begin{array}{c}BD=\sqrt{{\left(-3\right)}^2+{\left(1.5\right)}^2+{\left(-2.598\right)}^2}\\ =\sqrt{17.999604}\\ =4.243\text{ft}\end{array}$

Calculate the position of brace BE as shown below.

$BE={\text{r}}_E-{\text{r}}_B$

Substitute $\left(3\text{ft}\right)\text{i}+\left(3\text{ft}\right)\text{j}$ for ${\text{r}}_B$ and $\left(1.5\text{ft}\right)\text{j}+\left(2.598\text{ft}\right)\text{k}$ for ${\text{r}}_B$.

$\begin{array}{c}\overrightarrow{BE}=\left[\left(3\text{ft}\right)\text{i}+\left(3\text{ft}\right)\text{j}\right]-\left[\left(1.5\text{ft}\right)\text{j}+\left(2.598\text{ft}\right)\text{k}\right]\\ =\left(3\text{ft}\right)\text{i}+\left(1.5\text{ft}\right)\text{j}-\left(2.598\text{ft}\right)\text{k}\end{array}$

Calculate the length of brace BE as shown below.

$\begin{array}{c}BE=\sqrt{{\left(-3\right)}^2+{\left(1.5\right)}^2+{\left(-2.598\right)}^2}\\ =\sqrt{17.999604}\\ =4.243\text{ft}\end{array}$

Calculate the tension in brace BD $\left({\text{T}}_{BD}\right)$ as shown below.

${\text{T}}_{BD}=T_{BD}\frac{\overrightarrow{BD}}{BD}$

Substitute $-\left(3\text{ft}\right)\text{i}+\left(1.5\text{ft}\right)\text{j}-\left(2.598\text{ft}\right)\text{k}$ for $\overrightarrow{BD}$ and $4.243\text{ft}$ for $BD$.

$\begin{array}{c}{\text{T}}_{BD}=T_{BD}\frac{-3\text{i}+1.5\text{j}-2.598\text{k}}{4.243}\\ =T_{BD}\left(-0.707\text{i}+0.3535\text{j}-0.6123\text{k}\right)\end{array}$

Calculate the tension in brace BE $\left({\text{T}}_{BE}\right)$ as shown below.

${\text{T}}_{BE}=T_{BE}\frac{\overrightarrow{BE}}{BE}$

Substitute $\left(3\text{ft}\right)\text{i}+\left(1.5\text{ft}\right)\text{j}-\left(2.598\text{ft}\right)\text{k}$ for $\overrightarrow{BE}$ and $4.243\text{ft}$ for $BE$.

$\begin{array}{c}{\text{T}}_{BE}=T_{BE}\frac{3\text{i}+1.5\text{j}-2.598\text{k}}{4.243}\\ =T_{BE}\left(0.707\text{i}+0.3535\text{j}-0.6123\text{k}\right)\end{array}$

Apply the Equations of Equilibrium as shown below.

Apply the equilibrium equation of moment about point C and equate to zero for equilibrium.

$\begin{array}{l}{\displaystyle \sum M_C}=0\\ {\text{r}}_B\times {\text{T}}_{BD}+{\text{r}}_B\times {\text{T}}_{BE}+{\text{r}}_A\times \left(-75\text{j}\right)=0\end{array}$

Substitute $\left(5\text{ft}\right)\text{j}-\left(8.66\text{ft}\right)\text{k}$ for ${\text{r}}_A$, $\left(1.5\text{ft}\right)\text{j}+\left(2.598\text{ft}\right)\text{k}$ for ${\text{r}}_B$, $T_{BD}\left(-0.707\text{i}+0.3535\text{j}-0.6123\text{k}\right)$ for ${\text{T}}_{BD}$, and $T_{BE}\left(0.707\text{i}+0.3535\text{j}-0.6123\text{k}\right)$ for ${\text{T}}_{BE}$.

$\begin{array}{l}\left[\begin{array}{l}\left(1.5\text{j}+2.598\text{k}\right)\times \left(T_{BD}\left(-0.707\text{i}+0.3535\text{j}-0.6123\text{k}\right)\right)\\ +\left(1.5\text{j}+2.598\text{k}\right)\times \left(T_{BE}\left(0.707\text{i}+0.3535\text{j}-0.6123\text{k}\right)\right)+\left(5\text{j}-8.66\text{k}\right)\times \left(-75\text{j}\right)\end{array}\right]=0\\ T_{BD}\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 0& 1.5& 2.598\\ -0.707& 0.3535& -0.6123\end{array}\right|+T_{BE}\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 0& 1.5& 2.598\\ 0.707& 0.3535& -0.6123\end{array}\right|+\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 0& 5& -8.66\\ 0& -75& 0\end{array}\right|=0\\ \left[\begin{array}{l}{T}_{BD}\left(\text{i}\left(-0.9185-0.9184\right)-\text{j}\left(0+1.8368\right)+\text{k}\left(0+1.0605\right)\right)\\ +T_{BE}\left(\text{i}\left(-0.9185-0.9184\right)-\text{j}\left(0-1.8368\right)+\text{k}\left(0-1.0605\right)\right)\\ +\left(\text{i}\left(0-649.5\right)-\text{j}\left(0\right)+\text{k}\left(0\right)\right)\end{array}\right]=0\\ \left(\begin{array}{l}{T}_{BD}\left(-1.8369\text{i}-1.8368\text{j}+1.0605\text{k}\right)+\\ T_{BE}\left(-1.8369\text{i}+1.8368\text{j}-1.0605\text{k}\right)-649.5\text{i}\end{array}\right)=0\end{array}$

Multiply by $4.243$.

$\begin{array}{l}\left(\begin{array}{l}-1.8369T_{BD}\text{i}-1.8368T_{BD}\text{j}+1.0605T_{BD}\text{k}\\ -1.8369T_{BE}\text{i}+1.8368T_{BE}\text{j}-1.0605T_{BE}\text{k}-649.5\text{i}\end{array}\right)=0\\ \left\{\begin{array}{l}\left(-1.8369T_{BD}-1.8369T_{BE}-649.5\right)\text{i}+\left(-1.8368T_{BD}+1.8368T_{BE}\right)\\ +\left(1.0605T_{BD}-1.0605T_{BE}\text{k}\right)\end{array}\right\}=0\end{array}$

Resolve i, j, and k components as shown below.

Resolve j component.

$\begin{array}{l}-1.8368T_{BD}+1.8368T_{BE}=0\\ 1.8368T_{BD}=1.8368T_{BE}\\ T_{BD}=T_{BE}\end{array}$ (1)

Resolve i component.

$-1.8369T_{BD}-1.8369T_{BE}-649.5=0$

Substitute $T_{BE}$ for $T_{AD}$.

$\begin{array}{l}-1.8369T_{BE}-1.8369T_{BE}-649.5=0\\ -3.6738T_{BE}=649.5\\ T_{BE}=-176.79\text{lb}\\ T_{BE}=176.79\text{lb}\end{array}$

Hence, the tension in brace BE is $\underset{\_}{T_{BE}=176.8\text{lb}}$.

Calculate the tension in brace BD as shown below.

Substitute $176.8\text{lb}$ for $T_{BE}$ in Equation (1).

$T_{BD}=176.8\text{lb}$

Hence, the tension in brace BD is $\underset{\_}{T_{BD}=176.8\text{lb}}$.

Apply the Equations of Equilibrium as shown below.

Apply the equilibrium equation of forces in x-direction and equate to zero for equilibrium.

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ C_x+176.8\times \left(-0.707\right)+176.8\times 0.707=0\\ C_x=0\end{array}$

Apply the equilibrium equation of forces in y-direction and equate to zero for equilibrium.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ C_y+176.8\times 0.3535+176.8\times 0.3535-75=0\\ C_y+50=0\\ C_y=-50\text{lb}\end{array}$

Apply the equilibrium equation of forces in z-direction and equate to zero for equilibrium.

$\begin{array}{l}{\displaystyle \sum F_z}=0\\ C_z+176.8\times \left(-0.6123\right)+176.8\times \left(-0.6123\right)=0\\ C_z-216.5=0\\ C_z=216.5\text{lb}\end{array}$

Calculate the reaction at C as shown below.

$C=C_x\text{i}+C_y\text{j}+C_z\text{k}$

Substitute 0 for $C_x$, $-50\text{lb}$ for $C_y$, and $216.5\text{lb}$ for $C_z$.

$\begin{array}{c}C=\left(0\right)\text{i}+\left(-50\right)\text{j}+\left(216.5\right)\text{k}\\ =-\left(\text{50 lb}\right)\text{j}+\left(216.5\text{lb}\right)\text{k}\end{array}$

Therefore, the reaction at C is $\underset{\_}{C=-\left(\text{50 lb}\right)\text{j}+\left(216.5\text{lb}\right)\text{k}}$.