Chapter 4.2, Problem 4.71P

For the boom and loading shown, determine (a) the tension in cord BD, (b) the reaction at C.

Fig.P4.71

To determine

The tension in the cord $BD$.

Answer to Problem 4.71P

The tension in the cord $BD$ is $5.63\text{kips}$.

Explanation of Solution

Calculation:

If the rigid body subjected to three forces, such a body is commonly called a three-force body. If a two force body is in equilibrium the lines of action of the three forces must be either concurrent or parallel.

The figure 1 represents the free body diagram of the system.

Write the expression for the angle $\alpha$ from the given figure.

$\tan \alpha =\frac{EG}{AG}$ (I)

Here, $\alpha$ is the angle.

Write the expression for the angle $\beta$.

$\tan \beta =\frac{BH}{CF}$

Write the relations connecting the sides of the triangle.

$\frac{BF}{AG}=\frac{CF}{CG}$ (II)

$\frac{JE}{BH}=\frac{DJ}{DH}$ (III)

Figure 2 represents a force triangle as shown in the above figure with its interior angles computed from the known directions of the forces. Then use the law of sines to find the unknown forces.

Write the expression for the law of sines from the force triangle.

$\frac{T_{BD}}{\sin 94.76°}=\frac{3\text{kips}}{\sin 32.106°}=\frac{C}{\sin \left(53.13°\right)}$ (II)

Conclusion:

Substitute $12$ for $AG$, $32$ for $CF$, $48$ for $CG$ in equation II to find $BH$.

$\frac{BF}{12}=\frac{32}{48}$

$BF=8\text{in}$

$\begin{array}{c}BH=32-BF\\ =32-8\\ =24\text{in}\end{array}$

Substitute $24$ for $BH$, $48$ for $DJ$, $48$ for $CG$ in equation III to find $JE$.

$\begin{array}{l}\frac{JE}{24}=\frac{48}{32}\\ JE=36\text{in}\end{array}$

$\begin{array}{c}EG=JE-JG\\ =36-32\\ =4\text{in}\end{array}$

Substitute $4\text{in}$ for $EG$ ,and $48\text{in}$ for $AG$ in equation I to find $\alpha$.

$\begin{array}{c}\tan \alpha =\frac{4\text{in}}{48\text{in}}\\ =0.083\\ \alpha ={\tan }^{-1}\left(0.083\right)\\ =4.763°\end{array}$

Substitute $24\text{in}$ for $BH$ ,and $32\text{in}$ for $CF$ in equation I to find $\beta$.

$\begin{array}{c}\tan \beta =\frac{24\text{in}}{32\text{in}}\\ =0.75\\ \beta =\tan \left(0.75\right)\\ =36.86°\end{array}$

$\frac{T_{BD}}{\sin 94.76°}=\frac{3\text{kips}}{\sin 32.106°}=\frac{C}{\sin \left(53.13°\right)}$

On rearranging the above equation to find $T$.

$\begin{array}{c}{T}_{BD}\times 0.53=2.98\\ T_{{}_{BD}}=\frac{2.98}{0.53}\\ =5.63\text{N}\\ \text{=5}\text{.6}\text{N}\end{array}$

Therefore, the tension in the cord $BD$ is $5.63\text{kips}$.

To determine

The reaction at $C$.

Answer to Problem 4.71P

The reaction at $C$ is $4.52\text{kips}$.

Explanation of Solution

Calculation:

If the rigid body subjected to three forces, such a body is commonly called a three-force body. If a two force body is in equilibrium the lines of action of the three forces must be either concurrent or parallel.

Write the expression for the angle $\alpha$ from the given figure.

$\tan \alpha =\frac{EG}{AG}$ (I)

Here, $\alpha$ is the angle.

Write the expression for the angle $\beta$.

$\tan \beta =\frac{BH}{CF}$

Write the relations connecting the sides of the triangle.

$\frac{BF}{AG}=\frac{CF}{CG}$ (II)

$\frac{JE}{BH}=\frac{DJ}{DH}$ (III)

Figure 2 represents a force triangle as shown in the above figure with its interior angles computed from the known directions of the forces. Then use the law of sines to find the unknown forces.

Write the expression for the law of sines from the force triangle.

$\frac{T_{BD}}{\sin 94.76°}=\frac{3\text{kips}}{\sin 32.106°}=\frac{C}{\sin \left(53.13°\right)}$ (IV)

Conclusion:

Substitute $12$ for $AG$, $32$ for $CF$, $48$ for $CG$ in equation II to find $BH$.

$\frac{BF}{12}=\frac{32}{48}$

$BF=8\text{in}$

$\begin{array}{c}BH=32-BF\\ =32-8\\ =24\text{in}\end{array}$

Substitute $24$ for $BH$, $48$ for $DJ$, $48$ for $CG$ in equation III to find $JE$.

$\begin{array}{l}\frac{JE}{24}=\frac{48}{32}\\ JE=36\text{in}\end{array}$

$\begin{array}{c}EG=JE-JG\\ =36-32\\ =4\text{in}\end{array}$

Substitute $4\text{in}$ for $EG$ ,and $48\text{in}$ for $AG$ in equation I to find $\alpha$.

$\begin{array}{c}\tan \alpha =\frac{4\text{in}}{48\text{in}}\\ =0.083\\ \alpha ={\tan }^{-1}\left(0.083\right)\\ =4.763°\end{array}$

Substitute $24\text{in}$ for $BH$ ,and $32\text{in}$ for $CF$ to find $\beta$.

$\begin{array}{c}\tan \beta =\frac{24\text{in}}{32\text{in}}\\ =0.75\\ \beta =\tan \left(0.75\right)\\ =36.86°\end{array}$

$\frac{T_{BD}}{\sin 94.76°}=\frac{3\text{kips}}{\sin 32.106°}=\frac{C}{\sin \left(53.13°\right)}$

On rearranging the above equation to find $T$.

$\begin{array}{c}{T}_{BD}\times 0.53=2.98\\ T_{{}_{BD}}=\frac{2.98}{0.53}\\ =5.63\text{N}\\ \text{=5}\text{.6}\text{N}\end{array}$

$\begin{array}{c}3\text{kips}\times \text{sin}\left(53.13°\right)=C\times \sin \left(32.10°\right)\\ C=4.52\text{kips}\end{array}$

Therefore, reaction at $C$ is $4.52\text{kips}$.