Chapter 2.3, Problem 2.57P

For the cables of Prob. 2.44, find the value of α for which the tension is as small as possible (a) in cable BC, (b) in both cables simultaneously. In each case, determine the tension in each cable.

To determine

The value of the angle $\alpha$ for which the tension in the cable $BC$ is as much small as possible and the magnitude of the tension in both the cables $AC$ and $BC$.

Answer to Problem 2.57P

The value of the angle $\alpha$ for which the tension in the cable $BC$ is as much small as possible is $\underset{\_}{35°}$ and the magnitude of the tension in the two cables are $\underset{\_}{T_{AC}=4.91\text{kN}}$ and $\underset{\_}{T_{BC}=3.44\text{kN}}$.

Explanation of Solution

The arrangement of the system is given in Fig. P2.44. The tension in the cable $AC$ and $BC$ can be represented as $T_{AC}$ and $T_{BC}$ respectively. The force acting horizontally at the point $C$ is given as $6\text{kN}$.

The free body diagram of the arrangement shown in Fig. P2.44 is given in the Figure $1$.

Since the point $C$ is in equilibrium, a force triangle can be drawn using $T_{AC}$, $T_{BC}$, and the $6\text{kN}$ force. In order to have minimum tension in the cable $BC$, the angle between the cables $AC$ and $BC$ must be $90°$. The force triangle corresponding to the free body diagram is given in Figure 2.

The angle between the $6\text{kN}$ force and $T_{BC}$ in the force triangle is marked as $\beta$.

Write the expression for the angle $\alpha$ from Figure 2.

$\alpha =90°-\beta$

Apply law of sine to the force triangle in Figure 2.

$\frac{T_{AC}}{\sin \beta }=\frac{T_{BC}}{\sin 35°}=\frac{6\text{kN}}{\sin 90°}$ (I)

Conclusion:

From the force triangle the angle $\beta$ is obtained as,

$\begin{array}{c}\beta =180-90°-35°\\ =55°\end{array}$

Substitute $55°$ for $\beta$ to find the angle $\alpha$.

$\begin{array}{c}\alpha =90°-55°\\ =35°\end{array}$

Solve equation (I) for $T_{AC}$.

$T_{AC}=\frac{6\text{kN}}{\sin 90°}\left(\sin \beta \right)$ (II)

Substitute $55°$ for $\beta$ in equation (II) to find $T_{AC}$.

$\begin{array}{c}{T}_{AC}=\frac{6\text{kN}}{\sin 90°}\left(\sin 55°\right)\\ =4.91\text{kN}\end{array}$

Solve equation (I) for $T_{BC}$.

$\begin{array}{c}{T}_{BC}=\frac{6\text{kN}}{\sin 90°}\left(\sin 35°\right)\\ =3.44\text{kN}\end{array}$

Therefore, the value of the angle $\alpha$ for which the tension in the cable $BC$ is as much small as possible is $\underset{\_}{35°}$ and the magnitude of the tension in the two cables are $\underset{\_}{T_{AC}=4.91\text{kN}}$ and $\underset{\_}{T_{BC}=3.44\text{kN}}$.

To determine

The value of the angle $\alpha$ for which the tensions in both the cables $AC$ and $BC$ are as much small as possible and the magnitude of the tension in both the cables $AC$ and $BC$.

Answer to Problem 2.57P

The value of the angle $\alpha$ for which the tensions in both the cables $AC$ and $BC$ are as much small as possible is $\underset{\_}{55°}$ and the magnitude of the tension in the two cables are $\underset{\_}{T_{AC}=3.66\text{kN}}$ and $\underset{\_}{T_{BC}=3.66\text{kN}}$.

Explanation of Solution

The free body diagram corresponding to the given arrangement is given in Figure 1

The force triangle become isosceles when the tension in both the cables become the minimum, since the tensions will be equal on both the cables in such situation.

The force triangle corresponding to the minimum tension in both cables $AC$ and $BC$ is given in Figure 3.

Apply the law of sine to the force triangle given in Figure 3.

$\frac{T_{AC}}{\sin 35°}=\frac{T_{BC}}{\sin 35°}=\frac{6\text{kN}}{\sin 110°}$ (III)

Conclusion:

From inspection on the force triangle in Figure 3, the angle $\alpha$ is obtained as,

$\begin{array}{c}\alpha =90°-35°\\ =55°\end{array}$

Solve equation (III) for $T_{BC}$.

$\begin{array}{c}\frac{T_{BC}}{\sin 35°}=\frac{6\text{kN}}{\sin 110°}\\ T_{BC}=\frac{6\text{kN}}{\sin 110°}\left(\sin 35°\right)\\ =3.66\text{kN}\end{array}$

Solve equation (III) for $T_{AC}$ .

$\begin{array}{c}\frac{T_{AC}}{\sin 35°}=\frac{6\text{kN}}{\sin 110°}\\ T_{AC}=\frac{6\text{kN}}{\sin 110°}\left(\sin 35°\right)\\ =3.66\text{kN}\end{array}$

Therefore, the value of the angle $\alpha$ for which the tensions in both the cables $AC$ and $BC$ are as much small as possible is $\underset{\_}{55°}$ and the magnitude of the tension in the two cables are $\underset{\_}{T_{AC}=3.66\text{kN}}$ and $\underset{\_}{T_{BC}=3.66\text{kN}}$.