In Fig. 12-68, an 817 kg construction bucket is suspended by a cable Athat is attached at O to two other cables Band C, making angles$\mathit{\theta }\mathbf{1}\mathbf{}\mathbf{=}\mathbf{51}\mathbf{.}\mathbf{0}\mathbf{°}$and$\mathit{\theta }\mathbf{2}\mathbf{}\mathbf{=}\mathbf{66}\mathbf{.}\mathbf{0}\mathbf{°}$with the horizontal. Find the tensions in (a) cable A, (b) cable B, and (c) cable C. (Hint:To avoid solving two equations in two unknowns, position the axes as shown in the figure.)

${\text{θ}}_{\text{1}}=51°$

${\text{θ}}_{\text{2}}=66°$

$Massofbucketm=817kg$

Hint: To avoid solving two equations in two unknowns, position the axes as shown in the figure.

The bucket is tied to Cable A at point O. So from the free body diagram of the bucket, we can find the tension in Cable A.

Now as all the cables are attached at point O, they will be at equilibrium. So using equilibrium conditions, we can find the tension in Cable B and C.

Equations:

${\mathbf{\sum }}_{}^{}{\mathit{F}}_{\mathbf{x}}\mathbf{=}\mathbf{0}$

${\mathbf{\sum }}_{}^{}{\mathit{F}}_{\mathbf{y}}\mathbf{=}\mathbf{0}$

$\begin{array}{c}Weightofbucket=mas{s}_{bucket}\times g\\ =817kg\times 9.8m/s^2\\ =8.01\times {10}^{3}N\end{array}$

Free body diagram for the bucket:

At point A,

$T_A-weightofbucket=0$

$T_A=weightofbucket$

${\text{T}}_{\text{A}}=8.01\times {10}^{3}N$

Therefore the tension in Cable A,${\text{T}}_{\text{A}}=8.01\times {10}^{3}N$

Using the coordinates as described in the diagram.

We can say that cable A makes an angle${\theta }_2$with the negative y-axis, cable B makes an angle of$27°$with the positive y-axis and cable C is along the positive x-axis.

As the system is in equilibrium,the sum of the forces along the x and y axis should be equal to zero.

Therefore, we can write,

$\begin{array}{l}{\text{T}}_{\text{B}}\text{cos27}°-{\text{T}}_{\text{A}}\text{cos66}°=0\\ {\text{T}}_{\text{B}}\text{cos27}°={\text{T}}_{\text{A}}\text{cos66}°=\\ {\text{T}}_{\text{B}}=\frac{{\text{T}}_{\text{A}}\text{cos66}°}{\text{cos27}°}\\ T_B=3.65\times {10}^{3}N\end{array}$

The tension in Cable B,${\text{T}}_{\text{B}}=3.65\times {10}^3\text{\hspace{0.17em}N}$

Similarly, for the x-axis, we can write,

$\begin{array}{l}{\text{T}}_{\text{C}}+{\text{T}}_{\text{B}}\text{sin27}°-{\text{T}}_{\text{A}}\text{sin66}°=0\\ T_C={\text{T}}_{\text{A}}\text{sin66}°-{\text{T}}_{\text{B}}\text{sin27}°\\ T_C=(\text{8.01}\times {\text{10}}^{\text{2}}\text{\hspace{0.17em}N}\times \text{sin66}°-\text{3.65}\times {\text{10}}^{\text{3}}\text{\hspace{0.17em}N}\times \text{sin27}°)\\ T_C=5.66\times {10}^{3}N\end{array}$

The tension in Cable C is$T_C=5.66\times {10}^{3}N$