91Ó°ÊÓ

Given Data:

The angle of rope A with horizontal for part a and with vertical for part b are $\mathrm{Î\pm }=30°$ and $\mathrm{Î\pm }=60°$.

The angle of rope B with horizontal for part a and part b is $\mathrm{β}=45°$.

Tension in Cord:

The tension in rope A and rope B for part a can be found by applying Lami’s theorem. The tension in rope A and rope B for part b can be found by considering horizontal and vertical equilibrium.

The tension in rope A,$T_A$is given as:

$\frac{T_A}{\sin \left(90°+\mathrm{β}\right)}=\frac{T_A}{\sin \left(180°-\left(\mathrm{Î\pm }+\mathrm{β}\right)\right)}$

Here $\mathrm{Î\pm }$and $\mathrm{β}$are the angles of rope A and rope B from horizontal.

$T_c$ is the tension in rope C and its values is $w$by vertical equilibrium.

Substitute all the values in the above equation, and we get,

$$\begin{gathered} \frac{T_A}{\sin \left(90°+45°\right)}=\frac{w}{\sin \left(180°-\left(30°+45°\right)\right)} \\ {T}_A=0.732w \end{gathered}$$

The tension in rope B,$T_B$is given as:

$\frac{T_B}{\sin \left(90°+\mathrm{Î\pm }\right)}=\frac{T_c}{\sin \left(180°-\left(\mathrm{Î\pm }+\mathrm{β}\right)\right)}$

Substitute all the values in the above equation, and we get,

$$\begin{gathered} \frac{T_B}{\sin \left(90°+30°\right)}=\frac{w}{\sin \left(180°-\left(30°+45°\right)\right)} \\ {T}_B=0.897w \end{gathered}$$

Therefore, the tension in the ropes A, B, and C are $0.732w$, $0.897w$, and $w$respectively.

The horizontal equilibrium for the system of rope is given as:

$$\begin{gathered} T_A\sin \mathrm{Î\pm }-T_B\cos \mathrm{β}=0 \\ {T}_A\sin \mathrm{Î\pm }=T_B\cos \mathrm{β} \\ {T}_A\left(\sin 60°\right)=T_B\left(\cos 45°\right) \\ {T}_B=1.225T_A \end{gathered}$$ (1)

The horizontal equilibrium for the system of rope is given as:

$$\begin{gathered} T_A\cos \mathrm{Î\pm }=T_B\sin \mathrm{β}-T_C \\ {T}_A\left(\cos 60°\right)=\left(1.225T_A\right)\left(\sin 45°\right)-w \\ {T}_A=2.73w \end{gathered}$$

Substitute $T_A=2.73w$in the above equation (1), and we get,

$$\begin{gathered} T_B=1.225\left(2.73w\right) \\ {T}_B=3.35w \end{gathered}$$

$T_C$is the tension in rope C and its values is $w$by vertical equilibrium.

Therefore, the tension in the ropes A, B, and C is $2.73w$, $3.35w$, and $w$ respectively.