Question

Find the tension in each cord in Fig. E5.7 if the weight of the suspended object is w.

Short answer

(a) The tension in the ropes A, B, and C is $\mathbf{0}\mathbf{.}\mathbf{732}\mathbf{w}$, $\mathbf{0}\mathbf{.}\mathbf{897}\mathbf{w}$and $\mathbf{w}$.

(b) The tension in the ropes A, B, and C is $\mathbf{2}\mathbf{.}\mathbf{73}\mathbf{w}$, $\mathbf{3}\mathbf{.}\mathbf{35}\mathbf{w}$ and $\mathbf{w}$.

Step-by-step solution

The tension

Given Data:

The angle of rope A with horizontal for part a and with vertical for part b are $\alpha =30°$ and $\alpha =60°$.

The angle of rope B with horizontal for part a and part b is $\beta =45°$.

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.

Determine the tension in each rope by vertical and horizontal equilibrium(a)

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

$\frac{T_A}{\sin \left(90°+\beta \right)}=\frac{T_A}{\sin \left(180°-\left(\alpha +\beta \right)\right)}$

Here $\alpha$and $\beta$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:

$\frac{T_B}{\sin \left(90°+\alpha \right)}=\frac{T_c}{\sin \left(180°-\left(\alpha +\beta \right)\right)}$

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.

Determine the tension in each rope by vertical and horizontal equilibrium(b)

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

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

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

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

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.