Q7E - University Physics with Modern Physics

Question

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

Step-by-Step Solution

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 1: 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.

Step 2: 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.

Step 3: 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.