Question: A rope of negligible mass is stretched horizontally between two supports that are 3.44 m apart. When an object of weight 3160 N is hung at the center of the rope, the rope is observed to sag by 35.0 cm . What is the tension in the rope?

The distance between two ropes,$d=3.44\text{m}$

The weight attached at the center of the rope,$d=3.44\text{m}$

The amount of slagging of the rope,$l=35\text{cm}=0.35\text{m}$

Using the free body diagramand Newton’s second law and condition for equilibrium, youcan write a force equation for a vertical direction. Then, you can find the value of the angle made by the rope at a horizontal direction. Using the first force equation, you can find the tension in the rope. The law used is given below.

Newton’s second law:

$F_{net}=ma$

The free body diagram:

At equilibrium, by applying Newton’s second law, we get:

$2T\text{sin}\theta =W$

From the figure, we can write:

$$\begin{gathered} \text{tan}\theta =\frac{l}{\frac{d}{2}} \\ \text{tan}\theta =\frac{0.35}{1.72} \\ \theta ={\tan }^{-1}\frac{0.35}{1.72} \\ =11.5° \end{gathered}$$

Thus, we can write:

$$\begin{gathered} 2Ts\text{in}\theta =W \\ T=\frac{W}{2\text{sin}\theta } \\ =\frac{3160}{2\text{sin}\left(11.5°\right)} \\ =7.92\times {10}^3 \\ =7.92\text{kN} \end{gathered}$$

Therefore, the tension in the rope is 7.92 KN.