Question

3 kg of wet clothes are hung on the middle of a clothesline with posts 6 ft apart. The clothesline sags down by 3 feet. What is the total tension upon the clothesline?

Answer 1

Answer:

Tension in string equals 14.715 Newtons

Explanation:

The situation is represented in the figure attached below:

For equilibrium of the clothes along y- axis we have

$\sum F_{v}=0\\\\\Rightarrow 2Tcos(\theta )=W\\\\\therefore T=\frac{W}{2cos(\theta )}$

Applying values we get

$\sum F_{v}=0\\\\\Rightarrow 2Tcos(\theta )=W\\\\\therefore T=\frac{3\times 9.81}{2\times 1}=14.715N(\because cos(\theta )=\frac{3}{3}=1)$

Answer 2

Answer:

15√2 N.

Explanation:

Given that the mass of the wet clothes is, $m=3kg$

Here in the figure ABC the weight is hung from the middle then the sides AC and BC are equal will make their adjacent angle equal and the other angle which is ∠ACB will become 90°.

Now according the figure the tension is apply in two direction than by free body diagram we can solve newton equations.

$2Tcos\theta=mg\\T=\frac{mg}{2cos\theta}\\ T=\frac{3\times10}{2cos(45^{\circ})}\\ T=15\sqrt{2}N$

Therefore, the total tension upon the clothesline is, $T=15\sqrt{2}N$