Question

horizontal clothesline is tied between 2 poles, 14 meters apart. When a mass of 3 kilograms is tied to the middle of the clothesline, it sags a distance of 1 meters. What is the magnitude of the tension on the ends of the clothesline

Answer 1

Answer:

The tension is  $T = 103.96N$

Explanation:

The free body diagram of the question is shown on the first uploaded image From the question we are told that

           The distance between the two poles is $D =14 m$

          The mass tied between the two cloth line is  $m = 3Kg$

         The distance it sags is $d_s = 1m$

The objective of this solution is to obtain the magnitude of the tension on the ends of the  clothesline

Now the sum of the forces on the y-axis is zero assuming  that the whole system is at equilibrium

       And this can be mathematically represented as

                             $\sum F_y = 0$

 To obtain $\theta$ we apply SOHCAHTOH Rule

 So    $Tan \theta = \frac{opp}{adj}$

          $\theta = tan^{-1} [\frac{opp}{adj} ]$

            $= tan^{-1} [\frac{1}{7}]$

          $=8.130^o$

$=> \ \ \ \ \ \ \ \ 2T sin\theta -mg =0$

$=> \ \ \ \ \ \ \ \ T =\frac{mg}{2 sin\theta}$

$=> \ \ \ \ \ \ \ \ T = \frac{3 * 9.8 }{2 sin \theta }$

$=> \ \ \ \ \ \ \ \ T =\frac{29.4}{2sin(8.130)}$

$=> \ \ \ \ \ \ \ \ T = 103.96N$

             

                 

Answer 2

Answer:

T = 104N

Explanation:

We are to calculate the magnitude of the tension on the ends of the clothesline and to solve this, we need to draw a free body diagram to make it clearer.

Thus, i have attached the free body diagram.

Now, from the diagram, the angle made by the rope with the pole is given as α.

Now, we see that the rope sags by 1m. Since the weight is at midpoint, the point from the centre to the pole is 14/2 = 7m

Thus,using formula for finding angles in a right angle triangle, we have;

Tan α = opposite/adjacent = 1/7 =0.1429

Thus, α = tan^(-1)0.1429 = 8.133°

Now, to find the tension on the ends of the cloth line, we have to resolve the forces vertically.

Thus, still using the formula for finding angles in a right angle triangle, we have;

T sinα + T sinα = mg

i.e sum of upward forces = sum of dowward forces.

Thus,

2T(sin 8.133°) = 3 x 9.81

T(2 x 0.1415) = 29.43

T = 29.43/0.283 = 104 N