Question

A 65 kg students is walking on a slackline, a length of webbing stretched between two trees. the line stretches and sags so that line makes a 20 degree angle relative to the horizontal. what is the tension in the line?

Answer 1

Answer : Tension in the line = 936.7 N

Explanation :

It is given that,

Mass of student, m = 65 kg

The angle between slackline and horizontal, $\theta=20^0$

The two forces that acts are :

(i) Tension

(ii) Weight

So, from the figure it is clear that :

$mg=2T\ sin\ \theta$

$T=\dfrac{mg}{2\ sin\theta}$

$T=\dfrac{65\ kg\times 9.8\ m/s^2}{2(sin\ 20)}$

$T=936.7\ N$

Hence, this is the required solution.

Answer 2

The tension in the line is $\boxed{932.18{\text{ N}}}$.

Further Explanation:

Free body diagram is the graphical representation or illustration of all the forces applied on the body with their directions. This helps us to visualize all the applied forces on the body with their directions and make us possible to solve complex problems based on kinematics.

Tension is the force exerted by the wire, cable or string like object when someone pulls it.

The force in the downward direction is due to gravity called as weight. Its magnitude depends upon the mass of the body and expressed as $mg$. the term $g$ is called acceleration due to gravity and it has a constant value of $9.81{\text{ m/}}{{\text{s}}^2}$.

Given:

The mass of the student is $65\text{ kg}$.

The sag angle with respect to the horizontal is $20^\circ$.

Concept:

The student walking in the slack line exerts tension on the slack line ties with two trees. There is tension on the both side of the slack line. The horizontal component of tension both side of the slack line will be canceling each other effects and the vertical component of tension both side of the slack line is balanced by the weight of the student.

It can be expressed as:

$2T\sin\theta=mg$

Here, $T$ is the tension, $\theta$ is the sag angle, $m$ is the mass of object and $g$ is the acceleration due to gravity.

Substitute $20^\circ$  for $T$, $65\text{ kg}$ for $m$ and $9.81{\text{ m/}}{{\text{s}}^2}$ for $g$ in the above equation.

$\begin{aligned}2T\sin 20^\circ&=\left( {65{\text{ kg}}}\right)\left( {9.81{\text{ m/}}{{\text{s}}^2}} \right) \hfill\\T&=\frac{{\left({65{\text{ kg}}} \right)\left( {9.81{\text{ m/}}{{\text{s}}^2}}\right)}}{{2\sin 20^\circ}}\hfill\\&=932.18{\text{ N}}\hfill\\ \end{aligned}$

Therefore, the tension in the line is $\boxed{932.18{\text{ N}}}$.

Learn more:

1. Tension https://brainly.in/question/1796681

2. The tension in the string https://brainly.in/question/61401

3. Tension https://brainly.in/question/8509853

Answer Details:

Grade: College

Subject: Physics

Chapter: Kinematics

Keywords:

65 kg, student, walking, slackline, length, webbing, stretched, between, two, trees, sags, 20 deg, angle, relative, horizontal, tension, line, 932.18 N.