Question

Tarzan, who weighs 688 N , swings from a cliff at the end of a vine 18 m long (Figure). From the top of the cliff to the bottom of the swing, he descends by 3.2 m . The vine will break if the force on it exceeds 950 N . (a) Does the vine break? (b) If no, what is the greatest force on it during the swing? If yes, at what angle with the vertical does it break?

Short answer

1. No,thevine does not break.
2. The greatest force acting on the vine is $T=933\mathrm{N}$

Step-by-step solution

Given          

1. The weight of Tarzon is,$W=68s$
2. The length of the vine is,$l=18\mathrm{m}$
3. The vertical distance covered from the top of the cliff to the bottom of the swing is,$h=3.2\mathrm{m}$
4. The maximum limit of the force in the vine is,$T=950\mathrm{N}$

Determining the concept

Use the concept of energy conservation law and find the velocity at the lowest point. At the cliff, there is potential energy, and it is converted to kinetic energy during the swing. Use Newton’s second law and find the greatest tension in the vine. According to the law of energy conservation, energy can neither be created, nor be destroyed.

Formulae:

$$\begin{gathered} W=mg \\ U=mgh \\ K=\frac{1}{2}m{v}^2 \end{gathered}$$

where, K is kinetic energy, Uis potential energy, m is mass, v is velocity, g is an acceleration due to gravity, his height and W is work done.

(a) Determining if thevine break

The weight of Tarzan is,

$W=mg$

According to the energy conservation law, the potential energy is converted to the kinetic energy duringtheswing, and finally, the velocity at the lowest point,

$$\begin{gathered} mgh=\frac{1}{2}m{v}^2 \\ gh=\frac{1}{2}{v}^2 \\ {v}^2=2gh \end{gathered}$$

Examine the tension in the string at the lowest point. If it isn’t breaking, then it has greatest tension.

The free body diagram is shown at the lowest point. The centripetal force and tension are acting in upward direction and weight in downward direction. Use the sign convention according to the motion of the body. According to Newton’s second law,

$T-mg=\frac{m{v}^2}{r}$

Here,

$$\begin{gathered} r=l \\ =18m \\ T=\frac{{\mathrm{mv}}^2}{\mathrm{l}}+\mathrm{mg} \\ T=m\left(\frac{v^2}{l}+g\right) \\ T=\mathrm{m}\left(\frac{2gh}{l}+g\right) \\ T=\mathrm{mg}\left(\frac{2h}{l}+1\right) \\ T=688\mathrm{N}\left(\frac{2\times 3.2\mathrm{m}}{18\mathrm{m}}+1\right) \\ T=933\mathrm{N} \end{gathered}$$

Therefore, no, the vine does not break. The maximum limit of the force in the vine is $T=950\mathrm{N}$ but here the maximum tension in the vine at the lowest pointis $T=933\mathrm{N}$. It is less than the maximum limit.

(b) Determining thegreatest force acting on it during the swing if vine don’t break and the angle with the vertical if it breaks

The maximum force in the vine at the lowest point is,

$T=933\mathrm{N}$

Hence,the greatest force acting on the vine is,$T=933\mathrm{N}$

Therefore, the maximum tension in the vine at the lowest point can be found by using Newton’s second law and energy conservation law.