Chapter 6, Problem 134P

(a)

To determine

The tension in the vine just before it broke.

Answer to Problem 134P

The tension in the vine just before it broke is $1.64\text{kN}$.

Explanation of Solution

Find the equation for the length of the vine.

$\begin{array}{c}\sin {\theta }_i=\frac{\text{opposite}}{\text{hypotenuse}}\\ =\frac{5.00\text{m}}{L}\\ L=\frac{5.00\text{m}}{\sin {\theta }_i}\end{array}$ (I)

Here, ${\theta }_i$ is the initial angle between vine and the vertical and $L$ is the length of the vine.

Find the equation for the change in height.

$\begin{array}{c}\Delta y=y_f-y_i\\ =L\cos {\theta }_f-L\cos {\theta }_i\\ =L\left(\cos {\theta }_f-\cos {\theta }_i\right)\end{array}$ (II)

Here, ${\theta }_f$ is the final angle between vine and the vertical.

Use the law of conservation of energy to find the equation for the person T’s speed when the vine breaks.

$\begin{array}{c}\Delta K+\Delta U=0\\ \frac{1}{2}m\left(v_f^2-v_i^2\right)=-mg\Delta y\\ v_f^2-v_i^2=2g\left|\Delta y\right|\\ v_f-0=\sqrt{2gL\left(\cos {\theta }_f-\cos {\theta }_i\right)}\\ v_f=\sqrt{2gL\left(\cos {\theta }_f-\cos {\theta }_i\right)}\end{array}$ (III)

Here, $mg$ is the weight of person T, $g$ is the acceleration due to gravity, $\Delta K$ is the change in kinetic energy, $\Delta U$ is the change in potential energy, $v_i$ is the initial velocity of person T and $v_f$ is the final velocity of person T.

Find the equation for the tension.

$\begin{array}{c}\Sigma F=T-mg\cos {\theta }_f\\ \frac{m{v}_f^2}{L}=T-mg\cos {\theta }_f\\ \frac{mg{v}_f^2}{gL}=T-mg\cos {\theta }_f\\ T=mg\left(\frac{v_f^2}{gL}+\cos {\theta }_f\right)\end{array}$ (IV)

Conclusion:

Substitute $60°$ for ${\theta }_i$ in equation (I) to find $L$.

$\begin{array}{c}L=\frac{5.00\text{m}}{\sin 60°}\\ =5.77\text{m}\end{array}$

Substitute $5.77\text{m}$ for $L$, $9.80{\text{m/s}}^2$ for $g$, $60°$ for ${\theta }_i$ and $20°$ for ${\theta }_f$ in equation (III) to find $v$.

$\begin{array}{c}{v}_f=\sqrt{2\left(9.80{\text{m/s}}^2\right)\left(5.77\text{m}\right)\left(\cos 20°-\cos 60°\right)}\\ =7.05\text{m/s}\end{array}$

Substitute $5.77\text{m}$ for $L$, $9.80{\text{m/s}}^2$ for $g$, $7.05\text{m/s}$ for $v_f$, $900.0\text{N}$ for $mg$ and $20°$ for ${\theta }_f$ in equation (IV) to find $T$.

$\begin{array}{c}T=\left(900.0\text{N}\right)\left(\frac{{\left(7.05\text{m/s}\right)}^2}{\left(9.80{\text{m/s}}^2\right)\left(5.77\text{m}\right)}+\cos 20°\right)\\ =1.64\times {10}^3\text{N}\\ =\text{1}\text{.64}\text{kN}\end{array}$

Thus, the tension in the vine just before it broke is $1.64\text{kN}$.

(b)

To determine

If person T land on the other side of the river safely.

Answer to Problem 134P

Yes, person T lands on the other side of the river safely.

Explanation of Solution

Find the distance to the ground level at the moment of the vine breaking.

$\Delta y=8.00\text{m}-L\cos {\theta }_f$ (V)

Find the equation for the distance to the river edge.

$r=L\sin {\theta }_f$ (VI)

Here, $r$ is the distance to the river edge.

Find the quadratic equation in terms of time taken by person T to reach ground.

$\begin{array}{c}\Delta y=-v\sin {\theta }_f\Delta t-\frac{1}{2}g{\left(\Delta t\right)}^2\\ \frac{1}{2}g{\left(\Delta t\right)}^2+v\sin {\theta }_f\Delta t+\Delta y=0\end{array}$ (VII)

Find the equation for the horizontal distance travelled.

$\Delta x=v\cos {\theta }_f\Delta t$ (VIII)

Here, $\Delta x$ is the horizontal distance travelled.

Conclusion:

Substitute $5.77\text{m}$ for $L$ and $20°$ for ${\theta }_f$ in equation (V) to find $\Delta y$.

$\begin{array}{c}\Delta y=8.00\text{m}-\left(5.77\text{m}\right)\cos 20°\\ =2.58\text{m}\end{array}$

Substitute $5.77\text{m}$ for $L$ and $20°$ for ${\theta }_f$ in equation (VI) to find $r$.

$\begin{array}{c}r=\left(5.77\text{m}\right)\sin 20°\\ =1.97\text{m}\end{array}$

Substitute $-2.58\text{m}$ for $\Delta y$, $9.80{\text{m/s}}^2$ for $g$, $7.05\text{m/s}$ for $v_f$ and $20°$ for ${\theta }_f$ in equation (IV) to find $T$.

$\begin{array}{c}\frac{1}{2}\left(9.80{\text{m/s}}^2\right){\left(\Delta t\right)}^2+\left(7.05\text{m/s}\right)\sin 20°\Delta t-2.58\text{m}=0\\ 4.9{\text{m/s}}^2{\left(\Delta t\right)}^2+\left(2.41\text{m/s}\right)\Delta t-2.58\text{m}=0\end{array}$

Solve the quadratic equation to find the positive time interval.

$\begin{array}{c}\Delta t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{-2.41\text{m/s}\pm \left(-2.58\text{m}\right)}{2\left(4.9{\text{m/s}}^2\right)}\\ =0.52\text{s}\end{array}$

Substitute  $7.05\text{m/s}$ for $v_f$, $20°$ for ${\theta }_f$ and $0.52\text{s}$ for $\Delta t$ in equation (VIII) to find $\Delta x$.

$\begin{array}{c}\Delta x=\left(7.05\text{m/s}\right)\cos 20°\left(0.52\text{s}\right)\\ =3.4\text{m}\end{array}$

As the distance travelled in this time is greater than the width of the river.

$3.4\text{m}>1.97\text{m}$

Thus, the person T lands on the other side of the river safely.