Question

Tarzan, who weighs$\mathbf{820}\mathbf{}\mathbf{N}$ , swings from a cliff at the end of a $\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}$vine that hangs from a high tree limb and initially makes an angle of $\mathbf{22}\mathbf{.}{\mathbf{0}}^{\mathbf{°}}$with the vertical. Assume that an xaxis extends horizontally away from the cliff edge and a yaxis extends upward. Immediately after Tarzan steps off the cliff, the tension in the vine is$\mathbf{760}\mathbf{}\mathbf{N}$ . Just then, (a) what is the force on him from the vine in unit-vector notation and the net force on him (b) what is in unit-vector notation and as (c) what is a magnitude and (d) what is the angle relative to the positive direction of the xaxis? (e) What is the magnitude of Tarzan’s acceleration and (f) What is the angle of Tarzan’s acceleration just then?

Short answer

1. Force on Tarzan from the vine is$(285\mathrm{N})\mathrm{i}+(705\mathrm{N})\hat{j}$.
2. Net force on Tarzan in unit vector notation is$(285\mathrm{N})\hat{i}-(115\mathrm{N})\hat{j}$.
3. The magnitude of the net force is$307.3\mathrm{N}$.
4. Direction of the angle relative to the positive direction of the X-axis of thenet force is${22}^{°}$below to x-axis.
5. The magnitude of the acceleration of Tarzan is$3.67\mathrm{m}/{\mathrm{s}}^2$ .
6. Direction of acceleration of Tarzan is the same as force, ${22}^{°}$below the positive x-axis.

Step-by-step solution

Given data

• Weight of Tarzan$\left(W\right)$ is$820\mathrm{N}$ .
• The angle between the vine and vertical axis is ${22}^{°}.$

Understanding the concept of the net force

Newton’s second law states that the net force $\stackrel{\mathbf{\rightharpoonup }}{\mathit{F}}$ on a body with mass mis related to the body’s acceleration $\stackrel{\rightharpoonup }{\mathit{a}}$by,

$\stackrel{\mathbf{\rightharpoonup }}{\mathit{F}\mathbf{}}\mathbf{=}\mathit{m}\stackrel{\mathbf{\rightharpoonup }}{\mathit{a}}$

From the weight, we can find the mass of Tarzan. Using the free body diagram, we can resolve the force along the x and y axes and calculate the required quantities.

Draw the free body diagram

The given angle is $22°$with vertical. So, its angle with horizontal axis is $68°$.

In addition, if we resolve that tension along x and y-axis, we can get the component along x and y-axis are$760°\cos 68°$ and$760\sin 68°$ respectively.

Formulae:

$\mathrm{F}=\mathrm{ma}$

(i)

The magnitude of a force is,

$F=\sqrt{F_x^2+F_y^2}$

(ii)

The angle between the vector and the axis,

$\theta ={\tan }^{-1}\left(\frac{F_y}{F_x}\right)$ (iii)

(a) Calculate the force on Tarzan by vein in unit vector notation.

Resolving the components of the force by vain, we get

$$\begin{gathered} \stackrel{\rightharpoonup }{T}=(760\mathrm{N})\cos (68°)\hat{i}+(760\mathrm{N})\sin (68°)\hat{j} \\ =(285N)\hat{i}+(705N)\hat{j} \end{gathered}$$

Thus, the force on Tarzan from the vine is .

(b) Calculate the net force on Tarzan by vein in unit vector notation.

The net force on Tarzan is the resultant of the forces due to tension and weight of Tarzan. Therefore,

.$$\begin{gathered} {\stackrel{\rightharpoonup }{F}}_{Net}=\stackrel{\rightharpoonup }{T}+\stackrel{\rightharpoonup }{W} \\ =(285\mathrm{N})\hat{i}+(705\mathrm{N})\hat{j}-(820\mathrm{N})\hat{j} \\ =(285\mathrm{N})\hat{i}-(115\mathrm{N})\hat{j} \end{gathered}$$

So, the net force is,$=(285\mathrm{N})\hat{i}-(115\mathrm{N})\hat{j}$ .

(c) Calculate the magnitude of net force on Tarzan

As both the components of force acting on Tarzan are known, we can find the magnitude of force as,

$$\begin{gathered} F=\sqrt{F_x^2+F_y^2} \\ =\sqrt{{(285\mathrm{N})}^2+{(-115\mathrm{N})}^2} \\ =307.3\mathrm{N} \end{gathered}$$

Therefore, the magnitude of net force is $307.3\mathrm{N}$.

(d) Calculate the direction of angle relative to the positive direction of x axis.

Using the component of the force, we can find the angle relative to positive X axis,

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{F_y}{F_x}\right) \\ ={\tan }^{-1}\left(\frac{-115\mathrm{N}}{285\mathrm{N}}\right) \\ =-21.97° \\ =-22° \end{gathered}$$

That means net force is below to X axis$22°$.

(e) Calculate the magnitude of acceleration of Tarzan

From the weight, we can find the mass of Tarzan as,

$$\begin{gathered} W=mg \\ m=\frac{W}{g} \\ =\frac{820\mathrm{N}}{9.8\mathrm{m}/{\mathrm{s}}^2} \\ =3.67\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

The acceleration is calculated as,

$$\begin{gathered} a=\frac{F_{net}}{m} \\ =\frac{307.3N}{83.67kg} \\ =3.67\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the acceleration of Tarzan is$3.67\mathrm{m}/{\mathrm{s}}^2$ .

(f) Calculate the direction of acceleration of Tarzan.

The direction of acceleration is same as force, $21.{97}^{°}\approx {22}^{°}$below positive x axis.