91Ó°ÊÓ

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

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

$\stackrel{\mathbf{â‡\P Ä}}{\mathit{F}\mathbf{}}\mathbf{=}\mathit{m}\stackrel{\mathbf{â‡\P Ä}}{\mathit{a}}$

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.

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,

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

Resolving the components of the force by vain, we get

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

Thus, the force on Tarzan from the vine is .

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

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

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

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 $307.3\mathrm{N}$.

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

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

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

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$3.67\mathrm{m}/{\mathrm{s}}^2$ .

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