Bonzai!!!

Consider the motion of my friend.

She has three forces acting on her:- the 300N force exerted by me, the tension in the string, and gravity.

The weight of my friend is (9.8)*(60)N = 588 N

Let A be the angle of the rope with respect to its initial position when my friend is at her lowest position.

The vertical component of the tension will support my friends weight. Let T be the tension in the rope.

So, TcosA= 588N .....(1)

The horizontal component will balance the 300N force exerted by me.

So, TsinA= 300N ......(2)

Do (2)/(1).

tanA= 300/588 Or A~~ 27.03 degrees (calculation done using calculator)

So the angle made by the rope when the swing is at its lowest position is 27.03 degrees. Similarly the rope will swing another 27.03 degrees before returning to the highest position. So the rope will swing a total of 27.03*2 degrees= 54.06 degrees.

Now the radius of the circular path of the rope is 30 m, and the angle is 54.06 degrees.

Let x meter be the length of the arc.

We can conclude that x/{60*(pi)}= 54.06/360

Or x= 54.06 (pi) 60/360 ~~ 28.305

So the total distance my friend has traveled is 28.305 meter.

Note:- The conclusion that (length of arc)/(60pi)= Angle/(360) can be proved as follows.

The circumference of the circular path of the rope is 2 30 pi m= 60pi m The total angle in the circle is 360 degrees. So arc of length (60pi)m subtends 360 degrees at the center. If an arc of length x subtends angle Y at the center, then x/(60pi)= Y/360.

(Picture at http://postimage.org/image/u1wtsxf2v/)

At her starting point, the friend experience the vertical gravitational force $F_g$ , the horizontal tug force $F$ , and the force of the rope $T$ .

Since she is in static equilibrium, the angle formed by $T$ and the vertical must be $\tan^{-1} (\frac{F}{T})=\tan^{-1} (\frac{300\mathrm{N}}{60\mathrm{kg} \cdot 9.8\mathrm{m\cdot s^{-2}}} ) =0.472$ in radians.

Thus the distance traveled is just the arc length with two times that angle, which is $0.472 \cdot 2 \cdot 30 \mathrm{m} = 28.3 \mathrm{m}$ .

Construct a physics model, viewing the rope and the friend as a 30m long pendulum and a point mass of 60kg tied to the end of the pendulum. A horizontal force of 300N is exerted on the point mass.

Analyze the force acting on the point mass: Vertical: 300N Horizontal: 60kg * 9.8m/s^2 = 588N

Suppose the angle between the fully stretched rope and the horizontal = theta \tan \theta = 300/588

As \sin \theta = 0.5* displacement of the friend/length of the rope(30m) It can be calculated that displacement of the friend = 28.3

The vertical component of the tension acting in the rope is equal to the weight of the

friend i.e ,Tcosx=60g = 60*9.8 = 588N

The horizontal component of the tension is equal to the restoring force of 300N

Tsinx =300N

Tanx=300/588 = 0.5102

x = 27.3degrees

Arc/radius = x in radians

27.3degrees =3.14/180 *27.3

Arc/30 = 0.476

Arc = 14.28

Now since this is half the total distance travelled

so total distance travaled = 14.28*2 = 28.56 (approx)

if we draw the free body diagram of the of the rope with max tension (T) acting on it and let it has made an angular displacement \theta from its initial position. now if we break the forces acting on it into components then we will get the vertical component to be T\sin \theta and the horizontal to be T\cos \theta. now this horizontal component will be the max forcce with which the rope is pulled which is equal to 300N and the vertical component will be equal to the weight of the friend 60g N. solving this equations we get the angular displacement \theta of the rope. at full stretch when the friend let the rope go it will have a total angular displace ment of 2\theta. now from s=r\theta we will get the total distance covered. given r= the length of the rope, 30m.

If you pull your friend back with a horizontal force until they don't move anymore, then at that point they are in equilibrium. Therefore the y-component of the tension force T of the rope on your friend counteracts the gravitational force and the x-component counteracts your tugging. We then have $Tcos \theta=(9.8)(60)$ and $T sin \theta=300$ where $\theta$ is the angle the rope makes with the vertical. Solving for $\theta$ yields $\theta=0.472~rad$ . If L is the length of the rope, the total distance traveled by your friend before she lets go is $2L\theta=28.3~m$ .