How solve this !

A family decides to create a tire swing in their back yard for their son Ryan. They tie a nylon rope to a branch that is located 16 m above the ground, and adjust it so that tire swings 1 m above the ground. To make the ride more exciting, they construct a launch point that is 13 m above the ground, so that they don’t have to push Ryan all the time. You are their neighbor, and you are concerned that the ride might not be safe, so you calculate the maximum tension in the rope to see if it will hold.
a. where is the tension greatest?
b. Calculate the maximum tension in the rope, assuming that Ryan (mass 30 kg) starts from rest from his launch pad. Is it greater than the rated value of 750 N ?
c. Name two factors that may have been ignored in the above analysis, and describe whether they make the ride less safe or more safe ?

1

Answers

2015-04-11T21:13:31+05:30

This Is a Certified Answer

×
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
See the diagram.

The swing does not move along the rope, and remains tight with a tension T,
The forces along the rope are balanced. The tension supplies the centripetal force and counters the gravitational force.
    
           T = mg Cos Ф + m v²/L

         v = tangential instantaneous velocity of the swing.
         Ф = angle of the rope with the vertical
         m = mass of the swing with children on it.
         L = length of rope.

  Let the swing fall from an initial height  h₀ above the center position (its lowest position) and corresponding angle Ф₀:    Let us treat the lowest position ie., the center position as the height = 0 and so P.E. = 0.
                 h₀ = L (1 - Cos Ф₀)
                 So the initial PE = mg h₀  and K.E. = 0.
   Let the swing be in a  position at a height h above center position and the corresponding swing angle Ф.
      
                h = L (1 - Cos Ф)        and  h₀ = L (1 - Cos Ф₀)
         PE = mg h = m g L (1 - Cos Ф)
         KE = change in PE = mg (h₀ - h) = m g L (Cos Ф - CosФ₀)
         1/2 m v² = mg L (Cos Ф - Cos Ф₀)
           
   Thus the centripetal force = m v²/L = 2 mg (Cos Ф -  Cos Ф₀)  = 2 mg (h₀ - h) / L

   Thus the tension in the rope = T = mg (3 Cos Ф - 2 Cos Ф₀)
                   or,  = mg (L - h) / L +2 mg (h₀ - h) / L  =  mg [ L - 3 h + 2 h₀ ] / L
 
       we are given  L = 15 meters  and  h₀ = 12 meters
        
     The tension is maximum when h = 0, that is at the bottom of the swinging arc in the center position.

         T_maximum =  m g [ L + 2 h₀ ] / L = mg + 2 mg h₀ / L
======================
b.
   Tension maximum = 30 kg * 9.81 m/sec² [ 15 + 2 * 12 ] / 15 =  765.18 Newtons

   The tension is more than rated value of the rope = 750 Newtons by 15.18 Newtons.
The neigbour is right in suspecting that the rope is not  strong enough.

===============
c.  we assumed that the swing does not have weight.  We also assumed that the rope does not have weight.  These factors will add to the worry.  As these values are likely to increase the tension required in the wire more than the calculated value.

1 5 1