Complete Question
Captain Hook is fighting Peter Pan, and they are about to step onto a tightrope strung horizontally between two masts that are 16 m apart. When Pan and Hook are standing exactly halfway between the masts, the rope makes a 3 anglewith the horizontal. The rope has a diameter of 0.02 m and a Youngs modulus of 35 GPa.
What is the combined mass,in kilograms, of Peter Pan and Captain Hook?
Answer:
Their combined mass is ![m= 161.2kg](https://tex.z-dn.net/?f=m%3D%20161.2kg)
Explanation:
A sketch that describes the question is shown on the first uploaded image
From the question we are told that
The distance apart is
The angle the rope makes is ![\theta = 3^o](https://tex.z-dn.net/?f=%5Ctheta%20%3D%203%5Eo)
The diameter of the rope is ![d = 0.02m](https://tex.z-dn.net/?f=d%20%3D%200.02m)
The Young modulus is ![Y = 35Pa](https://tex.z-dn.net/?f=Y%20%3D%2035Pa)
From the diagram we see that the elongation of the rope can be mathematically evaluated as
And applying SOHCATOH rule
Substituting values
![x = \frac{8}{cos (3)}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B8%7D%7Bcos%20%283%29%7D)
![= 8.01m](https://tex.z-dn.net/?f=%3D%208.01m)
And ![\Delta L = \frac{16}{cos 3} -16](https://tex.z-dn.net/?f=%5CDelta%20L%20%3D%20%5Cfrac%7B16%7D%7Bcos%203%7D%20%20-16)
![\Delta L = 0.02196m](https://tex.z-dn.net/?f=%5CDelta%20L%20%3D%200.02196m)
The Tension on the rope can be mathematically represented as
![T = Y A * \frac{\Delta L}{L}](https://tex.z-dn.net/?f=T%20%3D%20Y%20A%20%2A%20%5Cfrac%7B%5CDelta%20L%7D%7BL%7D)
Where A is the area and is mathematically represented as
![A = \frac{\pi}{4} d^2](https://tex.z-dn.net/?f=A%20%3D%20%5Cfrac%7B%5Cpi%7D%7B4%7D%20d%5E2)
Substituting values
![A = \frac{\pi}{4} (0.02)^2](https://tex.z-dn.net/?f=A%20%3D%20%5Cfrac%7B%5Cpi%7D%7B4%7D%20%280.02%29%5E2)
Now Substituting values into the formula for the tension on the rope
![T = (35*10^9) * \frac{\pi}{4} (0.02)^2 * \frac{(0.02196)}{16}](https://tex.z-dn.net/?f=T%20%3D%20%2835%2A10%5E9%29%20%2A%20%5Cfrac%7B%5Cpi%7D%7B4%7D%20%280.02%29%5E2%20%2A%20%5Cfrac%7B%280.02196%29%7D%7B16%7D)
![=15093.4 N](https://tex.z-dn.net/?f=%3D15093.4%20N)
From the diagram we can mathematically evaluate the the weight of peter and hook as
![W = 2T sin \theta](https://tex.z-dn.net/?f=W%20%3D%202T%20sin%20%5Ctheta)
Where
Now substituting this into the equation and making m the subject
![m = \frac{2Tsin \theta}{g}](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7B2Tsin%20%5Ctheta%7D%7Bg%7D)
Substituting values
![m = \frac{2* 15093.4 sin(3)}{9.8}](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7B2%2A%2015093.4%20sin%283%29%7D%7B9.8%7D)
![m= 161.2kg](https://tex.z-dn.net/?f=m%3D%20161.2kg)
Note SOHCATOH is