Question

Captain Hook is ghting Peter Pan, and they are about to step onto a tightrope strung horizontally between two maststhat 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 1

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`

Step-by-step 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
`d_A = 16m`

The angle the rope makes is
`\theta = 3^o`

The diameter of the rope is
`d = 0.02m`

The Young modulus is
`Y = 35Pa`

From the diagram we see that the elongation of the rope can be mathematically evaluated as

`\Delta L = 2x - 16`

And applying SOHCATOH rule
`x = (8)/(cos \theta)`

Substituting values

`x = (8)/(cos (3))`

`= 8.01m`

And
`\Delta L = (16)/(cos 3) -16`

`\Delta L = 0.02196m`

The Tension on the rope can be mathematically represented as

`T = Y A * (\Delta L)/(L)`

Where A is the area and is mathematically represented as

`A = (\pi)/(4) d^2`

Substituting values

`A = (\pi)/(4) (0.02)^2`

Now Substituting values into the formula for the tension on the rope

`T = (35*10^9) * (\pi)/(4) (0.02)^2 * ((0.02196))/(16)`

`=15093.4 N`

From the diagram we can mathematically evaluate the the weight of peter and hook as

`W = 2T sin \theta`

Where
`W = mg`

Now substituting this into the equation and making m the subject

`m = (2Tsin \theta)/(g)`

Substituting values

`m = (2* 15093.4 sin(3))/(9.8)`

`m= 161.2kg`

Note SOHCATOH is

`Sin \theta = (opposite)/(hypotenuse)\\ Cos \theta = (adjacent )/(hypotenuse) \\Tan \theta = (opposite)/(adjacent)`