Physics, 30.03.2020 20:39 Blazingangelkl

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?

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Answer from: vannybelly83

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?

Their combined mass is m= 161.2kg

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 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 $x = \frac{8}{cos \theta}$

Substituting values

$x = \frac{8}{cos (3)}$

= 8.01m

And $\Delta L = \frac{16}{cos 3} -16$

$\Delta L = 0.02196m$

The Tension on the rope can be mathematically represented as

$T = Y A * \frac{\Delta L}{L}$

Where A is the area and is mathematically represented as

$A = \frac{\pi}{4} d^2$

Substituting values

$A = \frac{\pi}{4} (0.02)^2$

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

$T = (35*10^9) * \frac{\pi}{4} (0.02)^2 * \frac{(0.02196)}{16}$

=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

$m = \frac{2Tsin \theta}{g}$

Substituting values

$m = \frac{2* 15093.4 sin(3)}{9.8}$

m= 161.2kg

Note SOHCATOH is

$Sin \theta = \frac{opposite}{hypotenuse}\\ Cos \theta = \frac{adjacent }{hypotenuse} \\Tan \theta = \frac{opposite}{adjacent}$

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