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 $m= 161.2kg$

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

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