Question

(A) What is the maximum tension possible in a 1.00-millimeter-diameter nylon tennis racket string?

Answer 1

Complete Question

(A) What is the maximum tension possible in a 1.00- millimeter-diameter nylon tennis racket string?

(B) If you want tighter strings, what do you do to prevent breakage: use thinner or thicker strings? Why? What causes strings to break when they are hit by the ball?

The  tensile  strength of the nylon string is  $600*10^{6} \ N/m^2$

Answer:

A

   T = 471.3 \  N

B

 To prevent breakage the thickness of the string is increased

  String breakage when the racket hit the ball is as a result of the string not being thick enough to withstand the increase in tension

Explanation:

From the question we are told that

     The  diameter is  $d = 1.00 \ mm = 0.001 \ m$  

       The  tensile strength of the nylon string is $\sigma = 600 *10^{6} \ N/m^2$

  Generally the radius is mathematically evaluated as

     $r= \frac{d}{2}$

=>    $r = \frac{0.001}{2}$

=>     $r = 0.0005 \ m$

The cross sectional area is mathematically represented as

     $A = \pi r^2$

=>   $A = 3.142 * (0.005)^2$

=>    $A = 7.855*10^{-7}\ m^2$

Generally the tensile strength of nylon is mathematically represented as

      $\sigma = \frac{T}{ A }$

Where T is the tension on the maximum tension on the string

 So  

           $T = \sigma * A$

=>          $T = 600*10^{6} * 7.855*10^{-7}$

=>         $T = 471.3 \ N$

Form the equation above  we  see that

        $T \ \alpha \ A$

So if the tension is  increased to prevent breakage the thickness of the string is increased(i. e the cross-sectional  area )

String breakage when the racket hit the ball is as a result of the string not being thick enough to withstand the increase in tension