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?

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