The tibia is a lower leg bone (shin bone) in a human. The maximum strain that the tibia can experience before fracturing corresponds to a 1

Question

The tibia is a lower leg bone (shin bone) in a human. The maximum strain that the tibia can experience before fracturing corresponds to a 1 % change in length.
A. Young’s modulus for bone is about Y = 1.4 x 10 N/m². The tibia (shin bone) of a human is 0.35 m long and has an average cross-sectional area of 2.9 cm. What is the effective spring constant of the tibia?
B. If a man weighs 750 N, how much is the tibia compressed if it supports half his weight?
C. What is the maximum force that can be applied to a tibia with a cross-sectional area, A = 2.90 cm?

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niczorrrr 6 months 2021-08-08T14:07:51+00:00 1 Answers 5 views 0

Answers ( )

    0
    2021-08-08T14:09:21+00:00

    Answer:

    a

       k    =  11600000 N/m

    b

       \Delta  L  =  3.2323 *10^{-5} \ m

    c

      F =  3750.28 \  N  

    Explanation:

    From the question we are told that

        The Young modulus is  E =  1.4 *10^{10} \  N/m^2

         The length is  L  =  0.35 \ m

          The  area is  2.9 \ cm^2  =  2.9 *10^{-4} \ m ^2

       

    Generally the force acting on the tibia is mathematically represented as

           F =  \frac{E *  A  *  \Delta  L }{L}    derived from young modulus equation

    Now this force can also be mathematically represented as

          F =  k *  \Delta  L    

    So

         k    =  \frac{E *  A  }{L}

    substituting values

         k    =  \frac{1.4 *10^{10} *  2.9 *10^{-4}  }{ 0.35}

         k    =  11600000 N/m

        Since the tibia support half the weight then the force experienced by the tibia is  

            F_k  =  \frac{750 }{2}  =  375 \  N

     From the above equation the extension (compression) is mathematically represented as

              \Delta  L  =  \frac{ F_k  *  L  }{ A *  E }        

    substituting values

               \Delta  L  =  \frac{  375   *  0.35  }{ (2.9 *10^{-4}) *   1.4*10^{10} }

               \Delta  L  =  3.2323 *10^{-5} \ m

    From the above equation the maximum force is  

            F =  \frac{1.4*10^{10} *  (2.9*10^{-4})  *  3.233*10^{-5} }{ 0.35}  

             F =  3750.28 \  N  

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