Answer:
The tension in the tendon is 2.9kN
Explanation:
To solve the problem it is necessary to apply the concepts related to stress from the elongation of bodies by means of Young's Module.
The definition of Tension is given by
[tex]F = AY\frac{\Delta L}{L_0}[/tex]
Where,
A =Area
Y = Young's modulus
[tex]\Delta L =[/tex]Change in lenght
[tex]L_0 =[/tex] Initial Lenght
The change in Length for this case is given by
[tex]\Delta L[/tex] = 16.1-15 = 1.1cm
The area can be calculated as,
[tex]A = \pi r^2[/tex]
[tex]A = \pi (3*10^{-3})^2[/tex]
[tex]A = 2.827*10^{-5}m^2[/tex]
Therefore replacing the tension would be,
[tex]F = (2.827*10^{-5})(1.50×10^9)\frac{1.1*10^{-2}}{16*10^{-2}}[/tex]
[tex]F = 2915.34N[/tex]
Therefore the tension in the tendon is 2.9kN