The Achilles tendon, which connects the calf muscles to the heel, is the thickest and strongest tendon in the body. In extreme a
ctivities, such as sprinting, it can be subjected to forces as high as 13.0 times a person's weight. According to one set of experiments, the average area of the Achilles tendon is 78.2 mm^2 , its average length is 26 cm , and its average Young's modulus is 1474 MPa . How much tensile stress is required to stretch this muscle by 5.2% of its length? If we model the tendon as a spring, what is its force constant? If a 75 kg sprinter exerts a force of 13.0 times his weight on his Achilles tendon, by how much will it stretch?
1 answer:
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Answer:
12164.4 Nm
Explanation:
CHECK THE ATTACHMENT
Given values are;
m1= 470 kg
x= 4m
m2= 75kg
Cm = center of mass
g= acceleration due to gravity= 9.82 m/s^2
The distance of centre of mass is x/2
Center of mass(1) = x/2
But x= 4 m
Then substitute, we have,
Center of mass(1) = 4/2 = 2m
We can find the total torque, through the summation of moments that comes from both the man and the beam.
τ = τ(1) + τ(2)
But
τ(1)= ( Center of m1 × m1 × g)= (2× 470× 9.81)
= 9221.4Nm
τ(2)= X * m2 * g = ( 4× 75 × 9.81)= 2943Nm
τ = τ(1) + τ(2)
= 9221.4Nm + 2943Nm
= 12164.4 Nm
Hence, the magnitude of the torque about the point where the beam is bolted into place is 12164.4 Nm
