Answer:
see description
Explanation:
This is a problem where we have to assume we are in the linear region of the stress-strain curve for the antenna.
The physicist has a mass of 69 [Kg] and his equipment has a mass of 400 [Kg].
Now we just have to add those up, we end up with a mass of 469 [Kg].
The force exerted to the antenna (downwards) will be:
[tex]F = 469 [Kg] * 9.8 [m/s^2]=4956.2 [N][/tex]
To find the compression we have that the relationship between strain we take the equation:
[tex]\gamma = \frac{F*l_{o}}{A* \epsilon}[/tex]
where [tex]\gamma[/tex] is the young modulus, [tex]\epsilon[/tex] is the strain on the material, [tex]F[/tex] is the force exerted by the physicist+equipment, [tex]A[/tex] is the cross-section area of the cylinder, and [tex]l_{o}[/tex] is the initial longitude of the cylinder.
now the young modulus for steel is [tex]\gamma = 200*10^9 [Pa][/tex].
The cross section area for the cylinder is [tex]A = \pi *r^2 = \pi *(0.175)^2=0.0962[m^2][/tex]
The length of the antenna is [tex]l_{0}=610 [m][/tex]
we find [tex]\epsilon[/tex]:
[tex]\epsilon = \frac{F*l_{o}}{A* \gamma}=\frac{4956.2*610}{0.0962*200*10^9} = 0.000157 [m] = 0.157 [mm][/tex]
