Question

TV broadcast antennas are the tallest artificial structures on Earth. In 1987, a 69.0 kg physicist placed himself and 400 kg of equipment at the top of one 610 m high antenna to perform gravity experiments. By how much (in mm) was the antenna compressed, if we consider it to be equivalent to a steel cylinder 0.175 m in radius?

Answer 1

Answer:

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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:

$F = 469 [Kg] * 9.8 [m/s^2]=4956.2 [N]$

To find the compression we have that the relationship between strain we take the equation:

$\gamma = \frac{F*l_{o}}{A* \epsilon}$

where $\gamma$ is the young modulus, $\epsilon$ is the strain on the material, $F$ is the force exerted by the physicist+equipment, $A$ is the cross-section area of the cylinder, and $l_{o}$ is the initial longitude of the cylinder.

now the young modulus for steel is $\gamma = 200*10^9 [Pa]$.

The cross section area for the cylinder is $A = \pi *r^2 = \pi *(0.175)^2=0.0962[m^2]$

The length of the antenna is $l_{0}=610 [m]$

we find $\epsilon$:

$\epsilon = \frac{F*l_{o}}{A* \gamma}=\frac{4956.2*610}{0.0962*200*10^9} = 0.000157 [m] = 0.157 [mm]$