Question

TV broadcast antennas are the tallest artificial structures on Earth. In 1987, a 79.0 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 0.125 m in radius? (Assume a Young's modulus of 20 ⨯ 1010 Pa.)

Answer 1

`2.92* 10^(-4)` m

Step-by-step explanation:

`M` = mass of physicist and his equipment together = 79 + 400 = 479 kg

Force applied by weight of physicist and his equipment together on the antenna is given as

`F = Mg`

`F = (479)(9.8)`

`F = 4694.2` N

`L` = original length of the antenna = 610 m

`\Delta L` = compression in the length of the antenna

`r` = radius of the cylinder = 0.125 m

Area of cross-section of the cylinder is given as

`A = \pi r^(2)`

`A = (3.14) (0.125)^(2)`

`A = 0.0491` m²

`Y` = Young's modulus of steel = 20 x 10¹⁰ Pa

Young's modulus is given as

`Y = (FL)/(A\Delta L)`

`\Delta L = (FL)/(AY)`

Inserting the above values

`\Delta L = ((4694.2)(610))/((0.049)(20* 10^(10)))`

`\Delta L = 2.92* 10^(-4)` m