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

Answer:

$2.92\times 10^{-4}$ m

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

$Y = \frac{FL}{A\Delta L}$

$\Delta L = \frac{FL}{AY}$

Inserting the above values

$\Delta L = \frac{(4694.2)(610)}{(0.049)(20\times 10^{10})}$

$\Delta L = 2.92\times 10^{-4}$ m