Answer
The minimum diameter of the wire is $1.7~mm$
Work Step by Step
We can find the acrobat's weight:
$weight = mg = (55~kg)(9.80~m/s^2) = 539~N$
We can find the minimum radius:
$\frac{F}{A} = 2.5\times 10^8~N/m^2$
$A = \frac{F}{2.5\times 10^8~N/m^2}$
$\pi~r^2 = \frac{F}{2.5\times 10^8~N/m^2}$
$r = \sqrt{\frac{F}{(\pi)(2.5\times 10^8~N/m^2)}}$
$r = \sqrt{\frac{539~N}{(\pi)(2.5\times 10^8~N/m^2)}}$
$r = 8.28\times 10^{-4}~m$
$r = 0.828~mm$
Since the diameter is twice the radius, the minimum diameter of the wire is $1.7~mm$
