The support on the right is the same as the left.

A flexible cable is anchored at $2$ pins as shown in the figure. A load $P$ of $50\mathrm{.}5kN$ is applied in the middle of

the cable a horizontal distance $L$ of $5\mathrm{.}7m$ from each pin. The cable has a dip $D$ of $4\mathrm{.}0m.$ What is the cross

sectional area of the cable required to ensure a Factor of Safety of exactly $3\mathrm{.}5$ if the breaking stress of the

cable is $500$MPa $?$ Please provide your answer in $m{m}^2$ to three significant figures. You can neglect the

weight of the cable.