Answer
(a) $F = 2.2\times 10^6~N$
(b) $\theta = -27.3^{\circ}$
Work Step by Step
The force on the plane has a vertical component $F_y$ directed toward the ground and a horizontal component $F_x$ directed toward the center of the circle.
We can find the magnitude of $F_y$.
$F_y = m~a_y$
$F_y = (85000~kg)(12~m/s)$
$F_y = 1.02\times 10^6~N$
We can find the magnitude of $F_x$ which is equal to the centripetal force on the plane.
$F_x = \frac{mv^2}{r}$
$F_x = \frac{(85000~kg)(55~m/s)^2}{130~m}$
$F_x = 1.98\times 10^6~N$
(a) We can find the magnitude of the net force.
$F = \sqrt{(F_x)^2+(F_y)^2}$
$F = \sqrt{(1.98\times 10^6~N)^2+(1.02\times 10^6~N)^2}$
$F = 2.2\times 10^6~N$
(b) We can find the angle below the horizontal.
$tan(\theta) = \frac{F_y}{F_x}$
$\theta = arctan(\frac{1.02\times 10^6~N}{1.98\times 10^6~N})$
$\theta = -27.3^{\circ}$
