#### Answer

(a) $\theta = 20.9^{\circ}$
The truck does not need to go slower than the car.
(b) The normal force on the car is 11,800 N.
The normal force on the truck is 23,600 N.

#### Work Step by Step

(a) We can convert the speed to units of m/s.
$v = (65.0~mi/h)(\frac{1609~m}{1~mi})(\frac{1~h}{3600~s})$
$v = 29.0~m/s$
We can set up one force equation each for the horizontal direction and the vertical direction.
$F_N~sin(\theta) = \frac{mv^2}{r}$
$F_N~cos(\theta) = mg$
We can divide the first equation by the second equation.
$tan(\theta) = \frac{v^2}{gr}$
$\theta = arctan(\frac{v^2}{gr})$
$\theta = arctan(\frac{(29.0~m/s)^2}{(9.80~m/s^2)(225~m)})$
$\theta = 20.9^{\circ}$
The speed to safely round the curve does not depend on the mass of the vehicle, so the truck does not need to go slower than the car.
(b) We can find an expression for the normal force.
$F_N~cos(\theta) = mg$
$F_N = \frac{mg}{cos(\theta)}$
We can find the normal force on the car.
$F_N = \frac{mg}{cos(\theta)}$
$F_N = \frac{(1125~kg)(9.80~m/s^2)}{cos(20.9^{\circ})}$
$F_N = 11,800~N$
The normal force on the car is 11,800 N.
We can find the normal force on the truck.
$F_N = \frac{mg}{cos(\theta)}$
$F_N = \frac{(2250~kg)(9.80~m/s^2)}{cos(20.9^{\circ})}$
$F_N = 23,600~N$
The normal force on the truck is 23,600 N.