Answer :
Answer:
[tex]33.1^{\circ}[/tex]
Explanation:
Let's start by writing the equations of the forces along the two directions:
- Vertical:
[tex]N cos \theta = mg[/tex]
where
N is the normal reaction
[tex]\theta[/tex] is the angle between the road and the horizontal
(mg) is the weight of the car, with m being its mass and g the acceleration of gravity
- Horizontal:
[tex]N sin\theta = m \frac{v^2}{r}[/tex]
where
v is the speed of the car
r is the radius of the turn
Dividing the 2nd equation by the 1st one, we get:
[tex]tan \theta = \frac{v^2}{rg}[/tex]
In this problem:
[tex]r = 800 m[/tex] (radius of the turn)
[tex]v=160 mi/h = 71.5 m/s[/tex] is the speed
[tex]g=9.8 m/s^2[/tex]
Substituting, we find:
[tex]\theta= tan^{-1} (\frac{v^2}{rg})=tan^{-1}(\frac{(71.5)^2}{(800)(9.8)})=33.1^{\circ}[/tex]