Sure, let's go through this problem step-by-step to find the magnitudes of the normal force and the static frictional force that the ground exerts on the tires.
### Given Data:
1. Mass of the car, \( m = 1900 \) kg
2. Angle of the road above the horizontal, \( \theta = 10^\circ \)
3. Acceleration due to gravity, \( g = 9.81 \) m/s²
### (a) The Normal Force \((F_N)\):
The normal force is the component of the gravitational force perpendicular to the surface of the incline. It can be determined using the following formula:
[tex]\[ F_N = mg \cos(\theta) \][/tex]
First, we need to convert the angle from degrees to radians because trigonometric functions in physics typically use radians.
[tex]\[ \theta_{rad} = \frac{10^\circ \times \pi}{180^\circ} \][/tex]
Now, calculating the normal force:
[tex]\[ F_N = 1900 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times \cos(10^\circ) \approx 18355.83 \, \text{N} \][/tex]
So, the magnitude of the normal force is approximately:
[tex]\[ F_N \approx 18355.83 \, \text{N} \][/tex]
### (b) The Static Frictional Force \((F_s)\):
The static frictional force is the component of the gravitational force parallel to the surface of the incline. It can be determined using the following formula:
[tex]\[ F_s = mg \sin(\theta) \][/tex]
Using the angle in radians again:
[tex]\[ \theta_{rad} = \frac{10^\circ \times \pi}{180^\circ} \][/tex]
Now, calculating the static frictional force:
[tex]\[ F_s = 1900 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times \sin(10^\circ) \approx 3236.63 \, \text{N} \][/tex]
So, the magnitude of the static frictional force is approximately:
[tex]\[ F_s \approx 3236.63 \, \text{N} \][/tex]
### Summary:
(a) The magnitude of the normal force is approximately \( 18355.83 \, \text{N} \).
(b) The magnitude of the static frictional force is approximately [tex]\( 3236.63 \, \text{N} \)[/tex].
