A car \((m = 1700 \, \text{kg})\) is parked on a road that rises 15 degrees above the horizontal.
What are the magnitudes of:
a. The normal force
b. The static frictional force that the ground exerts on the tires?
Answer (3)
A) You need to work out the forces perpendicular to the road. So we can say: N or ma l f orce = ( 1700 × 9.8 ) × s in ( 15 ) ∴ N or ma l f orce = 4311.93 N ( 2 d . p . ) This is the normal force acting against the car.
B) This is very similar to the working out of part A. We can say that since the car is not moving, something is stopping the car from moving forwards. The friction is doing this. We can summarise that the force of the friction is acting parallel to the road. We can say: ( 1700 × 9.8 ) × cos ( 15 ) = 16092.32 N ( 2 d . p . ) The is the amount of friction is newtons that is stopping the car from moving forward.]
Hoped this wasn't too confusing, and that it helped!
When a car is parked on an incline, the gravitational force acts downwards, and this force can be broken down into two components: one perpendicular to the slope (which contributes to the normal force) and one parallel to the slope (which can potentially cause the car to slide down if not countered by static friction). To find the normal force and the static frictional force, we first need to calculate the weight of the car, which is given by Fg = mg (where m is the mass and g is the acceleration due to gravity, approximately 9.8 m/s2).
a. To find the normal force (N), we use the component of the car's weight that is perpendicular to the slope: N = mg cos(\u03b8), where \u03b8 is the angle of the slope. Since the car's mass is 1700 kg and the angle \u03b8 is 15 degrees, we have N = 1700 kg \u00d7 9.8 m/s2 \u00d7 cos(15\u00b0) = 16,165 N, approximately.
b. The static frictional force only acts if there is a component of weight parallel to the slope trying to move the car. This is given by mg sin(\u03b8), and it is the maximum force that static friction can counteract without the car slipping. Since the car is not moving, the actual static frictional force is equal to the component of gravitational force parallel to the incline up to the maximum possible static frictional force. The magnitude is f = mg sin(\u03b8) = 1700 kg \u00d7 9.8 m/s2 \u00d7 sin(15\u00b0) = 4,396 N, approximately.
The normal force acting on the car is approximately 16026 N, while the static frictional force that prevents the car from sliding down the incline is approximately 4304 N.
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