Calculating Normal and Frictional Forces on a Parked Car on an Inclined Road

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SUMMARY

The discussion focuses on calculating the normal and static frictional forces acting on a parked car with a mass of 1940 kg on an inclined road at an angle of 14.4 degrees. The normal force is determined by resolving the weight of the car into components, specifically the perpendicular force to the incline. The static frictional force is also influenced by the angle of the incline, which requires understanding the relationship between gravitational force and the incline's angle. Utilizing trigonometric functions is essential for accurate calculations.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Basic knowledge of trigonometry
  • Familiarity with force decomposition
  • Concept of static friction
NEXT STEPS
  • Study the decomposition of forces on inclined planes
  • Learn about static friction coefficients and their applications
  • Explore Newton's second law in two dimensions
  • Review examples of normal force calculations in physics
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This discussion is beneficial for physics students, educators, and anyone interested in understanding the mechanics of forces acting on objects on inclined surfaces.

luap12
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1. A car (m = 1940 kg) is parked on a road that rises 14.4 ° above the horizontal. What are the magnitudes of (a) the normal force and (b) the static frictional force that the ground exerts on the tires?



Homework Equations





3. I am not sure what I need to do here. Wouldn't the normal force just be the mass times gravity? Not sure what equations I should be using or how I should be incorporating the angle.
 
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luap12 said:
1. A car (m = 1940 kg) is parked on a road that rises 14.4 ° above the horizontal. What are the magnitudes of (a) the normal force and (b) the static frictional force that the ground exerts on the tires?



Homework Equations





3. I am not sure what I need to do here. Wouldn't the normal force just be the mass times gravity? Not sure what equations I should be using or how I should be incorporating the angle.

The normal force is the force exerted perpendicularly from the road on the car. You have to split the force of weight into one force that is pulling the object down the road at 14.4 degrees and one force perpendicular to the road. This image from SparkNotes could help you visualize the forces.
normal.gif
 
that works and makes sense now! Thanks!
 

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