Frictional Force & Normal forces

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Homework Help Overview

The problem involves a car parked on an inclined road, requiring the determination of the normal force and static frictional force acting on the tires. The context is rooted in the study of forces, particularly in relation to gravity and friction on an incline.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the forces acting on the car, including the normal force, gravitational force, and frictional force. There are questions about how these forces interact and whether the normal force is equal to the gravitational force or a component of it. Some participants attempt to clarify the relationship between the forces using free body diagrams.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the forces involved. Some have provided calculations for the normal and frictional forces based on the components of gravitational force, while others are questioning the assumptions made about net movement and force balance.

Contextual Notes

Participants are navigating the complexities of force components on an incline, with some uncertainty about the correct application of equations and the relationships between the forces. The discussion reflects a mix of attempts to clarify concepts and calculations without reaching a definitive conclusion.

onyxorca
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Homework Statement



A car (m = 1840 kg) is parked on a road that rises 18.7 ° 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



F=ma Fg=mg Ff=uN

The Attempt at a Solution



errr... what forces cancels out? is it FN=Fg=Ff or is it FN+Fg+Ff=0? i don't remember...
 
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On your free body diagram, at the angle of θ=18.7°, what are the forces acting? (After you split the weight of the car into components parallel to the road and perpendicular to the road)
 
frictional force, force of gravity, and normal force.
 
onyxorca said:
frictional force, force of gravity, and normal force.

So if there is no net movement perpendicular to the road, that is the magnitude of the normal force?
 
rock.freak667 said:
So if there is no net movement perpendicular to the road, that is the magnitude of the normal force?

the same as the gravitational force? 1 840 * 9.8 = 18 032 N?
 
onyxorca said:
the same as the gravitational force? 1 840 * 9.8 = 18 032 N?

No, the component of the weight perpendicular to the plane is equal to the normal force.
 
so that's equal to the y-component of the gravitational force?

18032 cos(18.7deg ) = 17080.10 N

and so frictional force is 18032sin18.7=5781.29 N ?
 
onyxorca said:
so that's equal to the y-component of the gravitational force?

18032 cos(18.7deg ) = 17080.10 N

and so frictional force is 18032sin18.7=5781.29 N ?

Yes but only because the net force in the x and y directions are zero.
 

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