Frictional Force & Normal forces

In summary, a car parked on a road with a 18.7 ° incline will experience a normal force and a static frictional force from the ground. The normal force is equal to the perpendicular component of the weight of the car, which is equal to 18032 cos(18.7deg ) = 17080.10 N. The static frictional force is equal to the parallel component of the weight, which is equal to 18032sin18.7=5781.29 N. Both forces are equal because the net force in the x and y directions are zero.
  • #1
onyxorca
20
0

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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  • #2
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)
 
  • #3
frictional force, force of gravity, and normal force.
 
  • #4
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?
 
  • #5
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?
 
  • #6
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.
 
  • #7
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 ?
 
  • #8
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.
 

FAQ: Frictional Force & Normal forces

1. What is frictional force?

Frictional force is the force that opposes the motion of two surfaces in contact with each other. It is caused by the roughness of the surfaces and the interlocking of their microscopic irregularities.

2. How is frictional force calculated?

Frictional force can be calculated by multiplying the coefficient of friction (a value that represents the roughness of the surfaces) by the normal force (the force exerted by one surface on the other in a direction perpendicular to the surfaces).

3. What factors affect the magnitude of frictional force?

The magnitude of frictional force is affected by the roughness of the surfaces, the weight of the objects in contact, and the force pushing the surfaces together. It also depends on the type of material the surfaces are made of.

4. How does frictional force affect the motion of objects?

Frictional force can either slow down or prevent the motion of objects. It can also cause objects to move in a different direction than the applied force, known as sideways or lateral friction.

5. What is the difference between frictional force and normal force?

Frictional force is the force that opposes motion, while normal force is the force exerted by one surface on the other in a direction perpendicular to the surfaces. Frictional force is caused by the contact between two surfaces, while normal force is the result of the weight of an object pressing down on a surface.

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