1. FRICTION

2. Friction- a force that resists the motion of an object
Acts parallel to the surface and opposite the direction of the motion.
Dependent on the type of contact surfaces.
Independent of contact area.
Equal to applied force when object is at rest or traveling at a constant velocity.
Fnet

3. TYPES OF FRICTION
Static friction(starting friction)- Friction that is produced by surface projections and bonding.
Kinetic friction(sliding friction)- friction an object has while in motion.
Rolling friction-one object rolls over another
Air resistance- friction produced by the air pushing against something in free fall.

4. Ff = µ FN FORMULA Ff = frictional force (N) FN = normal force (N)
µ-Coefficient of friction
*(µ) Has No Units*
*See reference table*

5. ExA. A block of wood rests on a wood desktop
ExA. A block of wood rests on a wood desktop. If it has a mass of 10kg, what is the minimum force required to move it? FN = Fg = mg FN = 10kg(9.81m/s2) FN = 98.1N Ff = µ FN = .42(98.1N) = 41.2N
FN= 98.1N
Ff= 41.2N
F= 41.2N
Fg= 98.1N
All forces are balanced if the object is at rest or traveling
at a constant velocity.
The value for static friction is used because its at rest.

6. exB-If a 60N force is applied to the block, what will be its acceleration? Fnet = 60N – 41.2N Fnet = 18.8N a = Fnet/m = 18.8N/10kg a = 1.88 m/s2
Ff= 41.2N
F = 60N

7. FORCE VECTOR DIAGRAMS STATIONARY OBJECT: Fg- weight Fg = mg Fn
Fn – normal force- force
pushing surfaces
together Fn = Fg Fn Fg

8. Applied Parallel forceFn
Fp - Parallel force
Ff – frictional force
Fn – normal force
Fg- weight- Fn Ff Fp Fg
At constant velocity Fp = Ff

9. Example
A cross country skier has a mass of 67kg and is sliding on waxed ski’s across a flat snow covered surface. If they have a constant velocity,
a) what is their normal force?
B) what is the frictional force?
FN=Fg=mg
FN= 67kg(9.81m/s2)
FN= N
Ff=µFN
Ff= .05(657.27N)
Ff= 32.82N

10. Applied force at an angle
Fa – applied force
Fp – parallel force
Fp = Fa cosΘ = Ff
Fy = FasinΘ
Fn = Fg - Fy Fn Fa Fy Fp Ff
Fg = mg

11. A worker pulls a 40kg crate across the floor at a constant velocity
A worker pulls a 40kg crate across the floor at a constant velocity. If he applies 214N of force along a rope held at 30 degrees with the ground, find Fg , Fp, Fy , Fn , and μ.
Fp = Fa cosθ = Ff
Fp = 214N cos 30
Fp = 185.3N
Fg = mg
F = 40kg(9.81m/s2)
Fg =392.4N
Fy = Fa sin θ
Fy = 214N sin 30
Fy = 107N

12. Fn = Fg – Fy
Fn = 392.4N – 107N
Fn = 285.4N
μ = Ff / Fn
μ = 185.3N/ 285.4N
μ = 0.65

13. Sliding object on an incline
Fg = mg (always perpendicular
to the ground) Ff Fn
Fn = Fg cosθ (perpendicular to
the surface) Ѳ
Fp = Ff ( both are parallel to
the surface)
Fp = Fg sinθ Fp Fn Fg

14. Ex: A 20kg wooden box is sliding down an incline of 30 degrees at a constant velocity. Find Fg , Fp , Fn and μ.
Fg = m g
Fg = 20kg(9.81m/s2)
Fg = 196N θ Fg Fn Fp
Fn = Fg cosѲ
Fn= 196N cos 30
Fn = 170N
Fp = Fg sin Ѳ
Fp = 196N sin 30
Fp= 98N
µ = Ff / Fn = 98N/170N = .576

15. Pulling an object up an incline
Fnet = Fp – Ff
So Fp = Fnet + Ff
Fparallel
Ffriction
Fp ≠Ff

16. ExA: A net force of 35N is required to move a 10kg crate up an incline of 20 degrees at a constant velocity. What is the parallel force applied to the block?
Ff = Fgsinϴ
Ff=10kg(9.81m/s2)sin 20°
Ff =33.5N
Fp=Fnet+ Ff
Fp= 35N N
Fp = 68.5N

17. exB: What is the coefficient of sliding friction between the crate and the incline?
µ = Ff /Fn Fn= Fgcosϴ Fg=mg µ = Ff = 33.5N mg cosϴ 10kg(9.81m/s2)cos 20° µ = 33.5N / 92.18N µ = .36

18. COOL MATH TRICK Sin Ѳ = opp/hyp cos Ѳ= adj/hyp
µ = Ff = Fgsin Ѳ = opp/hyp = opp/adj
Fn Fg cos Ѳ adj/hyp
µ = tan Ѳ
*Only works for objects sliding down an incline*