Week. Student will: Incorporate Force of Friction into calculation Static Friction Kinetic Friction.

Week

Friction

Objective

Student will: Incorporate Force of Friction into

calculationStatic FrictionKinetic Friction

Cornell Notes (1/3)Friction: The force exerted by a surface as an object moves across it or tries to move across it.

Force of Friction = Coefficient of Friction x Normal Force

Force: Newton (N):no units

Normal Force: Newton (N)

Questions1) How is friction

define?

Know how to calculate Normal Force

Cornell Notes (2/3)Two Kinds Of Friction Coefficient

s

Coefficient of static

friction ()

Use if two surfaces are not moving

Coefficient of kinetic friction ()

Use if two surfaces

are moving

Cornell Notes (1/5)

Example: FrictionThe Flash is pulling a car, that weighs 1,200 kg, to a nearby gas station and makes the car accelerate at 10 m/s2. Assume friction and the coefficient of kinetic friction "µk" is 0.9. Compute the Flash’s applied force on the car.

Cornell Notes (2/5)

The Flash is pulling a car, that weighs 1,200 kg, to a nearby gas station and makes the car accelerate at 10 m/s2. Assume friction and the coefficient of kinetic friction ("µ)" is 0.9. Compute the Flash’s applied force on the car.

Given:m = 1200 kg = 10 m/s2

µk = 0.9 = 9.8 m/s2

Unknown: = ?

Steps1) Define

Cornell Notes (3/5)Choose an equation or situation:

Rearrange the equation to isolate the unknown:

2) Plan

F⃗net=m a⃗

F⃗app=m a⃗

F⃗app+(−F⃗ fr )m

= a⃗

F⃗app−μm g⃗

m= a⃗

F⃗ fr=μ F⃗ N⃗FN=m g⃗

F⃗app− μm g⃗= a⃗m

F⃗app= a⃗m+μm g⃗F⃗app+(−μ F⃗ fr )

m=a⃗

Cornell Notes (4/5)Substitute the values into the equation and solve

3) Calculate

F⃗app=22584

F⃗app=22584 N

F⃗app=10 ∙1200+0.9∙1200 ∙9.8

F⃗app=10 ∙1200+0.9∙1200 ∙9.8

F⃗app= a⃗m+μm g⃗

Cornell Notes (5/5)

The Flash’s applied force is 22,584 N.

4) Evaluate

Cornell Notes (1/5)

Example: Coefficients of FrictionWhen Colossus pushes a 2000 kg shipping box resting on

a road, he exerts a force of 1800 N to start it sliding. Once it is sliding, he uses a force of only 1500 N to keep the shipping box moving with constant speed. What are the coefficients of static and kinetic friction between the shipping box and the road?

Cornell Notes (2/5)Given:

m = 2000 kg 1800 N1500 N = 0 N = 0 N

Unknown:

Steps1) Define

When Colossus pushes a 2000 kg shipping box resting on a road, he exerts a force of 1800 N to start it sliding. Once it is sliding, he uses a force of only 1500 N to keep the shipping box moving with constant speed. What are the coefficients of static and kinetic friction between the shipping box and the road?

Cornell Notes (3/5)Choose an equation or situation:

Rearrange the equation to isolate the unknown:

2) Plan

Fn = mg because it is

on a horizontal surface

�⃗� 𝑎𝑝𝑝 . 𝑠− �⃗� 𝑓𝑟 . 𝑠=𝑚�⃗�s�⃗� 𝑎𝑝𝑝 .𝑘−𝐹 𝑓𝑟 .𝑘=𝑚�⃗�𝑘

�⃗� 𝑎𝑝𝑝 . 𝑠− �⃗� 𝑓𝑟 . 𝑠=0�⃗� 𝑎𝑝𝑝 . 𝑠=�⃗� 𝑓𝑟 .𝑠

�⃗� 𝑎𝑝𝑝 . 𝑠=𝑢𝑠 ∙ �⃗�𝑁

�⃗�𝑎𝑝𝑝 . 𝑠

𝑚�⃗�=𝑢𝑠

�⃗� 𝑎𝑝𝑝 . 𝑠=𝑢𝑠 ∙𝑚�⃗�

�⃗� 𝑎𝑝𝑝 .𝑘−𝐹 𝑓𝑟 .𝑘=0�⃗� 𝑎𝑝𝑝 .𝑘=�⃗� 𝑓𝑟 .𝑘

�⃗� 𝑎𝑝𝑝 .𝑘=𝑢𝑘 ∙𝐹 𝑁

�⃗�𝑎𝑝𝑝 .𝑘

𝑚�⃗�=𝑢𝑘

�⃗� 𝑎𝑝𝑝 .𝑘=𝑢𝑘 ∙𝑚�⃗�

�⃗� 𝑎𝑝𝑝 . 𝑠− �⃗� 𝑓𝑟 . 𝑠=𝑚�⃗� s⃗𝐹 𝑎𝑝𝑝 .𝑘−𝐹 𝑓𝑟 .𝑘=𝑚�⃗�𝑘

Cornell Notes (4/5)Substitute the values into the equation and solve

3) Calculate

.091743119266

.092

.076452599388

.076

18002000 ∙9.8

=𝑢𝑠

Coefficient of frictions do not have units

15002000 ∙9.8

=𝑢𝑘

�⃗�𝑎𝑝𝑝 . 𝑠

𝑚�⃗�=𝑢𝑠

�⃗�𝑎𝑝𝑝 .𝑘

𝑚�⃗�=𝑢𝑘

Cornell Notes (5/5)

The coefficient of static friction is .092 and coefficient of kinetic friction is .076.

4) Evaluate