1. Aim: How do we explain the force of friction?

2. Friction
When an object is in contact with a surface, the surface exerts a contact force on the object. The contact force that acts parallel to the surface, opposing the motion of an object is called friction.

3. What does friction depend on?
We will investigate what factors friction depends on.

4. Static Friction vs Kinetic Friction
Static friction- the force that must be overcome to set an object in motion. Static friction acts on objects at rest Kinetic Friction- the force of friction that acts on an object that is moving.

5. Turn and Talk Question
Which is greater the static friction or the kinetic friction? (Is it easier to get an object to move or keep it moving once already in motion?) The static friction is greater than the kinetic friction

6. Equations for Friction
Ff static =< μs FN Ff kinetic = μk FN Ff = force of friction FN= normal force μs = coefficient of static friction μk = coefficient of kinetic friction

7. Freebody Diagram (Note: Fa is the applied force)
The notation for some of these forces is different than what we use.
Identify the following forces:
N=Normal Force
W= Weight
Fs=Static Friction

8. More questions on the freebody diagram
If the block is pushed across the floor,
What must be the relationship between the normal force acting on the block and the weight of the block? They are equal
What must be the relationship the applied force and the friction force if the object is to accelerate? The applied force has to be greater than friction

9. Does friction depend on weight or the normal force?

10. Sample Friction Problem
What force must one exert on a 25 kg wooden box on a wooden floor in order to make it move?
Fg =mg= 25kg(9.8 m/s2)=245 N
Ff =µs FN =0.42 (245 N)=102.9N

11. Sample Friction Problem
2. What force must be applied to move a 30 kg steel at rest crate across a stainless steel floor? Fg =mg=30 kg (9.8m/s2 )=294 N Ff=µsFN=0.74(294N)= N

12. Sample Friction Problems
3. A 60 kg skier (who wears waxed skis) slides to a stop across the snow. What is the force of kinetic friction acting on the skier? Fg=mg=60kg(9.8m/s2)=588N Ff=µkFN=(0.05)(588N)=29.4N

13. 4. A car of mass 700 kg (in neutral) is being pushed across a dry asphalt road with a force of 7500 N.
Calculate the weight of the car.
Fg=mg=700kg(9.8m/s2)=6860N
b) Calculate the friction force acting against the car. (Note: The tires are rubber)
Ff=µkFN=0.67(6860N)=4596.2N
c) Calculate the net force acting on the car. Fnet= =2263.8N
d) Does the car accelerate? If so, how much?
Fnet=ma
2263.8N=700kg(a)
a=3.2m/s2

14. Thought Question
Which way is easier to move the child in the sled? Explain why.
It is easier to move the child as shown in B because there is less friction due to the fact that the normal force is smaller than in situation A