### click - Uplift Education

```FRICTION
WHAT DO YOU REMEMBER ABOUT FRICTION?
WHAT DO YOU REMEMBER ABOUT FRICTION?



Friction opposes motion
Friction is dependent on the texture of the surfaces
Friction is dependent on normal force
motion
friction
WHAT DO YOU REMEMBER ABOUT FRICTION?



Friction opposes motion
Friction is dependent on the texture of the surfaces
Friction is dependent on normal force
Ffr = m Fn
μ is called coefficient of friction
 m has no units
 depends on characteristics of both surfaces
 Higher m = rougher surface / more friction
motion
friction
Note: Friction does NOT
depend on the surface
area of contact
The coefficient of friction is different for when an object is at rest
and when it is moving.
• μs = coefficient of static friction (object at rest)
• μk = coefficient of kinetic friction (object moving)
Static friction is greater than kinetic friction - its harder to start
an object moving than it is to keep it moving.
surface-on-surface
μs
μk
hook velcro-on-fuzzy velcro
>6.0
>5.9
avg tire-on-dry pavement
0.9
0.8
grooved tire-on-wet pavement
0.8
0.7
smooth tire-on-wet pavement
0.5
0.4
metal-on-metal (lubricated)
0.1
0.05
steel-on-ice
0.1
0.05
steel-on-Teflon
0.05
0.05
FRICTION PROBLEMS – WE DO
A 28 kg crate initially at rest on a horizontal floor
requires a 75 N horizontal force to set it in motion. Find
the coefficient of static friction between the crate and
the floor.
What do we do first?
FRICTION PROBLEMS – WE DO
A 28 kg crate initially at rest on a horizontal floor requires a
75 N horizontal force to set it in motion. Find the coefficient
of static friction between the crate and the floor.
Remember our strategy:
1) Draw a free body diagram
2) Identify all variables
3) Identify relevant equations
4) Solve!
FRICTION PROBLEMS – WE DO
A 28 kg crate initially at rest on a horizontal floor
requires a 75 N horizontal force to set it in motion. Find
the coefficient of static friction between the crate and
the floor.
μs = 0.27
FRICTION PROBLEMS – WE DO
A force of 40.0 N accelerates a 5.0-kg block at 6.0 m/s2
along a horizontal surface.
a. How large is the frictional force?
b. What is the coefficient of friction?
FRICTION PROBLEMS – WE DO
A force of 40.0 N accelerates a 5.0-kg block at 6.0 m/s2
along a horizontal surface.
a. How large is the frictional force?
b. What is the coefficient of friction?
Ff = 10 N
μk = 0.2
FRICTION PROBLEMS – WE DO
2) A 12 kg suitcase is pushed with a force of 38 N to the
left. If the coefficient of kinetic friction between the
suitcase and the floor is 0.3, how far will the suitcase
move after 5 sec?
2.8 m
FRICTION PROBLEMS – YOU DO
FOLLOW THE STEPS! Make a plan before you plug in
numbers! Be able to explain your reasoning!
1)
A 30 kg crate requires a 53 N force to keep it moving
at 1 m/s. Find the coefficient of kinetic friction.
mk = 0.18
2)
You need to move a 105-kg sofa to a different
location in the room. It takes a 403-N force to start
the sofa moving. What is the coefficient of static
friction between the sofa and the carpet?
Question:
How does the weight of a person in an elevator
depend on the motion of that elevator?
What will the scale show if the elevator is
1. at rest or moving with constant speed
2. speeding up
3. slowing down
Newton’s 3. law:
Force with which the person acts on the scale (reading of the scale) is
equal to the normal force on the person.
So, if we find normal force we know the
reading of the scale, so called APPARENT WEIGHT
Let’s assume that elevator is moving upward, and let this be positive direction.
1. draw free body diagram 2. apply Newton’s 2. law : Fnet = ma
+
Fn
1. elevator is at rest or moving with constant speed
mg
Fn – mg = ma = 0
→ Fn = mg
apparent weight = weight
Fn
2. elevator is speeding up: a is positive
Fn – mg = ma
mg
→ Fn = mg + ma
apparent weight > weight
Fn
mg
3. elevator is slowing down: a is negative
Fn – mg = - ma
→ Fn = mg - ma
apparent weight < weight
the scale would show less, and you would feel lighter
YOU DO
You are riding in an elevator holding a spring scale with a
1kg mass suspended from it. You look at the scale and
see that it reads 9.3 N. What, if anything, can you
conclude about the elevator’s motion at that time?
EXIT TICKET
1) Solve the elevator problem for an elevator
traveling downward. YOU MUST EXPLAIN YOUR
REASONING USING NEWTON’S 2nd LAW
2) Write down 2 things you learned friction
problems.
```