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

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

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.

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 diagram2) Identify all variables3) Identify relevant equations4) Solve!5) Check your work!

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

CHECK YOUR UNDERSTANDING

CHECK YOUR UNDERSTANDING

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.

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?

mk = 0.18

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

mgLLLLLLLLLLLLLL Fn – mg = ma = 0 → Fn = mg

apparent weight = weight

+

1. elevator is at rest or moving with constant speed

2. elevator is speeding up: a is positive

mgLLLLLLLLLLLLLL Fn – mg = ma → Fn = mg + ma

apparent weight > weightthe scale would show more, and you would feel heavier

3. elevator is slowing down: a is negative

mgLLLLLLLLLLLLLL Fn – mg = - ma → Fn = mg - ma

apparent weight < weightthe scale would show less, and you would feel lighter

Fn

Fn

Fn

CHECK YOUR UNDERSTANDING

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.