Friction Relationships

Ranking Task

Rank the friction on the 1.0 kg from greatest to least

rampfriction
Friction graphics

Friction
Fill In The Blanks

Phonebook Friction
- Mythbusters

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Download the file below.

Mythbusters_-_Phone_Book_Friction

Friction Problems

*Remember
Fn = Weight = mg
on horizontal surface

Ex) A 5.0 kg Steel block is resting on a *horizontal table*. The coefficient of static friction (us) is 0.75 and the uk is 0.57.

a) What minimum force is needed to start this block moving (fs)?

↓

fs = Fnus = Fgus = mgus

= 5.0 kg(9.8 m/s2)(.75)

= 37. N

b) What is the frictional force on this object as it moves?

↓

fk = mguk

= 5.0 kg(9.8 m/s2)(.57)

= 28 N

c) What force must be applied to the object to keep it moving at constant velocity? (Hint: Fnet = 0)

At Constant velocity

↓

F applied = friction force

= 28 N

Friction Questions

Coefficients of Friction & F = ma

Remember ...

Static Friction    fs = UsFn

Kinetic Friction  fk = UkFn

On Horizontal Surface

fs = usFg   (force to start object moving)

fk = ukW    AND    W = mg

Example 1

A 1000. N car skids on a wet concrete road. If the road is horizontal, what is the friction force on the car?

w = 1000. N     Uk = .58     fk = ?

Uktable

↓

fk = UkFg = .58 x 1000. N

= 5.8 x 102 N

Example 2

A skier is being pulled along a horizontal surface at constant speed with a force of 50. N. What is the weight of the skier?

At constant speed:
fk = Applied Force

fk = 50. N     w = ?

Uktable

uk = .05

↓

fk = ukFg

50. N = .05[Fg]

w = 1000 N

Example 3

A 10. kg block of wood sliding on a horizontal wooden table is brought to rest. What is the force on the block of wood that caused it to stop?

The force of kinetic friction
stops the block.

fk = ?

Uktable

Uk = .30

↓

fk = Fnuk = Fguk = mguk

= (10. kg)(9.8 m/s2).30

= 29. N

Example 4

A 53. kg block, slowed by friction, has an acceleration of -0.10 m/s2. What is the force of friction on the block?

m = 53. kg     a = -.10 m/s2

↓

fk = F = ma

fk = F = 53. kg(-.10 m/s2)

= -5.3 N

Example 5

A force of 8.0 N gives a 3.0 kg mass an acceleration of 2.0 m/s2 to the right. What is the friction on the block?

Fapplied = 8.0 N Right     m = 3.0 kg     a = 2.0 m/s2 Right

Fnet = ma = 3.0 kg(2.0 m/s2)

= 6.0 N

What friction force will produce a net force of 6.0 N?

↓

friction = 2 N left

Extension: Friction on a Ramp

How does the magnitude of the friction force on a box on a ramp change when the angle of the ramp increases?

↓

Friction decreases because the
normal force decreases

Review - Which law of motion best explains this?

School Blocks YouTube? Download the file below.

LiquidNitrogen_pingpong

↓

3rd Law of Motion
Action/Reaction

Review

Friction Video
Georgia Pacific Broadcasting

Check Your Understanding

5 quick questions — pick an answer to see if you’re right.

1. A 5.0 kg block rests on a horizontal table with μs = 0.75. What minimum force is needed to start it moving? (g = 9.8 m/s²)

2. The same 5.0 kg block has μk = 0.57. What applied force keeps it moving at constant velocity? (g = 9.8 m/s²)

3. A skier is pulled along level snow at constant speed with a 50. N force. If μk = 0.05, what is the skier's weight?

4. An 8.0 N force gives a 3.0 kg block an acceleration of 2.0 m/s² to the right. What is the friction force?

5. A box slides down a ramp. As the ramp angle is increased, the kinetic friction force on the box…

NYSSLS Regents Practice · Dynamics · Solving Problems

Moving a Crate in a Warehouse

A worker pushes a 50.-kg wooden crate across a level concrete floor. The coefficient of kinetic friction between the crate and the floor is 0.30.

What is the magnitude of the kinetic friction force on the sliding crate?

▶ Want a challenge? Open 6 more Regents-style questions  (harder · mixed formats)

Challenge score: 0 / 6

Support or Refute · Justify

C1. A student makes the following claim.

Claim: To slide the crate at constant velocity twice as fast, the worker must push twice as hard.

Place a check mark in either the Support or Refute box. Justify your response using a mathematical relationship.

Multiple choice

C2. The worker turns the crate onto a side with twice the contact area. Assuming the same surfaces, what happens to the friction force?

Substitute values

C3. The worker pushes horizontally with 250 N. Substitute numerical values into a mathematical representation that can be used to determine the crate’s acceleration. Then calculate the acceleration.

a = Fpush − Ffm = N − N kg = m/s2
Determine the value

C4. The crate is sliding when the worker stops pushing. Determine the magnitude of the crate’s acceleration as friction slows it down.

a = m/s2
Multiple choice

C5. To make the crate easier to slide, the company is considering several changes. Which change would reduce the friction force the most?

Constructed response

C6. The worker can either push down on the crate at an angle or pull up on a rope attached to it at the same angle, using the same force. Explain which choice results in less friction, using the normal force in your answer.