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# Topic 2.2 Extended A – Friction PowerPoint PPT Presentation

tension. T. T. T. T. friction. f. f. f. f. Force. f. T. Time. Topic 2.2 Extended A – Friction.  Recall that friction acts opposite to the intended direction of motion, and parallel to the contact surface.

Topic 2.2 Extended A – Friction

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tension

T

T

T

T

friction

f

f

f

f

Force

f

T

Time

## Topic 2.2 ExtendedA – Friction

Recall that friction acts opposite to the intended direction of motion, and parallel to the contact surface.

Suppose we begin to pull a crate to the right, with gradually increasing force.

We plot the applied force, and the friction force, as functions of time:

kinetic

friction

static

friction

static

kinetic

fk

tension

friction

Force

Time

static

kinetic

## Topic 2.2 ExtendedA – Friction

fs,max

Observe the graph of the friction force.

During the static friction phase, the static friction forcefs matches exactly the applied force.

fs increases until it reaches a maximum value fs,max.

The friction force then almost instantaneously decreases to a constant value fk, called the kinetic frictionforce.

Make note of the following general properties of the friction force:

fk= a constant

fk< fs,max

0≤ fs≤ fs,max

## Topic 2.2 ExtendedA – Friction

So what causes friction?

People in the manufacturing sector who work with metals know that the more you smoothen and polish two metal surfaces, the more strongly they stick together if brought in contact.

In fact, if suitably polished in a vacuum, they will stick so hard that they cannot be separated.

We say that the two pieces of metal have been cold-welded.

surface 1

surface 1

surface 1

surface 2

surface 2

surface 2

cold welds

## Topic 2.2 ExtendedA – Friction

At the atomic level, when two surfaces come into contact, small peaks on one surface bind with small peaks on the other surface, in a process similar to cold welding.

Applying the initial sideways force, all of the cold welds oppose the motion.

Then suddenly, the cold welds break, and new peaks contact each other and cold weld.

But if the surfaces remain in relative sliding motion, fewer welds have a chance to form.

Of course, the friction force depends on what materials the two surfaces are made of.

We define the unitless constant called the coefficient of frictionμ.

μdepends on the two surface materials.

fs ≤ μsN

static friction

fk = μkN

kinetic friction

## Topic 2.2 ExtendedA – Friction

Since there are two types of friction, static and kinetic, every pair of materials will have two coefficients of friction, μs and μk.

In addition to the "roughness" or "smoothness" of the materials, the friction force depends on the normal force N.

The harder the two surfaces are squished together (this is what the normal force measures) the more cold welds can form.

Here are the relationships between the friction force f, the coefficient of friction μ, and the normal force N:

y

N

x

fs

mg

mgsin15°

mgcos15°

μs=

FBD, coin

## Topic 2.2 ExtendedA – Friction

One might ask how you can find the coefficient of friction between two materials:

Here is one way:

A piece of wood with a coin on it is raised on one end until the coin just begins to slip. The angle the wood makes with the horizontal is θ = 15°.

15°

What is the coefficient of static friction?

θ = 15°

∑Fx= 0

∑Fy= 0

N – mgcos15° = 0

fs – mgsin15° = 0

fs = mgsin15°

N = mgcos15°

fs = μsN

= tan15°

= 0.268

mgsin15° = μsmgcos15°

Thus the coefficient of static friction between the metal of the coin and the wood of the plank is 0.268.

y

N

x

fk

mg

FBD, coin

## Topic 2.2 ExtendedA – Friction

Now suppose the plank of wood is long enough so that you can lower it to the point that the coin keeps slipping, but no longer accelerates (v = 0).

12°

If this new angle is 12°,what is the coefficient of kinetic friction?

θ = 12°

∑Fx= 0

∑Fy= 0

N – mgcos12° = 0

fk – mgsin12° = 0

fk = mgsin12°

N = mgcos12°

fk = μkN

μk= tan12°

= 0.213

mgsin12° = μkmgcos12°

Thus the coefficient of kinetic friction between the metal of the coin and the wood of the plank is 0.213.

a

y

N

x

fk

mg

FBD, coin

## Topic 2.2 ExtendedA – Friction

If the plank of wood is now raised to 16°, what is the coin’s acceleration?

∑Fy= 0

N – mgcos16° = 0

N = mgcos16°

∑Fx= -ma

16°

fk – mgsin16° = -ma

fk = mgsin16° - ma

fk = μkN

mgsin16° - ma = μkmgcos16°

ma = mgsin16° - μkmgcos16°

a = (sin16° - μkcos16°)g

a = [ 0.276 – 0.213(0.961)](10)

a = 0.7 m/s2

y

30°

x

F

F

FBD, crate

N

N

30°

a

a

f

f

mg

mg

## Topic 2.2 ExtendedA – Friction

Since friction is proportional to the normal force, be aware of problems where an applied force increases or diminishes the normal force.

A 100-n crate is to be dragged across the floor by an applied force F of 60 n, as shown. The coefficients of static and kinetic friction are 0.75 and 0.60, respectively.

What is the acceleration of the crate?

Static friction will oppose the applied force until it is overcome.

y

30°

x

F

FBD, crate

N

a

f

mg

## Topic 2.2 ExtendedA – Friction

Determine if the crate even begins to move.

Thus, find the maximum value of the static friction, and compare it to the horizontal applied force:

The horizontal applied force is just

= 60cos30°

Fcos30°

= 51.96 n.

The maximum static friction force is

fs,max=μsN

=0.75N

The normal force is found from...

N + Fsin30°- mg =0

N + 60sin30°- 100=0

N =70

fs,max=0.75(70)

fs,max=52.5 n

Our analysis shows that the crate will not even begin to move!

y

30°

x

F

FBD, crate

N

a

f

mg

## Topic 2.2 ExtendedA – Friction

If someone gives the crate a small push (of how much) it will “break” loose.

What will its acceleration be then?

The horizontal applied force is still

= 60cos30°

Fcos30°

= 51.96 n.

The kinetic friction force is

fk=μkN

=0.60N

The normal force is still

N =70.

fk=0.60(70) = 42 n.

Thus

The crate will accelerate.

Fcos30°- f =ma

51.96- 42=(100/10)a

a = 0.996 m/s2

D

D

## Topic 2.2 ExtendedA – Friction

THE DRAG FORCE AND TERMINAL SPEED

A fluid is anything that can flow.

We generally think of a fluid as a liquid, like water.

But air is also a fluid.

And under certain conditions, solids can act like fluids.

Whenever there is a relative velocity between a fluid and a body, a drag force D is experienced.

That drag force always points in the direction of the relative velocity of the fluid.

If a boat is moving through still water, it feels a drag force opposite to its motion.

If a boat is moving against a current, it feels a drag force in the direction of the current.

Think of the drag force as a fluid friction force.

D

D

Some fluid densities

D

D

## Topic 2.2 ExtendedA – Friction

THE DRAG FORCE AND TERMINAL SPEED

The drag force depends on many things.

D is proportional to the cross-sectional areaA. This is how much of the body actually cuts into the fluid.

D is proportional to the cross-sectional area A

D is proportional to the fluiddensityρ. This is how much mass the fluid has per unit volume.

D is proportional to the fluid density ρ

ρwater = 1000 kg/m3

ρair = 1.2 kg/m3

D

D

D

D

## Topic 2.2 ExtendedA – Friction

THE DRAG FORCE AND TERMINAL SPEED

The drag force depends on many things.

D is proportional to the drag coefficientC. This is a unitless, experimentally derived quantity that represents how aerodynamic a body is.

D is proportional to the drag coefficient C

0 < C < 1

The more aerodynamic the body, the smaller the drag coefficient C.

D is proportional to the square of the relative velocityv, of the body and the fluid.

D is proportional to the SQUARE of the relative velocity v

1

2

The drag force

D = CρAv2

## Topic 2.2 ExtendedA – Friction

THE DRAG FORCE AND TERMINAL SPEED

Putting it all together we have…

Note the factor of 1/2 in the formula. This is found from experiment and theory beyond the scope of this course.

If v doubles, note that D quadruples.

## Force and Motion4-6 Friction

1

2

1

2

1

2

D30 = CρAv302

= (0.8)(1.2)(4)(13.42)

= (0.8)(1.2)(4)(31.32)

v70 =

70 mph

1

30 mi

h

1 m

3.28 ft

5280 ft

mi

1 h

3600 s

1

2

v30 =

D70 = CρAv702

13.4 m/s

30 mph

D70

D30

1881

345

=

## Topic 2.2 ExtendedA – Friction

THE DRAG FORCE AND TERMINAL SPEED

Suppose a minivan has a cross-sectional area of 4 m2 and an experimentally-established drag coefficient of 0.8. Compare the drag forces on the van at 30 mph and 70 mph.

= 13.4 m/s

= 31.3 m/s

= 345 n

= 1881 n

D70 = 5.45 D30

= 5.45

This is why it is more economical to drive at lower speeds.

This is also why you can only go so fast on a bicycle.

At first, v = 0.

"A female Blue Whale weighing 190 metric tonnes (418,877lb) and measuring 27.6m (90ft 5in) in length was caught in the Southern Ocean on 20 March 1947."

mg

Then, as v increases, so does D.

Guinness World Records. Falkland Islands Philatelic Bureau. 2 March 2002.

D

mg

v

y

y

y

v reaches a maximum value, called terminal speed. D = mg.

D

vterminal

mg

## Topic 2.2 ExtendedA – Friction

THE DRAG FORCE AND TERMINAL SPEED

Suppose a blue whale suddenly materializes high above the ground.

1 metric ton is 1000 kg.

Obviously, it will begin to fall.

At first v = 0, so the drag is also 0.

Because of the freefall acceleration, v increases. Thus D increases.

But as D increases, the acceleration decreases…

…until D = mg, at which time a is 0 and v stops changing.

v has reached its maximum value, called terminal speed vterminal.

1

2

CρAvt2 = mg

terminal speed

2mg

CρA

vt =

## Topic 2.2 ExtendedA – Friction

THE DRAG FORCE AND TERMINAL SPEED

At terminal speed vt, D = mg. Thus

D = mg

2mg

CρA

vt =

ellipse A = πab

a

2(190000)(10)

(0.60)(1.2)(115)

=

b

27 m

## Topic 2.2 ExtendedA – Friction

THE DRAG FORCE AND TERMINAL SPEED

So how do we estimate the terminal speed of the whale?

We need the mass: m = 190(1000 kg) = 190000 kg.

We need the drag coefficient: C = 0.60 (an estimate).

We need the cross-sectional area A:

- We will assume that the whale is approximately ellipsoid in shape:

- Since the whale is 27 m long, a simple estimate makes out the central diameter to be d = 27/5 = 5.4 m.

- Thus A = π(5.4/2)(27/2) = 115 m2.

= 214 m/s