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Topic 2.2 Extended A – FrictionPowerPoint Presentation

Topic 2.2 Extended A – Friction

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### Topic 2.2 ExtendedA – Friction

### Topic 2.2 ExtendedA – Friction

### Topic 2.2 ExtendedA – Friction

### Topic 2.2 ExtendedA – Friction

### Topic 2.2 ExtendedA – Friction

### Topic 2.2 ExtendedA – Friction

### Topic 2.2 ExtendedA – Friction

### Topic 2.2 ExtendedA – Friction

### Topic 2.2 ExtendedA – Friction

### Topic 2.2 ExtendedA – Friction

### Topic 2.2 ExtendedA – Friction

### Topic 2.2 ExtendedA – Friction

### Topic 2.2 ExtendedA – Friction

### Topic 2.2 ExtendedA – Friction

### Topic 2.2 ExtendedA – Friction

### Force and Motion4-6 Friction

### Topic 2.2 ExtendedA – Friction

### Topic 2.2 ExtendedA – Friction

### Topic 2.2 ExtendedA – Friction

### Topic 2.2 ExtendedA – Friction

T

T

T

T

friction

f

f

f

f

Force

f

T

Time

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

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

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 2

surface 2

surface 2

cold welds

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

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:

N

x

fs

mg

mgsin15°

mgcos15°

μs=

FBD, coin

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.

N

x

fk

mg

FBD, coin

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.

y

N

x

fk

mg

FBD, coin

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

30°

x

F

F

FBD, crate

N

N

30°

a

a

f

f

mg

mg

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.

30°

x

F

FBD, crate

N

a

f

mg

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!

30°

x

F

FBD, crate

N

a

f

mg

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

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

Some fluid densities

D

D

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

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

2

The drag force

D = CρAv2

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.

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

=

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

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.

2

CρAvt2 = mg

terminal speed

√

2mg

CρA

vt =

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

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

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