Lecture 10 Goals Describe Friction

Physics 201: Lecture 10, Pg 1

Lecture 10 GoalsGoals

Describe Friction

Problem solving with Newton’s 1st, 2nd and 3rd Laws

Forces in circular and curvelinear motion


Another experiment: A modified Atwood’s machine

Two blocks, m1 & m2, are connected by a massless frictionless string/pulley on the table as shown. The table surface is frictionless and little g acts downward.What is the acceleration of the horizontal block? Requires two FBDs andNewton’s 3rd Law.

m1

m2

m2g

N

m1g

T

T

Mass 2

Fx = m2a2x = -T

Fy = 0 = N – m2g

Mass 1

Fy = m1a1y= T – m1gCorrelated motion:

|a1y| = | a2x| ≡ a

If m1 moves up moves m2 right


Another experiment: A modified Atwood’s machine

Two blocks, m1 & m2, are connected by a massless frictionless string/pulley on the table as shown. The table surface is frictionless and little g acts downward.What is the acceleration of the horizontal block.

Requires two FBDs andNewton’s 3rd Law.

m1

m2

m2g

N

m1g

T

T

Mass 2

B. Fx = m2 a = -T

Fy = 0 = N – m2g

Mass 1

A. Fy = m1 a= T – m1gSubbing in T from B into A

m1 a= - m2 a – m1g

m1 a + m2 a = m1g

a = m1g / (m1 + m2 )


A “special” contact force: Friction

What does it do? It opposes motion (velocity, actual or that which

would occur if friction were absent!) How do we characterize this in terms we have learned?

Friction results in a force in a direction opposite to the direction of motion (actual or, if static, then “inferred”)!

maFFAPPLIED

ffFRICTION mgg

NN

i

j


If no acceleration

No net force So frictional force just equals applied force Key point: It is conditional!

FFAPPLIED

ffFRICTION mgg

NN

ii

j j


Friction...

Friction is caused by the “microscopic” interactions between the two surfaces:


Friction: Static frictionStatic equilibrium: A block with a horizontal force F applied,

FBD

mF

fs

N

mg

Fx = 0 = -F + fs fs = F

Fy = 0 = - N + mg N = mg As F increases so does fs


Static friction, at maximum (just before slipping)

Equilibrium: A block, mass m, with a horizontal force F applied, Direction: A force vector to the normal force vector N N and the

vector is opposite to the direction of acceleration if were 0. Magnitude: fS is proportional to the magnitude of N

fs = s N F

m fs

N

mg


Kinetic or Sliding friction (fk < fs)Dynamic equilibrium, moving but acceleration is still zero

As F increases fk remains nearly constant (but now there acceleration is acceleration)

Fm

1

FBD

fk

N

mg

Fx = 0 = -F + fk fk = F

Fy = 0 = - N + mg N = mg v

fk = k N


Model of Static Friction (simple case) Magnitude:

f is proportional to the applied forces such that

fs ≤ s N s called the “coefficient of static friction”

Direction: Opposite to the direction of system acceleration if were 0


Sliding Friction

Direction: A force vector to the normal force vector N N and the vector is opposite to the velocity.

Magnitude: ffk is proportional to the magnitude of N N

ffk = k N N ( = Kmg g in the previous example)

The constant k is called the “coefficient of kinetic friction” Logic dictates that S > K for any system


Coefficients of Friction

Material on Material s = static friction k = kinetic friction

steel / steel 0.6 0.4

add grease to steel 0.1 0.05

metal / ice 0.022 0.02

brake lining / iron 0.4 0.3

tire / dry pavement 0.9 0.8

tire / wet pavement 0.8 0.7


Forces at different angles

Case 1 Case 2

F

mg

N

Case1: Downward angled force with frictionCase 2: Upwards angled force with frictionCases 3,4: Up against the wallQuestions: Does it slide?

What happens to the normal force?What happens to the frictional force?

mg

Cases 3, 4

mg

N NF

F

ff

ff

ff


Many Forces are Conditional Notice what happens if we change the direction of the applied

force The normal force can increase or decrease

Let a=0

F

fF mg

N > mgi

j


An experimentTwo blocks are connected on the table as shown. Thetable has unknown static and kinetic friction coefficients.Design an experiment to find S

m1

m2

m2g

N

m1g

T

T

Mass 2

Fx = 0 = -T + fs = -T + S N

Fy = 0 = N – m2g

fS

Mass 1

Fy = 0 = T – m1g

T = m1g = S m2g S = m1/m2


3rd Law : Static Friction with a bicycle wheel You are pedaling hard

and the bicycle is speeding up.

What is the direction of the frictional force?

You are breaking and the bicycle is slowing down



Static Friction with a bicycle wheel

You are pedaling hard and the bicycle is speeding up.


Hint…you are accelerating to the right

a = F / m

Ffriction, on B from E is to the right

Ffriction, on E from,B is to the left


Force pairs on an Inclined plane

Forces on the block (static case)

NormalForce

Friction Force

f= N

x

yForces on the plane by block


The inclined plane coming and going (not static):the component of mg along the surface > kinetic friction

Putting it all together gives two different accelerations, ax = g sin ± uk g cos . A tidy result but ultimately it is the

process of applying Newton’s Laws that is key.

Fx = max = mg sin ± uk N

Fy = may = 0 = -mg cos + N


Forces with rotation about a fixed axisKey steps

Identify forces (i.e., a FBD)

Identify axis of rotation

Circular motion implies ar & Fr = mar

Apply conditions (position, velocity &acceleration)


Example The pendulum

Consider a person on a swing:

When is the tension equal to the weight of the person + swing?

(A) At the top of the swing (turnaround point)(B) Somewhere in the middle(C) At the bottom of the swing(D) Never, it is always greater than the weight

(E) Never, it is always less than the weight

axis of rotation

(A)(B)

(C)


Example Gravity, Normal Forces etc.

at bottom of swing vt is max

Fr = m ac = m vt2 / r = T - mg

T = mg + m vt2 / r

T > mg

at top of swing vt = 0

Fr = m 02 / r = 0 = T – mg cos T = mg cos

T < mg

axis of rotation

vt

mg

T

mg

T

y

x


Conical Pendulum (Not a simple pendulum) Swinging a ball on a string of length L around your head

(r = L sin ) axis of rotation Fr = mar = T sin

Fz = 0 = T cos – mg so T = mg / cos (> mg)

mar = (mg / cos ) (sin )

ar = g tan vT2/r

vT = (gr tan )½ Period:T= 2 r / vT =2 (r cot

/g)½ = 2 (L cos /g)½

r

L


Conical Pendulum (very different) Swinging a ball on a string of length L around your head

axis of rotation

Period:t = 2 r / vT =2 (r cot /g)½ = 2 (L cos / g )½ = 2 (5 cos / 9.8 )½ = 4.38 s = 2 (5 cos / 9.8 )½ = 4.36 s = 2 (5 cos 1/ 9.8 )½ = 4.32 s r

L


A match box car is going to do a loop-the-loop of radius r.

What must be its minimum speed vt at the top so that it can manage the loop successfully ?

Another example of circular motionLoop-the-loop 1


To navigate the top of the circle its tangential velocity vT must be such that its centripetal acceleration at least equals the force due to gravity. At this point N, the normal force, goes to zero (just touching).

Loop-the-loop 1

Fr = mar = mg = mvt2/r

vt = (gr)1/2

mg

vT


The match box car is going to do a loop-the-loop. If the speed at the bottom is vB, what is the normal force, N, at that point?

Hint: The car is constrained to the track.

Loop-the-loop 2

mg

v

N

Fr = mar = mvB2/r = N - mg

N = mvB2/r + mg


Once again the car is going to execute a loop-the-loop. What must be its minimum speed at the bottom so that it can make the loop successfully?

This is a difficult problem to solve using just forces. We will skip it now and revisit it using energy considerations later on…

Loop-the-loop 3


Orbiting satellites vt = (gr)½

Net Force: ma = mg = mvt

2 / r gr = vt

2

vt = (gr)½

The only difference is that g is less because you are further from the Earth’s center!


Geostationary orbit

The radius of the Earth is ~6000 km but at 36000 km you are ~42000 km from the center of the earth.

Fgravity is proportional to 1/r2 and so little g is now ~10 m/s2 / 50 vT = (0.20 * 42000000)½ m/s = 3000 m/s

At 3000 m/s, period T = 2 r / vT = 2 42000000 / 3000 sec = = 90000 sec = 90000 s/ 3600 s/hr = 24 hrs

Orbit affected by the moon and also the Earth’s mass is inhomogeneous (not perfectly geostationary)

Great for communication satellites

(1st pointed out by Arthur C. Clarke)


Recap

Assignment: HW5

For Tuesday: Read Chapter 7.1-7.4

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Lecture 10 Goals Describe Friction