Physics 207: Lecture 8, Pg 1 Lecture 8 l Goals: Solve 1D & 2D motion with friction Utilize Newton’s 2 nd Law Differentiate between Newton’s 1 st,

Physics 207: Lecture 8, Pg 1

Lecture 8

Goals:Goals:

Solve 1D & 2D motion with friction

Utilize Newton’s 2nd Law

Differentiate between Newton’s 1st, 2nd and 3rd Laws

Begin to use Newton’s 3rd Law in problem solving


Friction with no acceleration

No net force So frictional force just cancels applied force

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,

As F increases so does fs

Fm

FBD

fs

N

mg


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

Fm

FBD

fs

N

mg

Fx = 0 = -F + fs

fs = F

Fy = 0 = - N + mg

N = mg

Force applied

Fri

ctio

nF

orce


Static friction, at maximum (just before slipping)

fS is proportional to the magnitude of N

fs = s N

Fm fs

N

mg

Force applied

Fri

ctio

nF

orce


Model of Static Friction

Magnitude:

f is proportional to the applied forces such that

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

Direction: If just a single “applied” force, friction is in opposite direction


Kinetic (fk < fs)

Dynamic equilibrium, moving but acceleration is still zero

F

m

FBD

fk

N

mg

Fx = 0 = -F + fk fk = F

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

fk = k N


Model of Sliding Friction

Direction: to the normal force vector N N and

opposite to the velocity.

Magnitude: ffk is proportional to the magnitude of N N

ffk = k NN

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


Sliding friction (fk < fs) but now |a| > 0

A change in velocity

As F increases fk remains nearly constant

(but now there is acceleration)

Fm

FBD

fk

N

mg

Fx = -F + fk = net Force

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

fk = k N


Acceleration, Inertia and Mass The tendency of an object to resist any attempt to

change its velocity is called Inertia Mass is that property of an object that specifies how

much resistance an object exhibits to changes in its velocity (acceleration)

If mass is constant then

If force constant

Mass is an inherent property of an object Mass is independent of the method used to measure it Mass is a scalar quantity The SI unit of mass is kg

netFa

ma 1||

|a|

m


ExerciseNewton’s 2nd Law

A. increasingB. decreasingC. constant in timeD. Not enough information to decide

An object is moving to the right, and experiencing a net force that is directed to the right. The magnitude of the force is decreasing with time (read this text carefully).

The speed of the object is


1st: Frictionless experiment (with a ≠ 0)

Two blocks are connected on the table as shown. The

table is frictionless. Find the acceleration of mass 2.

Requires two FBDsT

Mass 2

Fx = m2ax = -T

Fy = 0 = N – m2g

m1

m2

m2g

N

m1g

T

Mass 1

Fy = m1ay = T – m1g

Notice ay = ax = a

Eliminate T

m1a + m2a = m1

a = m1 / (m2+m1)g


Experiment with friction (with a ≠ 0)

Two blocks, of m1 & m2 , are connected on the table as shown. The table has unknown static and kinetic friction coefficients.

Given an a, find K.

Similar but now with friction.

T

Mass 2

Fx = m2a = -T + fk = -T + k N

Fy = 0 = N – m2g

m1

m2

m2g

N

m1g

T

fk

Mass 1

Fy = m1a = T – m1g

T = m1g + m1a = k m2g – m2a k = (m1(g+a)+m2a)/m2g


To repeat, net force acceleration

In physics: A force is an action which causes an object to

accelerate (translational & rotational)

This is Newton’s Second Law

zz

yy

xx

maF

maF

maF

amFF

0net


Home Exercise Newton’s 2nd Law

A. A

B. B

C. D

D. F

E. G

A mass undergoes motion along a line with velocities as given in the figure below. In regards to the stated letters for each region, in which is the magnitude of the force on the mass at its greatest?


Remember: Forces are Conditional

Notice what happens if we change the direction of the applied force

The normal force can increase or decrease Here the normal force exceeds mg

Let a=0

F

fF mg

N F sin +mg

ii

j j F sin


Forces at different angles

Case 1 Case 2

F

mg

N

Case1: Downward angled force with friction

Case 2: Upwards angled force with friction

Cases 3,4: Up against the wall

Questions: Does it slide?

What happens to the normal force?

What happens to the frictional force?

mg

Cases 3, 4

mg

NN

F

F

ff

ff

ff


Example (non-contact)

Consider the forces on an object undergoing projectile motion

FB,E = - mB g

EARTH

FE,B = mB g

FB,E = - mB g

FE,B = mB g


Gravity

Newton also recognized that gravity is an attractive, long-range force between any two objects.

When two objects with masses m1 and m2 are separated by distance r, each object “pulls” on the other with a force given by Newton’s law of gravity, as follows:


Cavendish’s Experiment

F = m1 g = G m1 m2 / r2

g = G m2 / r2

If we know big G, little g and r then will can find m2 the mass of the Earth!!!


Example (non-contact)

FB,E = - mB g

EARTH

FE,B = mB g

FB,E = - mB g

FE,B = mB g

Compare: g = G m2 / 40002 g’ = G m2 / (4000+40)2

g / g’ = / (4000+40)2 / 40002 = 0.98

Question: By how much does g change at an altitude of 40 miles? (Radius of the Earth ~4000 mi)


Recap

Assignment: HW4, (Chapter 6 & 7 due 10/4)

For Monday finish Chapter 7

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Physics 207: Lecture 8, Pg 1 Lecture 8 l Goals: Solve 1D & 2D motion with friction Utilize Newton’s 2 nd Law Differentiate between Newton’s 1 st,