Physics 207: Lecture 8, Pg 1 Lecture 8 l Goals: (Finish Chap. 6 and begin Ch. 7) Solve 1D & 2D...

Physics 207: Lecture 8, Pg 1

Lecture 8 Goals: (Finish Chap. 6 and begin Ch. 7)Goals: (Finish Chap. 6 and begin Ch. 7)

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

Assignment: HW4, (Chap. 6 & 7, due 10/5)

Finish reading Chapter 7

1st Exam Wed., Oct. 7th from 7:15-8:45 PM Chapters 1-7

in room 2103 Chamberlin Hall


Net force gives rise to 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

and mass plays a role


Mass We have an idea of what mass is from everyday life. In physics:

Mass (in Phys 207) is a quantity that specifies how much inertia an object has

(i.e. a scalar that relates force to acceleration)(Newton’s Second Law)

Mass is an inherent property of an object. Mass and weight are different quantities; weight is

usually the magnitude of a gravitational (non-contact) force.

“Pound” (lb) is a definition of weight (i.e., a force), not a mass!


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 object’s surroundings 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


Exercise Newton’s 2nd Law

A. AB. BC. CD. DE. G

A 10 kg 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?


Now: Back to forces that oppose motion


Static and Kinetic Friction Friction is a model force that exists between objects & it is

conditional At Static Equilibrium: A block, mass m, with a horizontal force F

applied, Direction:

1. Force vector to the normal force vector N N 2. 2. Opposite to the direction of acceleration if were 0.

Magnitude: f is proportional to the applied forces such that

fs ≤ s N

s called the “coefficient of static friction”


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

As F increases so does fs

Fm

1

FBD

fs

N

mg

Fx = 0 = -F + fs fs = F

Fy = 0 = - N + mg N = mg


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


Sliding Friction: Modeling

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


An experiment (with a ≠ 0)Two blocks are connected on the table as shown. Thetable has unknown static and kinetic friction coefficients.Design an experiment to find K.

Non-equilibrium: Set m2 and adjust m1 to find regime where a ≠ 0

Requires two FBDs

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


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


Inclined plane with “Normal” and Frictional Forces

1. Static Equilibrium Case2. Dynamic Equilibrium (see 1)

3. Dynamic case with non-zero acceleration

Block weight is mg

NormalForce

Friction Force

“Normal” means perpendicular

mg cos

f

x

y

mg sin

F = 0

Fx= 0 = mg sin – f

Fy= 0 = –mg cos + N

with mg sin = f ≤ S N

if mg sin > S N, must slide

Critical angle s = tan c


Inclined plane with “Normal” and Frictional Forces

1. Static Equilibrium Case

2. Dynamic Equilibrium Friction opposite velocity

(down the incline)

mg

NormalForce

Friction Force

“Normal” means perpendicular

mg cos

fK

x

y

mg sin

F = 0

Fx= 0 = mg sin – fk


fk = k N = k mg cos

Fx= 0 = mg sin – k mg cos

k = tan (only one angle)

v


Inclined plane with “Normal” and Frictional Forces3. Dynamic case with non-zero acceleration

Result depends on direction of velocity

Weight of block is mg

NormalForce

Friction ForceSliding Down

vmg sin fk

Sliding Up

Fx= max = mg sin ± fk


fk = k N = k mg cos

Fx= max = mg sin ± k mg cos

ax = g sin ± k g cos


Velocity and acceleration plots:

Notice that the acceleration is always down the slide and that, even at the turnaround point, the block is always motion although there is an infinitesimal point at which the velocity of the block passes through zero.

At this moment, depending on the static friction the block may become stuck.


Recap

Assignment: HW4, (Chapter 6 & 7 due 10/5)

Finish Chapter 7

Physics 207: Lecture 8, Pg 1 Lecture 8 l Goals: (Finish Chap. 6 and begin Ch. 7) Solve 1D & 2D...

Documents

Transcript of Physics 207: Lecture 8, Pg 1 Lecture 8 l Goals: (Finish Chap. 6 and begin Ch. 7) Solve 1D & 2D...