In reference to Newton’s 3 rd Law

Physics 207: Lecture 7, Pg 1

"Professor Goddard does not know the relation between action and reaction and the need to have something better than a vacuum against which to react. He seems to lack the basic knowledge ladled out daily in high schools." New York Times editorial, 1921, about Robert Goddard's revolutionary rocket work.

"Correction: It is now definitely established that a rocket can function in a vacuum. The 'Times' regrets the error." New York Times editorial, July 1969.

In reference to Newton’s 3rd Law


Lecture 7 Goals:Goals:

Identify the types of forces

Use a Free Body Diagram to solve 1D and 2D problems with forces in equilibrium and non-equilibrium (i.e., acceleration) using Newton’ 1st and 2nd laws.

Distinguish static and kinetic coefficients of friction

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

Assignment: HW4, (Chapters 6 & 7)

Read Chapter 7

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


Example Non-contact Forces

All objects having mass exhibit a mutually attractive force (i.e., gravity) that is distance dependent

At the Earth’s surface this variation is small so little “g” (the associated acceleration) is typically set to 9.80 or 10. m/s2

FB,G


Contact (i.e., normal) Forces

Certain forces act to keep an object in place. These have what ever force needed to balance

all others (until a breaking point).

FB,T


No net force No acceleration

FB,T Normal force is always to a surface

0

0

0net

y

x

F

F

amFF

FB,G

(Force vectors are not always drawn at contact points)

mgN

NmgFy

0

y


No net force No acceleration

0net amFF

If zero velocity then “static equilibrium” If non-zero velocity then “dynamic equilibrium”

Forces are vectors

321net FFFamFF


High Tension

A crane is lowering a load of bricks on a pallet. A plot of the position vs. time is

There are no frictional forces Compare the tension in thecrane’s wire (T) at the point it contacts the pallet to the weight (W) of the load (bricks+ pallet)

A: T > W B: T = W C: T< W D: don’t knowH

eigh

t

Time


Analyzing Forces: Free Body DiagramA heavy sign is hung between two poles by a rope at

each corner extending to the poles.

Eat at Mickey’s

A hanging sign is an example of static equilibrium (depends on observer)

What are the forces on the sign and how are they related if the sign is stationary (or moving with constant velocity) in an inertial reference frame ?


Free Body Diagram

Step two: Sketch in force vectorsStep three: Apply Newton’s 2nd Law (Resolve vectors into appropriate components)

mg

T1T2Eat at Mickey’s

T1

T2

mg

Step one: Define the system


Free Body Diagram

Eat at Bucky’s

T1 T2

mg

Vertical : y-direction 0 = -mg + T1 sin1 + T2

sin2

Horizontal : x-direction 0 = -T1 cos1 + T2 cos2


Scale Problem You are given a 1.0 kg mass and you hang it

directly on a fish scale and it reads 10 N (g is 10 m/s2).

Now you use this mass in a second experiment in which the 1.0 kg mass hangs from a massless string passing over a massless, frictionless pulley and is anchored to the floor. The pulley is attached to the fish scale.

What does the string do? What does the pulley do? What does the scale now read?

1.0 kg

10 N

?

1.0 kg


Pushing and Pulling Forces String or ropes are examples of

things that can pull You arm is an example of an object

that can push or push


Examples of Contact Forces:A spring can push


A spring can pull


Ropes provide tension (a pull)

In physics we often use a “massless” rope with opposing tensions of equal magnitude


Moving forces around

Massless strings: Translate forces and reverse their direction but do not change their magnitude

(we really need Newton’s 3rd of action/reaction to justify)

Massless, frictionless pulleys: Reorient force direction but do not change their magnitude

string

T1 -T1

T1 -T1

T2

-T2| T1 | = | -T1 | = | T2 | = | T2 |


Scale Problem You are given a 1.0 kg mass and you hang it

directly on a fish scale and it reads 10 N (g is 10 m/s2).

Now you use this mass in a second experiment in which the 1.0 kg mass hangs from a massless string passing over a massless, frictionless pulley and is anchored to the floor. The pulley is attached to the fish scale.

What force does the fish scale now read?

1.0 kg

10 N

?

1.0 kg


What will the scale read?

A 5 NB 10 NC 15 ND 20 NE something else


Scale Problem Step 1: Identify the system(s).

In this case it is probably best to treat each object as a distinct element and draw three force body diagrams. One around the scale One around the massless pulley (even

though massless we can treat is as an “object”)

One around the hanging mass

Step 2: Draw the three FBGs. (Because this is a now a one-dimensional problem we need only consider forces in the y-direction.)

?

1.0 kg


Scale Problem

Fy = 0 in all cases 1: 0 = -2T + T ’ 2: 0 = T – mg T = mg 3: 0 = T” – W – T ’ (not useful here)

Substituting 2 into 1 yields T ’ = 2mg = 20 N(We start with 10 N but end with 20 N)

?

1.0 kg

? 1.0 kg

-mg

T

-T -T

-T ’

T”

W

1:2:3:T ’


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

ii

j j


Friction...

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


Static and Kinetic Friction Friction exists between objects and its behavior has been

modeled.

At Static 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: f is proportional to the applied forces such that

fs ≤ s N

s called the “coefficient of static friction”


Friction...

Force of friction acts to oppose motion: Parallel to a surface Perpendicular to a NNormal force.

maFF

ffF mgg

NN

ii

j j


Sliding Friction: Quantitatively

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


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

What is the direction of the frictional force?


Recap

Assignment: Soon….HW4 (Chapters 6 & 7, due 10/5) Read through first half of Chapter 7

In reference to Newton’s 3 rd Law

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