Monday, Sept. 18, 2002PHYS 1443-003, Fall 2002 Dr. Jaehoon Yu 1 PHYS 1443 – Section 003 Lecture #5...

Monday, Sept. 18, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu

1

PHYS 1443 – Section 003Lecture #5

Monday, Sept. 18, 2002Dr. Jaehoon Yu

1. Newton’s Laws of Motion2. Application of Newton’s Laws3. Friction4. Newton’s laws and its use in uniform and non-uniform

circular motion

Today’s homework is homework #6, due 1am, next Wednesday!!

Remember the first term exam on Monday, Sept. 30!!


2

Uniform Circular Motion• A motion with a constant speed on a circular path.

– The velocity of the object changes, because the direction changes

– Therefore, there is an acceleration

vi

vf

v

rvi

vf

rr2

r1

The acceleration pulls the object inward: Centripetal Acceleration

t

v

tt

vva

if

if

Average Acceleration r

r

v

v

Angle is

r

v

r

vv

r

v

t

raa

ttr

2

00limlim

Instantaneous Acceleration

Is this correct in dimension?

What story is this expression telling you?

r

rvv

r

r

t

va


3

Non-uniform Circular Motion• Motion not on a circle but through a curved path

– Requires both tangential (at) and radial acceleration (ar)

a r

a t

a

ar

at

a

dt

vdat

r

var

2

ar

at

a

Tangential Acceleration:

Radial Acceleration:

r

x

y

Or

rr

v

dt

vdaaa tr

2

Total Acceleration:


4

Relative Velocity and AccelerationThe velocity and acceleration in two different frames of references can be denoted, using the formula in the previous slide:

O

Frame S

r’

O’

Frame S’v0

v0t

r

tvrr 0' Galilean transformation equation

0' vvv

What does this tell you?

The accelerations measured in two frames are the same when the frames move at a constant velocity with respect to each other!!!

The earth’s gravitational acceleration is the same in a frame moving at a constant velocity wrt the earth.

0'

vdt

rd

dt

rd

0' vvv

dt

vd

dt

vd

dt

vd 0'

constant is when ,' 0vaa


5

Newton’s First Law and Inertial FramesGalileo’s statement on natural states of matter: Any velocity once imparted to a moving body will be rigidly maintained as long as the external causes of retardation are removed!!This statement is formulated by Newton into the 1st law of motion (Law of Inertia): In the absence of external forces, an object at rest remains at rest and an object in motion continues in motion with a constant velocity. What does this statement tell us? 1. When no force is exerted on an object, the

acceleration of the object is 0. 2. Any isolated object, the object that do not interact

with its surrounding, is either at rest or moving at a constant velocity.

3. Objects would like to keep its current state of motion, as long as there is no force that interferes with the motion. This tendency is called the Inertia.

A frame of reference that is moving at constant velocity is called an Inertial Frame


6

Newton’s Second Law of Motion

amFi

i

The acceleration of an object is directly proportional to the net force exerted on it and inversely proportional to the object’s mass. How do we write the above statement in a mathematical expression?

xi

ix maF Since it’s a vector expression, each component should also satisfy:From the above vector expression, what do you conclude the dimension and unit of force are?

]][[]][[ 2 LTMam22 /]][[]][[][ smkgLTMamForce

The dimension of force is The unit of force in SI is See Table 5.1 for lbs to

kgm/s2 conversion.

yi

iy maF zi

iz maF

lbssmkgN4

1/11 2


7

Free Body Diagrams• Diagrams of vector forces acting on an object A great tool to solve a problem using forces or using dynamics1. Select a point on an object and w/ information given2. Identify all the forces acting only on the selected object3. Define a reference frame with positive and negative axes specified4. Draw arrows to represent the force vectors on the selected point5. Write down net force vector equation6. Write down the forces in components to solve the problems No matter which one we choose to draw the diagram on, the results should be the same, as long

as they are from the same motion

MWhich one would you like to select to draw FBD?What do you think are the forces acting on this object?

Gravitational forcegMF G A force supporting the object exerted by the floor

Me

Which one would you like to select to draw FBD?What do you think are the forces acting on this elevator?

NF

gMF G

Gravitational force The force pulling the elevator (Tension)

mWhat about the box in the elevator? Gravitational

forceNormal force

NF

TF

gMF G gmF GB

TF

gMF G

NF

gmF BG

NF


8

Acceleration Vector a

F

F

Example 5.1Determine the magnitude and direction of acceleration of the puck whose mass is 0.30kg and is being pulled by two forces, F1 and F2, as shown in the picture, whose magnitudes of the forces are 8.0 N and 5.0 N, respectively.

NFF x 0.460cos0.8cos11

xxxx maNFFF 7.87.40.421

Components of F1

Components of F2

Components of total force F

2/293.0

7.8sm

m

Fa xx Magnitude and

direction of acceleration a

NFF y 9.660sin0.8sin11

NFF x 7.420cos0.5cos 222

NFF y 7.120sin0.5sin 222

yyyy maNFFF 2.57.19.621

2/173.0

2.5 sm

m

Fa yy 222 /341729 sma

3029

17tantan 11

x

y

a

a

2/1729 smjijaiaa yx


9

Gravitational Force and Weight

Since weight depends on the magnitude of gravitational acceleration, g, it varies depending on geographical location.

The attractive force exerted on an object by the Earth

Gravitational Force, Fg

gmamF G

Weight of an object with mass M is MggMFW G

By measuring the forces one can determine masses. This is why you can measure mass using spring scale.

Actual unit of weight is in the unit of force but the unit of mass is commonly used in place of force.


10

Newton’s Third Law (Law of Action and Reaction)If two objects interact, the force, F12, exerted on object 1 by object 2 is equal in magnitude and opposite in direction to the force, F21, exerted on object 1 by object 2. F12

F21

2112 FF

The action force is equal in magnitude to the reaction force but in opposite direction. These two forces always act on different objects. What is the reaction force to the force of a free fall object?

The force exerted by the ground when it completed the motion.

Stationary objects on top of a table has a reaction force (normal force) from table to balance the action force, the gravitational force.


11

Example 5.3 A large man and a small boy stand facing each other on frictionless ice. They put their hands together and push against each other so that they move apart. a) Who moves away with the higher speed and by how much?

2112 FF

b) Who moves farther while their hands are in contact?

M m

F12

F21=-F12

Given in the same time interval, since the boy has higher acceleration and thereby higher speed, he moves farther than the man.

22

22

1ta

m

Mtv

m

Mtatvx MxMxfbxbxfb

MaMF 21

2112 FF tatavv MxMxMxiMxf

Mxfbxbxbxibxf vm

Mta

m

Mtatavv Mx

masses theof ratio by the if mMvv Mxfbxf

FFF 2112

bamF 12

MMxMxfb xm

Mtatv

m

Mx

2

2

1

bxx maF 12 012 byy maF

Mxx MaF 21 021 Myy MaF

FFF 2112 Mxbx am

M

m

Fa


12

Some Basic Information

Normal Force, n:

When Newton’s laws are applied, external forces are only of interest!!

Why? Because, as described in Newton’s first law, an object will keep its current motion unless non-zero net external forces are applied.

Tension, T:

Reaction force that balances gravitational force, keeping objects stationary.Magnitude of the force exerted on an object by a string or a rope.A graphical tool which is a diagram diagram of external forces on an objectof external forces on an object and is extremely useful analyzing forces and motion!! Drawn only on the object.

Free-body diagram


13

Applications of Newton’s Laws

M

Suppose you are pulling a box on frictionless ice, using a rope.

T

What are the forces being exerted on the box?Gravitational force: FgNormal force: nTension force: T

n= -Fg

TFree-body diagram

Fg=MgTotal force: F=Fg+n+T=T

xx MaTF

If T is a constant force, ax, is constant

tM

Tvtavv xixxixf

0ygy ManFF

M

Tax

0ya

2

2

1t

M

Ttvxxx xiif

n= -Fg

Fg=Mg

T


14

Example 5.4A traffic light weighing 125 N hangs from a cable tied to two other cables fastened to a support. The upper cables make angles of 37.0o and 53.0o with the horizontal. Find the tension in the three cables.

321 TTTF

Free-bodyDiagram

53o37o

x

y

T1

37o

T2

53o

T3

053sin37sin 21 mgTT

053cos37cos 21 TT 754.037cos

53cos221 TTT

NTT 12525.137sin754.053sin 22

NTTNT 4.75754.0 ;100 212

03

1

i

iixx TF

03

1

i

iiyy TF

Newton’s 2nd law

x-comp. of net force

y-comp. of net force


15

Example 5.6A crate of mass M is placed on a frictionless inclined plane of angle . a) Determine the acceleration of the crate after it is released.

amnFF g

Free-bodyDiagram

x

yM

da

Fg

nn

F= -MgSupposed the crate was released at the top of the incline, and the length of the incline is d. How long does it take for the crate to reach the bottom and what is its speed at the bottom?

22 sin2

1

2

1tgtatvd xix

0cos mgnFnMaF gyyy

sinMgFMaF gxxx

singax x

y

sin

2

g

dt

sin2sin

2sin dg

g

dgtavv xixxf

Monday, Sept. 18, 2002PHYS 1443-003, Fall 2002 Dr. Jaehoon Yu 1 PHYS 1443 – Section 003 Lecture #5...

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Transcript of Monday, Sept. 18, 2002PHYS 1443-003, Fall 2002 Dr. Jaehoon Yu 1 PHYS 1443 – Section 003 Lecture #5...