Announcements – 2/1/10

Announcements – 2/1/10

•  Google “empty browser cache” •  Quiz tomorrow (Driscoll). •  Sample quiz, lecture notes are at website •  Please show a reasonable amount of your

work/calculations on quizzes/homework, etc.

•  Midterm is next Tuesday, 2/9/10

We know about: Units and Measurements, Motion in 1, 2 & 3 dimensions, How to describe these motions with Vectors.

Now: We’ll consider how to cause and change motions, and the relationships between Forces and motion.

Chapter 5: Force and Motion I-a

•  Force F –  is the interaction between objects –  is a vector –  causes acceleration – Net force: vector sum of all the forces on an

object. response of the object

“Environment”

Force Examples

•  Gravitational •  Friction •  tension •  spring •  Normal

•  electrostatic •  magnetic •  Nuclear •  Centripetal •  Centrifugal •  police

•  etc.......

Isaac Newton (1642 – 1727) and his laws

Newton’s first law: If no force acts on a body, then the body’s velocity cannot change, that is, the body can not accelerate

“A body in motion remains in motion, a body at rest remains at rest”

– if at rest, remains at rest – if moving, continues moving with same velocity – a Force is required to produce a change (acceleration) in motion – no equation associated with this Law – At 1st glance this law appears to not be true, but more careful consideration will always show a hidden Force.

http://en.wikipedia.org/wiki/Isaac_Newton

Mass

•  Symbol: m •  SI Base unit: kg (by the way, the English unit for mass is the “slug”)

•  Scalar quantity

Definition: The mass of an object is a measure of its “resistance” to being accelerated.

Query: Is one’s mass the same on the (surface) of moon and on the earth? Is one’s weight?

(gravity on the moon is ~ 1/6th the earth’s) There is a difference between mass and weight

• Reference frame is “inertial” if Newton’s three laws of motion hold. In contrast, reference frames in which Newton’s law are not obeyed are labeled “non-inertial”

• Newton believed that at least one inertial reference frame R always exists. Frame R' moving with constant velocity with respect to R is also an inertial reference frame. Areference frame R" which accelerates with respect to R is a non-inertial frame.

The earth rotates so it is accelerating with respect to an inertial reference frame. But we often make an approximation that the earth is an inertial reference frame. This works well for most small scale phenomena, but not soo well for large scale phenomena such as global wind systems.

An Inertial reference frame is one in which Newton’s laws hold

Force: The concept of force was tentatively defined as a push or pull exerted on an object. We can define a force exerted on an object quantitatively by measuring the acceleration it causes using the following procedure

We place an object of mass m = 1 kg on a frictionless surface and measure the acceleration a that results from the application of a force F. The force is adjusted so that a = 1 m/s2.

We then say (DEFINE) that F = 1 newton (symbol: N)

(5-3)

Before starting, set up coordinate system and free body diagram.

x

y

0

Only the forces are important, not the actual body!

F3

F2

F1

Example: An object experiences three forces as shown in the figure. What is the object’s acceleration (a) in unit vector notation and as (b) a magnitude and a direction

(a) Break the problem into x- and y-components

x

y

0

F3

F1

F2

F1 = (41N cos 60o) i + (-41N sin 60o) j F2 = (55N) i + (0) j F3 = (32N cos 30o) i + (32N sin 30o) j

Fnet = (103.2N) i + (-19.51N) j


Tan Ø = = tan-1Ø = -10.7o

(b) Find and use

x

y

0

F3

F1

F2

= Fnet = (103.2N)2 + (-19.51N)2 = 105.0N Fnet

Fnet

Fnet j Fnet i

tan Ø=

Fnet j Fnet i

103.2N -19.51N = -0.189


Consider a traffic light of m = 4 kg held by one rope which in turn is supported by two other ropes as shown with angles q1 = 30o q2 = 45o , Which of the three ropes has the greater tension?

q1=30o q2=45o

1 2

3

Consider a traffic light of m = 4 kg held by one rope which in turn is supported by two other ropes as shown with angles q1 = 30o q2 = 45o , Which of the three ropes has the greater tension?

q1=30o q2=45o

1 2 3

1) rope 1 2) rope 2 3) rope 3 4) All ropes have the same tension

Ropes 1 & 2 help share the weight, but rope 3 carries all the weight.

Consider a traffic light of m = 4kg held by one rope which in turn is supported by two other ropes as shown with angles q1 = 30o q2 = 45o , Which of the three ropes has the greater tension?

q1=30o q2=45o

1 2

3

What is this Force?

The gravitational force near the surface of a very large object (i.e., the Earth):

Fg = m g

Weight (gravitational force) W = Fg = m g

–  g varies with altitude, planet, etc.... –  weight and mass are different, e.g. 72 kg ball, same mass on earth and moon, weighs 706 N on earth but 118 N on the moon

Newton’s Second Law Newton’s second law: The net force on a body is equal to the product of the body’s mass and the acceleration of the body

F = m a

F : vector sum of all the forces that act on that body Fx = m ax F y = m ay F z = m az

Units: 1 N = (1 kg)(1 m/s2) = 1 kg . m/s2

“Special” Forces

• The normal force: N –  When a body presses against a surface, the surface deforms and pushes on the body with a normal force N that is perpendicular to the surface –  N does not always equal mg

• Friction: f –  is the resistance force on a body when the body slides or attempts to slide along a surface –  Always opposes motion

• Tension: T – When a cord is attached to a body and pulled taut, the cord pulls on the body with force T

CP 5-5: The body in fig (c) has a weight of 75 N, Is T equal to, greater than, or less than 75 N when the body is moving upward (a) at constant speed (b) at increasing speed (c) at decreasing speed?

Newton’s Third Law

– They do not cancel each other since they act on different bodies (different perspective)

Newton’s third law: When two bodies interact, the forces on the bodies from each other are always equal in magnitude and opposite in direction

FAB = - FBA

Dog + Car From the picture, did the car decelerate, velocity remain constant, or accelerate? Assume the dog was peacefully sitting on the seat before this event.

1.  car decelerated 2.  car’s velocity remained constant 3.  car accelerated

If a sports car and a semi-truck collide head-on, which vehicle experiences the greater force?

1.  car 2.  truck 3.  both experience the same

force

Remember that a modern semi-truck has a mass of 20 cars!

If a sports car and a semi-truck collide head-on, which vehicle experiences the greater acceleration?

Remember that a modern semi-truck has a mass of 20 cars!

1.  car 2.  truck 3.  both experience the same

acceleration

Sample problem 5-5. M = 3.3 kg, m = 2.1 kg, frictionless surface H falls as S accelerates to the right (a) What is the acceleration of S?

Fg = mg

T

Acceleration a links the two masses together

mg - Ma = ma

mg - T = ma

T = Ma

Forces on m T = Ma

unbalanced forces on M

Newton’s Laws of Motion

•  Newton’s first law: If no force acts on a body, then the body’s velocity cannot change, that is, the body can not accelerate.

•  Newton’s second law: The net force on a body is equal to the product of the body’s mass and the acceleration of the body SF = m a

•  Newton’s third law: When two bodies interact, the forces on the bodies from each other are always equal in magnitude and opposite in direction

FAB = - FBA

General scheme for solving Newton’s law problems -  Isolate the objects in the problem -  For each object of interest, identify all the

external forces on that object and draw a free-body diagram for this object

-  establish a convenient coordinate system for each object and find the component of the forces along those axes.

-  Apply Newton’s 2nd law in the x and y directions for each object. ( e.g. ∑Fx= max, ∑Fy = may)

-  Solve the resulting set of equations.

Assume F = 20 N, surface frictionless. What is the acceleration ?

T1 = m1a = (10kg)(1.0 m/s2) = 10N

How about T1?

F = ( m1 + m2 + m3+ m4)a = (20kg)a = 20N ==> a = 20N/20kg = 1.0 m/s2

T1 T3

How about T3? T3 = (10+3+5)(1) = 18 N

How do we approach this problem? Where are the axes?

Sample problem 5-4. M = 3.3 kg, m = 2.1 kg, frictionless surface H falls as S accelerate to the right (a) What is the acceleration of S?

T

T = Ma

How do we approach this problem? Two separate (free body - FB) problems, linked by the rope (T) and acceleration

FB 1

FB 2

For FB 1: Fnet, x = Max, Fnet, y = May, Fnet, z = Maz

No acceleration in y or z: For y: FN = FgS For z: no forces

T = Max

F = m a Fg = mg

Sample problem 5-4. M = 3.3 kg, m = 2.1 kg, frictionless surface H falls as S accelerate to the right (a) What is the acceleration of S?

T

T = Ma FB1

FB 2

For FB 2: Only have y components

Fnet, y = may = T – FgH = T - mg Fg = mg

Fg = -mg k Since H accelerated in –y direction: Substitute –a for ay Thus, T = mg – ma

But we know T = Ma from FB1:

Solving Two eqns With two unknowns >>>

x

y

Consider m1 = 4 kg, m2 = 2 kg, What is the acceleration of either mass if the inclined plane is frictionless?

T T

How do we approach this problem?

m1 = 4 kg, m2 = 2 kg. What is the acceleration of either mass if the inclined plane is frictionless?

How do you set up this problem?

x

y Horizontal - Vertical

--or--

slanted

Easiest because of direction of motion

First set up the coordinate system, but how? Which Coordinate system?

m1g m2g

gravity still points down

m1 = 2m2. What is the acceleration of either mass if the inclined plane is frictionless?

m1g

N T

m2g

T gravity still points down

Normal force is perpendicular to the plane

m1g, m2g

Tension is along the rope


FB1

FB2

N T Set-up the free body diagram(s):

m2g

T

Ø

T = Ti + 0 jN = 0i + Nj

m1g = m1gsinθ −i( ) + m1gcosθ − j( )

Evaluate the force components m1g


FB1

FB2 For FB1

N T

m2g

T

Ø

x : Fxi∑ = T − m1gsinθ = m1a

y : Fyi∑ = N − m1gcosθ = 0

Now apply Newton’s Second Law to each body

a

m1g

Which direction(s) will the acceleration be in?

F = m a

FB1

FB2 For FB1, use x/y coordinate system

Compare forces that pull on chord: i.e. m1g sinØ vs. m2g They are balance for m1 = 4 kg, m2 = 2 kg, and sinØ = 0.5

N T

m2g

T

Ø

Continue with Newton’s Second Law to each body

a

m1g

If m1 > 2 m2 What is the acceleration now?

F = m a

FB1

FB2 For FB2, first write eqn wrt to its own x/y axes:

m2 : m2a = T − m2g

y

But then account for the fact that FB1 and FB2 will experience the same a, and an a in FB2 coordinates corresponds to –a in FB1 coordinates (substitute –a for a)

m2 : m2a = −T + m2g

N T

m2g

T

Ø

a

m1g

Now we have a full set of equations

FB1

FB2 x : Fxi∑ = T − m1gsinθ = m1a

y : Fyi∑ = N − m1gcosθ = 0m2 : − T + m2g = m2a

Eliminate T (equations are generic)

F

ao mo

F

aX mX

Mass: Mass is an intrinsic characteristic of a body that automatically comes with the existence of the body. But what is it exactly? It turns out that mass of a body is the characteristic that relates a force F applied on the body and the resulting acceleration a.

Consider that we have a body of mass mo = 1 kg on which we apply a force F = 1 N. According to the definition of the newton , F causes an acceleration ao = 1 m/s2. We now apply F on a second body of unknown mass mX which results in an acceleration aX . The ratio of the accelerations is inversely proportional to the ratio of the masses

Thus by measuring aX we are able to determine the mass mX of any object.

Fnet

a m

Newton’s Second Law

The results of the discussions on the relations between the net force Fnet applied on an object of mass m and the resulting acceleration a can be summarized in the following statement known as: “Newton’s second law” The net force on a body is equal to the

product of the body’s mass and its acceleration

In equation form Newton’s second law can be written as:

The above equation is a compact way of summarizing three separate equations, one for each coordinate axis:

Fg

In this section we describe some characteristics of forces we will commonly encounter in mechanics problems

The Gravitational Force: It is the force that the earth exerts on any object (in the picture a cantaloupe) It is directed towards the center of the earth. Its magnitude is given by Newton’s second law.

y

y g W

mg

Weight: The weight of a body is defined as the magnitude of the force required to prevent the body from falling freely.

Note: The weight of an object is not its mass. If the object is moved to a location where the acceleration of gravity is different (e.g. the moon where gm = 1.7 m/s2) , the mass does not change but the weight does.

Contact Forces: As the name implies these forces act between two objects that are in contact. The contact forces have two components. One that is acting along the normal to the contact surface (normal force) and a second component that is acting parallel to the contact surface (frictional force)

Normal Force: When a body presses against a surface, the surface deforms and pushes on the body with a normal force perpendicular to the contact surface. An example is shown in the picture to the left. A block of mass m rests on a table.

Note: In this case FN = mg. This is not always the case.

Friction: If we slide or attempt to slide an object over a surface, the motion is resisted by a bonding between the object and the surface. This force is known as “friction”. More on friction in chapter 6

(5-7)

Tension: This is the force exerted by a rope or a cable attached to an object Tension has the following characteristics: 1.  It is always directed along the rope 2.  It is always pulling the object 3.  It has the same value along the rope. (for example between points A and B)

The following assumptions are made: a.  The rope has negligible mass compared to the mass of the object it pulls b.  The rope does not stretch c.  If a pulley is used as in fig.(b) and fig.(c), we assume that the pulley is massless

and frictionless. A B

Newton’s Third Law:

When two bodies interact by exerting forces on each other, the forces are equal in magnitude and opposite in direction

Applying Newton’s Laws / Free body Diagrams Part of the procedure of solving a mechanics problem using Newton’s laws is drawing a free body diagram. This means that among the many parts of a given problem we choose one which we call the “system”. Then we choose axes and enter all the forces that are acting on the system and omitting those acting on objects that were not included in the system.

Recipe for the application of Newton’s law’s of motion

1.  Choose the system to be studied

2.  Make a simple sketch of the system

3.  Choose a convenient coordinate system

4.  Identify all the forces that act on the system. Label them on the diagram

5.  Apply Newton’s laws of motion to the system

Announcements – 2/1/10

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