Welcome back to Physics 211

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# Welcome back to Physics 211 - PowerPoint PPT Presentation

Welcome back to Physics 211. Today’s agenda: More on Newton’s Laws Free Body Diagrams. Newton’s Second Law. Second Law: Σ F on object =F net = m a of object where F net is the vector sum of all external forces on the object considered m = (inertial) mass.

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Presentation Transcript

### Welcome back to Physics 211

Today’s agenda:

More on Newton’s Laws

Free Body Diagrams

PHY211 Fall 2011 Lecture 6-2

Newton’s Second Law
• Second Law:
• ΣFon object =Fnet= m aof object
• where Fnet is the vector sum of all external forces on the object considered
• m = (inertial) mass

PHY211 Fall 2011 Lecture 6-2

Illustration of Newton’s Second Law

• Pull cart with constant force as displayed on force-meter
• How does cart respond?
• Fon object = m aof object
• Lifting mass using cotton twine with different accelerations

PHY211 Fall 2011 Lecture 6-2

Equilibrium

An object on which the net force is zero is said to be in equilibrium. The object might be at rest in static equilibrium, or it might be moving along a straight line with constant velocity in dynamic equilibrium. Both are identical from a Newtonian perspective because the net force and the acceleration are zero.

PHY211 Fall 2011 Lecture 6-2

EXAMPLE 6.2 Towing a car up a hill

QUESTION:

Draw picture

Draw free body diagram

Add up forces (sum = zero in equilibrium)

PHY211 Fall 2011 Lecture 6-2

Using Newton’s Second Law
• The essence of Newtonian mechanics can be expressed in two steps.
• The forces on an object determine its acceleration a = Fnet/m, and
• The object’s trajectory can be determined by using the equations of kinematics.

PHY211 Fall 2011 Lecture 6-2

EXAMPLE 6.3 Speed of a towed car

QUESTION:

Draw picture

Draw free body diagram

Add up forces (sum = ma not in equilibrium)

PHY211 Fall 2011 Lecture 6-2

Newton’s first law can be applied to

static equilibrium.

inertial equilibrium.

dynamic equilibrium.

both A and B.

both A and C.

PHY211 Fall 2011 Lecture 6-2

The coefficient of static friction is

smaller than the coefficient of kinetic friction.

equal to the coefficient of kinetic friction.

larger than the coefficient of kinetic friction.

not discussed in this chapter.

PHY211 Fall 2011 Lecture 6-2

The force of friction is described by

the law of friction.

the theory of friction.

a model of friction.

the friction hypothesis.

PHY211 Fall 2011 Lecture 6-2

Relationship between Mass, Weight and Gravity

Mass

• scalar quantity that describes an object’s inertia
• describes the amount of matter in an object
• intrinsic property of an object tells us something about the object, regardless of where the object is, what it’s doing, or whatever forces may be acting on it

PHY211 Fall 2011 Lecture 6-2

Gravity

Somewhat more loosely, gravity is a force that acts on mass. 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.

PHY211 Fall 2011 Lecture 6-2

Gravity cont.

where G = 6.67 x 10 -11 Nm2/kg2

near Earth

PHY211 Fall 2011 Lecture 6-2

Static Friction

The box is in static equilibrium, so the static friction must exactly balance the pushing force:

PHY211 Fall 2011 Lecture 6-2

Static Friction

• An object remains at rest as long as fs < fs max.
• The object slips when fs = fs max.
• A static friction force fs > fs max is not physically   possible.

where the proportionality constant μs is called the coefficient of static friction.

PHY211 Fall 2011 Lecture 6-2

Kinetic Friction

The kinetic friction force is proportional to the magnitude of the normal force.

where the proportionality constant μk is called the coefficient of kinetic friction.

PHY211 Fall 2011 Lecture 6-2

Block on Incline

f

N

W

q

PHY211 Fall 2011 Lecture 6-2

What if  > tan-1ms ?

• depends on the type of surfaces of the objects
• depends on the normal force that the objects exert on each other
• does not depend on the surface area where the two objects are touching
• does not depend on the speed with which one object is moving relative to the other

fk = μkn

PHY211 Fall 2011 Lecture 6-2

f

What if  > tan-1 μs ?

n

W

• Block begins to slide
• Resolve along plane:

μkWcosθ - Wsinθ= ma

• Or:

a = g(μkcosθ - sinθ)

q

PHY211 Fall 2011 Lecture 6-2

Example Problem #43
• A baggage handler drops your 10 kg suitcase onto a conveyor belt running at 2.0 m/s. The materials are such that μs=0.50 and μk=0.30. How far is your suitcase dragged before it is riding smoothly on the belt?

PHY211 Fall 2011 Lecture 6-2

Example Problem #43
• A baggage handler drops your 10 kg suitcase onto a conveyor belt running at 2.0 m/s. The materials are such that μs=0.50 and μk=0.30. How far is your suitcase dragged before it is riding smoothly on the belt?

PHY211 Fall 2011 Lecture 6-2

FHW6 Due F 10/7/11
• Chapter 5 text E&P – 6, 13, 14, 18, 26, 2
• WHW7 Chapter 5 text E & P 30, 43, 46, 47, 53, 56
• FHW7 Ch. 6 E&P 1, 5, 18, 20, 21, 29, 37, 46

PHY211 Fall 2011 Lecture 6-2