Concept of Force and Newton’s Laws of Motion

8.01W02D2

Today’s Reading Assignment:

Young and Freedman Sections 4.1-4.6, 5.1-5.3

Review: Relatively Inertial Reference Frames

Two reference frames with the

zero relative acceleration

One moving object has different

position vectors in different frames

Law of addition of velocities

Acceleration in either reference frame is the same

1 2= +r R rrr r

rv

1=

rV + r

v2

d dt= =A V 0r rr

rV =d

rR dt

ra

1=

rA +

ra2 =

ra2

Concept Question: Relatively Inertial Reference Frames

Suppose Frames 1 and 2 are relatively inertial reference frames.

1)An object that is at rest in Frame 2 is moving at a constant velocity in reference Frame 1.

2)An object that is accelerating in Frame 2 has the same acceleration in reference Frame 1.

3)An object that is moving at constant velocity in Frame 2 is accelerating in reference Frame 1.

4)An object that is accelerating in Frame 2 is moving at constant velocity in reference Frame 1.

5)Two of the above6)None of the above

C.Q. Answer: Newton’s First Law

Answer: (5). Both (1) and (2) are correct.

An object that is at rest in Frame 2 is moving at a constant velocity in reference Frame 1. The accelerations are the same in relatively inertial reference frames because the relative velocity of the two reference frames is constant hence the relative acceleration of the two reference frames is zero.

Inertial Mass Inertial mass as a ‘quantity of matter’

Standard body with mass ms and SI units [kg]

Mass of all other bodies will be determined relative to the mass of our standard body.

Apply the same action to the standard body and an unknown body

Define the unknown mass in terms of ratio

Mass ratio is independent of the method used to produce the uniform accelerations.

mu

ms

=as

au

Definition of Force

Force is a vector quantity

The magnitude of the total force is defined to be

The direction of the vector sum of all the forces on a body is the same as the direction of the acceleration.

The SI units for force are newtons (N):

total m≡F ar r

rFtotal ≡m

ra

1 N ≡1kg⋅m⋅s−2

Superposition Principle Apply two forces and on a body,

the total force is the vector sum of the two forces:

Notation: The force acting on body 1 due to the interaction between body 1 and body 2 is denoted by

1Fr

2Fr

total1 2= +F F F

r r r

3 31 32total = +F F F

r r r

Example: The total force exerted on body 3 due to the interactions with bodies 1 and 2 is:

12Fr

Examples of Forces• Gravitation

• Electric and magnetic forces

• Elastic forces (Hooke’s Law)

• Frictional forces: static and kinetic friction, fluid resistance

• Contact forces: normal forces and static friction

• Tension and compression

Newton’s First Law Every body continues in its state of rest, or of uniform motion in a right line, unless it is

compelled to change that state by forces impressed upon it.

Newton’s First Law in relatively inertial reference frames:

If the vector sum of forces acting on an object at rest in Frame 2 is zero then the vector sum of forces acting on the object moving at constant speed in Frame 1 is also zero

rF

i=

r0∑ ⇒ r

v =constant

Newton’s Second LawThe change of motion is proportional to the motive

force impresses, and is made in the direction of the right line in which that force is impressed,

, , ,1 1 1

, , . N N N

x i x y i y z i zi i i

F ma F ma F ma= = =

= = =∑ ∑ ∑

When multiple forces are acting,

In Cartesian coordinates:

1

.N

ii

m=

=∑F ar r

.m=F ar r

Newton’s Third Law

To every action there is always opposed an equal reaction: or, the mutual action of two bodies upon each other are always equal, and directed to contrary parts.

Action-reaction pair of forces cannot act on same body; they act on different bodies.

1,2 2,1=−F Fr r

Force Law: Newtonian Induction

• Definition of force has no predictive content.

• Need to measure the acceleration and the mass in order to define the force.

• Force Law: Discover experimental relation between force exerted on object and change in properties of object.

• Induction: Extend force law from finite measurements to all cases within some range creating a model.

• Second Law can now be used to predict motion!

• If prediction disagrees with measurement adjust model.

Empirical Force Law: Hooke’s Law

Consider a mass m attached to a spring

Stretch or compress spring by different amounts produces different accelerations

Hooke’s law:

Direction: restoring spring to equilibrium

Hooke’s law holds within some reasonable range of extension or compression

| | k l= ΔFr

Table Problem: Hooke’s Law

Consider a spring with negligible mass that has an unstretched length 8.8 cm. A body with mass 150 g is suspended from one end of the spring. The other end (the upper end) of the spring is fixed. After a series of oscillations has died down, the new stretched length of the spring is 9.8 cm. Assume that the spring satisfies Hooke’s Law when stretched. What is the spring constant?

Force Law: Gravitational Force near the Surface of the Earth

Near the surface of the earth, the gravitational interaction between a body and the earth is mutually attractive and has a magnitude of

where is the gravitational mass of the body and g is a positive constant.

grav gravm g=Fr

m

grav

g ; 9.81 m ⋅s−2

Force Laws: Contact Forces Between Surfaces

The contact force between two surfaces is denoted by the vector

Normal Force: Component of the contact force perpendicular to surface and is denoted by

Friction Force: Component of the contact force tangent to the surface and is denoted by

Therefore the contact force can be modeled as a vector sum,normalsurface hand ≡F N

r r

,tangentsurface hand ≡F f

rr

rC ≡

rN +

rf

rF

surface,handtotal ≡

rC

Concept Question: Car-Earth Interaction

Consider a car at rest. We can conclude that the downward gravitational pull of Earth on the car and the upward contact force of Earth on it are equal and opposite because

1. the two forces form a third law interaction pair.2. the net force on the car is zero.3. neither of the above.4. unsure

Concept Question: Normal Force

Consider a person standing in an elevator that is accelerating upward. The upward normal force N exerted by the elevator floor on the person is

1. larger than

2. identical to

3. smaller than

the downward force of gravity on the person.

Kinetic Friction The kinetic frictional force fk is proportional to the normal force, but independent of surface area of contact and the velocity.

The magnitude of fk is

where µk is the coefficients of friction.

Direction of fk: opposes motion

fk=μkN

Static Friction Varies in direction and magnitude depending on applied forces:

Static friction is equal to it’s maximum value

0 ≤ fs ≤ fs,max =μsN

f

s,max=μsN

Tension in a Rope

The tension in a rope at a distance x from one end of the rope is the magnitude of the action-reaction pair of forces acting at that point ,

left,right right,left( ) ( ) ( )T x x x= =F Fr r

Table Problem: Tension in a Rope Between Trees

Suppose a rope is tied rather tightly between two trees that are separated by 30 m. You grab the middle of the rope and pull on it perpendicular to the line between the trees with as much force as you can. Assume this force is 1000 N, and the point where you are pulling on the rope is 1 m from the line joining the trees. What is the magnitude of the tension in the rope?

Concept of System: Reduction

• Modeling complicated interaction of objects by isolated a subset (possible one object) of the objects as the system

• Treat each object in the system as a point-like object

• Identify all forces that act on that object

Free Body Diagram

• Represent each force that is acting on the object by an arrow on a free body force diagram that indicates the direction of the force

• Choose set of independent unit vectors and draw them on free body diagram.

• Decompose each force in terms of vector components.

• Add vector components to find vector decomposition of the total force

iFr

1 2T = + +⋅⋅⋅F F F

r r r

1, 2,T T T

x x xF F F= + +⋅⋅⋅

, , ,ˆ ˆ ˆ

i i x i y i zF F F= + +F i j kr

1, 2,T T T

y y yF F F= + +⋅⋅⋅

1, 2,T T T

z z zF F F= + +⋅⋅⋅

Newton’s Second Law: Strategy

• Treat each object in the system as a point-like object

• Identify all forces that act on that object, draw a free body diagram

• Apply Newton’s Second Law to each body

• Find relevant constraint equations

• Solve system of equations for quantities of interest

Worked Example: Pulley and Inclined Plane

A block of mass m1, constrained to move along a plane inclined at angle ϕ to the horizontal, is connected via a massless inextensible rope that passes over a massless pulley to a bucket to which sand is slowly added. The coefficient of static friction is μs. Assume the gravitational constant is g. What is the mass of the bucket and sand (m2) just before the block slips upward?

Solution: Pulley and Inclined Plane

Sketch and coordinate system

Free body force diagrams

Newton’s Second Law

i1

:T −m1gsinφ−μsN =0

j1 : N −m1gcosφ =0

Newton’s Second LawObject 1:

Object 2: j2

:T −m2g=0

Constraint Conditions

ra

1=

r0

ra2 =

r0

fs = fs,max =μsN

T ≡T1,r =T2,r

Simplification: m2g −m1gsinφ−μsm1gcosφ =0

Solution: m2=m1(sinφ+ μscosφ)

Concept Question: Varying Tension in String

A block of mass m1, constrained to move along a plane inclined at angle ϕ to the horizontal, is connected via a massless inextensible rope that passes over a massless pulley to a bucket to which sand is slowly added. The coefficient of static friction is μs. Assume the gravitational constant is g. What happens to the tension in the string just after the block begins to slip upward?

1. Increases2. Decreases3. Stays the same4. oscillates5. I don’t know6. None of the above

Table Problem: Two Blocks and Two Pulleys

Two blocks 1 and 2 of mass m1 and m2 respectively are attached by a string wrapped around two pulleys as shown in the figure. Block 1 is accelerating to the right on a fricitonless surface. You may assume that the string is massless and inextensible and that the pulleys are massless. Find the accelerations of the blocks and the tension in the string connecting the blocks.

Lecture Demo: Block with Pulley and Weight on Incline B24

http://scripts.mit.edu/~tsg/www/index.php?page=demo.php?letnum=B%24&show=0

Table Problem: Painter on Platform

A painter pulls on each rope with a constant force F. The painter has mass m1 and the platform has mass m2.

a. Draw free body force diagrams for the platform and painter. Is there something wrong with this picture?

b. Find the acceleration of the platform.

Source: why you need to know physics

Next Reading Assignment: W02D3

Young and Freedman (Review) 5.1-5.3

Appendix: Newton’s Second Law

Detailed Problem Solving Strategy

Methodology for Newton’s 2nd LawI. Understand – get a conceptual grasp of the

problem

Sketch the system at some time when the system is in motion.

Draw free body diagrams for each body or composite bodies:

Each force is represented by an arrow indicating the direction of the force

Choose an appropriate symbol for the force

II. Devise a Plan Choose a coordinate system:

• Identify the position function of all objects and unit vectors.

• Include the set of unit vectors on free body force diagram.

Apply vector decomposition to each force in the free body diagram:

Apply superposition principle to find total force in each direction:

ˆ ˆ ˆ( ) ( ) ( )i x i y i z iF F F= + +F i j kr

( ) ( )( ) ( )( ) ( )

total

1 2

total

1 2

total

1 2

ˆ :

ˆ :

ˆ :

x x x

y y y

z z z

F F F

F F F

F F F

= + +

= + +

= + +

i

j

k

L

L

L

II. Devise a Plan: Equations of Motion

• Application of Newton’s Second Law

• This is a vector equality; the two sides are equal in magnitude and direction.

total1 2 .m= + +⋅⋅⋅=F F F a

r r r r

( ) ( )( ) ( )( ) ( )

1 2

1 2

1 2

ˆ :

ˆ :

ˆ :

x x x

y y y

z z z

F F ma

F F ma

F F ma

+ + =

+ + =

+ + =

i

j

k

L

L

L

II. Devise a Plan (cont’d)

Analyze whether you can solve the system of equations

• Common problems and missing conditions.

• Constraint conditions between the components of the acceleration.

• Action-reaction pairs.

• Different bodies are not distinguished.

Design a strategy for solving the system of equations.

III. Carry Out your Plan

Hints:

Use all your equations. Avoid thinking that one equation alone will contain your answer!

Solve your equations for the components of the individual forces.

IV. Look Back

• Check your algebra

• Substitute in numbers

• Check your result

• Think about the result: Solved problems become models for thinking about new problems.