Vectors & Concept of Force

8.01

W02D1 Fall 2006

Coordinate System

1. An origin as the reference point

2. A set of coordinate axes with scales and labels

3. Choice of positive direction for each axis

4. Choice of unit vectors at each point in space

Coordinate system: used to describe the position of a point in space and consists of

Cartesian Coordinate System

Vector

A vector is a quantity that has both direction and magnitude. Let a vector be denoted by the symbol

The magnitude of

is denoted by

rA

rA

|rA |≡A

Vector Addition

rA

rBLet and be two vectors. Define a

new vector ,the “vector addition” of and by the geometric construction shown in either figure

rC =

rA +

rB

rA

rB

Summary: Vector Properties

Addition of Vectors

1. Commutativity

2. Associativity

3. Identity Element for Vector Addition such that

4. Inverse Element for Vector Addition such that

Scalar Multiplication of Vectors

1. Associative Law for Scalar Multiplication

2. Distributive Law for Vector Addition

3. Distributive Law for Scalar Addition

4. Identity Element for Scalar Multiplication: number 1 such that

1rA =

rA

(b+ c)rA =b

rA + c

rA

c (rA +

rB) =c

rA + c

rB

b(crA) =(bc)

rA =(cb

rA) =c(b

rA)

rA +

rB =

rB +

rA

(rA +

rB) +

rC =

rA + (

rB +

rC)

r0

rA +

r0 =

r0 +

rA =

rA

−rA

rA + −

rA( ) =

r0

Application of Vectors

(1) Vectors can exist at any point P in space.

(2) Vectors have direction and magnitude.

(3) Vector Equality: Any two vectors that have the same direction and magnitude are equal no matter where in space they are located.

Vector Decomposition

Choose a coordinate system with an origin and axes. We can decompose a vector into component vectors along each coordinate axis, for example along the x,y, and z-axes of a Cartesian coordinate system. A vector at P can be decomposed into the vector sum,

rA =

rA x +

rA y +

rA z

Unit Vectors and Components

The idea of multiplication by real numbers allows us to define a set of unit vectors at each point in space withComponents:

(i, j, k)

| i |=1, |j |=1, |k |=1

rA

x=Ax i,

rA y =Ay j,

rA z =Az k

rA =Ax i + Ay j + Az k

rA =(Ax,Ay,Az)

Vector Decomposition in Two Dimensions

Consider a vector

x- and y components:

Magnitude:

Direction:

rA =(Ax,Ay,0)

Ax =Acos(θ), Ay =Asin(θ)

A = Ax

2 + Ay2

Ay

Ax

=Asin(θ)Acos(θ)

=tan(θ)

θ =tan−1(Ay / Ax)

Vector Addition

Vector Sum:

Components

rA =Acos(θA) i + Asin(θA) j

rB =Bcos(θB) i + Bsin(θB) j

rC =

rA +

rB

C

x=Ax + Bx, Cy =Ay + By

Cx =Ccos(θC ) =Acos(θA)+ Bcos(θB)Cy =Csin(θC ) =Asin(θA) + Bsin(θB)

rC =(Ax + Bx) i + (Ay + By) j =Ccos(θC ) i +Csin(θC ) j

Example: Displacement Vector

At 2 am one morning a person runs 250 m along the infinite corridor at MIT from Mass Ave to the end of Building 8, turns right at the end of the corridor and runs 178 m to the end of Building 2, and then turns right and runs 30 m down the hall.

1. Construct a vector diagram that represents this motion. Indicate your choice of unit vectors.

2. What is the direction and magnitude of the straight line between start and finish?

Example: Displacement Vector Solution

Total displacement:

Magnitude:

Direction:

Δrr =Δ

rr1 + Δ

rr2 + Δ

rr3

=250 m j +178 mi +−30 mj

=178 mi + 220 mj

Δ

rr =((178 m)2 + (220 m)2 )1/ 2 =283m

θ =tan−1((rΔr)y / (

rΔr)x)

=tan−1(220 m/ 178 m) =51.0o

Concept of Force • Force is a vector quantity

• The magnitude of the total force is defined to be

F = (mass) x (magnitude of the acceleration)

• The direction of the total force on a body is the

same as the direction of the acceleration.

• The SI units for force are newtons (N):

1 N = 1kg . m/s2

rFtotal ≡m

ra

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

rF

1 rF

2

total1 2= +F F F

r r r

rF

3total =

rF31 +

rF32

Example: The total force exerted on m3 by m1 and m2 is:

rF

12

Force Law

• 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

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.

rF

grav=mgrav g

m

grav

g =9.80665 m⋅s−2

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

Concept Question: Hooke’s Law

Two identical springs with spring constant k are attached to each other. A block of mass m is suspended from the lower spring. What is the equivalent spring constant keq of the system of two springs? Note the block displaces the same amount in the two figures.

1) keq = k

2) keq = k/2

3) keq = 2k

4) an unknown relation; there is not enough information.

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.

rF

1,2=−

rF2,1

rF

1,2≡ force on 1 due to interaction between 1and 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

rF

surface,handnormal ≡

rN

rF

surface,handtangent

rf

rC ≡

rN +

rf

rF

surface,handtotal ≡

rC

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

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 an interaction pair.2. the net force on the car is zero.3. neither of the above.4. unsure

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

Concept Question: Tension in Rope

You are trying to pull a rock resting on the ground with a heavy rope (the rope has non-zero mass). Just before the rock slips and starts to move, the magnitude of the tension in the rope is

1) greater than the magnitude of the pulling force?

2) equal to the magnitude of the pulling force?

3) less than the magnitude of the pulling force?

4) Not enough information is given to answer.

Static Equilibrium for Forces

(1) The sum of the forces acting on a body at rest is zero

total 1 2 ...= + + =F F F 0rr r r

Measuring Forces: Statics

The science of statics investigates how the forces can act is such a way, and is not concerned with the motions produced by the individual forces. With this in mind, we can measure forces using a statics procedure.

We shall choose a spring that satisfies Hooke’s Law for spring forces between

0.2 N and 10 N.

Linear Calibration of Scale

apply a standard force by suspending a body of mass 0.102 kg

gravitation force acting on body is mg = 1.0 N

spring stretches and exerts a force pulling the body up

mark off 1 N on the scale at the position of the reference point on spring

A second body of twice the mass is now suspended from the spring and we label the position of the reference point by 2.0 N . Continue scale linearly.

rF

spring=

rFgrav

Force Measurements

Attach an unknown body to the spring and measure the spring force.

Use Newton’s Third Law, deduce the force that the spring exerts on the body.

We are measuring the force with which the spring pulls the body, which is equal by Newton’ s Third Law to the

force with which the body pulls the spring.

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

For each object in static equilibrium the sum of individual forces acting on object is zero

1. 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

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

3. Decompose each force in terms of vector components.

4. Use free body diagram to add vector components to find vector decomposition of the total force which is equal to zero (static equilibrium)

5. Set each component of the total force equal to zero, for example the x-component of the total force is zero

rFT =Fx

T i + FyT j + Fz

T k =r0

rFT =

rF1 +

rF2 +⋅⋅⋅=

r0

F

xT =F1,x

T + F2,xT +⋅⋅⋅

Methodology for Newton’s 2nd Law: Statics

I. Understand – get a conceptual grasp of the problem

Think about the problem.

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

Choose a coordinate system and identify the position function of all objects.

Quantify the constraint conditions.

Draw free body diagrams for each body in the problem.

II. Devise a Plan

Draw free body diagrams for each body:

• Include the set of unit vectors

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

• Choose an appropriate symbol for the force

Choosing directions for the forces

• If you solve for a force and find that the force is negative then the force points in the opposite direction as your choice of direction for the force.

II. Devise a Plan: Static Equilibrium of Forces

• Application of Static Equilibrium of Forces

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

rF total =

rF1 +

rF2 +⋅⋅⋅=

r0

i : Fx( )

1+ Fx( )

2+ ... =0

j : Fy( )1+ Fy( )

2+ ... =0

k : Fz( )1+ Fz( )

2+ ... =0

II. Devise a Plan - (con’t)

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

II. Devise a Plan (con’t)

Analyze whether you can solve the system of equations

• Common problems and missing conditions.

• 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.

Group Problem: Hooke’s Law and Free Body Diagram

The spring in configuration (a) is stretched a distance x . The same spring in configuration (b) will stretch a distance

1.2 x2. x3. x/24.05.Not sure