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CHAPTER 2 FORCE AND MOTION. OBJECTIVES. Ability to understand and define scalar and vector quantity. Ability to understand the concept of vector addition, subtraction & components and applying the analytical component method.

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CHAPTER 2 FORCE AND MOTION

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Chapter 2 force and motion

CHAPTER 2

FORCE AND MOTION


Objectives

OBJECTIVES

  • Ability to understand and define scalar and vector quantity.

  • Ability to understand the concept of vector addition, subtraction & components and applying the analytical component method.

  • Ability to understand and distinguish between speed, velocity and acceleration

  • Ability to apply motion equation based on physical situations.

  • Ability to understand the Newton’s Law and its application.


Subtopics

SUBTOPICS

  • Scalars & Vectors

  • Speed, Velocity & Acceleration

  • Motion Equation

  • Newton’s Law

  • Force From Newton’s Law


Chapter 2 force and motion

SCALARS & VECTORS


Vectors

VECTORS

2 vectors are the same if :

(a) magnitude a = magnitude b

|a| = |b|

(b) a and b parallel or same direction


Vector addition

VECTOR ADDITION

1)

R = A + B ?

2)

3)


Chapter 2 force and motion

the sum of vectors is independentof the order

in which the vectors are added,

as long as we maintain their length and direction


Chapter 2 force and motion

VECTORS SUBSTRACTION

R = A – B = A + (–B)

The –ve of a vector is represented by an arrow of the

same length as the original vector, but pointing in the

opposite direction


Chapter 2 force and motion

Vector components

Magnitudes of components

Phase

VECTOR COMPONENTS

  • component of a vector is the influence of that vector in a

  • given direction

  • component method: Most widely used analytical method

  • for adding multiple vectors


Chapter 2 force and motion

UNIT VECTOR

  • Unit vector has a magnitude of unity, or one, and thereby simply indicates a vector’s direction.


Chapter 2 force and motion

y

y

F = F1 + F2

Fy2

F = F1 + F2

Fy2

Fy = Fy1 + Fy2

F2

Fx2

F1

Fy1

Fx2

Fx1

Fy1

Fx = Fx1 + Fx2

Fx1

x

x

VECTORS ADDITION BY COMPONENTS

  • resolve the vectors into rectangular vector components and adding the components for each axis independently


Chapter 2 force and motion

(a) Resolve the vectors into their x- and y-components.

(b) Add all of the x-components and all of the y-components together vectorally to obtain the x- and y-components Cx and Cy respectively


Chapter 2 force and motion

EXAMPLE 1

You are given two displacement vectors:

1) Awith magnitude of 6.0m in the direction of

45o below the + x-axis, and

2) B, which has an x – component of +2.5m

and a y-component of +4.0m.

Find a vector Cso that A + B + C equals a vectorD that has magnitude of 6.0m in the + y-direction.


Chapter 2 force and motion

SOLUTION

  • A = 6.0m, 45o below the + x-axis (4th quadrant)

  • Bx = (2.5m)x

  • By = (4.0m)y

  • Find C such that A + B + C = D = (+6.0m) y


Chapter 2 force and motion

  • Calculate x – and y – components separately:

    x-components:

    Ax + Bx + Cx = Dx

    4.24m + 2.5m + Cx = 0

    ∴Cx = - 6.74m

    y-components:

    Ay + By + Cy = Dy

    - 4.24m + 4.0m + Cy = 6.0m

    ∴Cy = +6.24m

  • So, C = (-6.74m) x + (6.24m) y

  • We may also express the results in magnitude-angle form:

    Magnitude:

    Phase:


Chapter 2 force and motion

EXAMPLE 2

For the vector shown in Figure above determine;


Chapter 2 force and motion

DISTANCE & DISPLACEMENT


Chapter 2 force and motion

SPEED & VELOCITY


Chapter 2 force and motion

SPEED

SI Unit: m/s

VELOCITY

SI Unit: m/s


Chapter 2 force and motion

EXAMPLE 3

A jogger jogs from one end to the other of a straight

300m track in 2.50 min and then jogs back to the

starting point in 3.30 min. What was the jogger’s

average velocity

(a) in jogging to the far end of the track

(b) coming back to the starting point, and

(c) for total jog

3.3 minutes

300m

2.5 minutes


Chapter 2 force and motion

SOLUTION

Given :

Δx1= 300mΔt1= 2.50 min x 60 s = 150 s

Δx2 = -300m Δt2 = 3.30 min x 60 s = 198 s

a)

b)

c)


Chapter 2 force and motion

ACCELERATION

- rate of change of velocity.

SI Unit: meters per second squared (m/s2).


Chapter 2 force and motion

EXAMPLE 4

A couple of sport-utility vehicle (SUV) are traveling

at 110km/h on a PLUS highway. The drives sees

an accident in the distance and slows down to

55km/hin 10s. What is the average acceleration of the SUV?


Chapter 2 force and motion

SOLUTION

  • Change velocities to SI unit.

    1km/h = 0.278 m/s

  • v0 = 110kmh-1 x (0.278ms-1/1kmh-1) = 30.5m/s

    v= 55kmh-1 x (0.278ms-1/1kmh-1) = 15.3m/s

    t = 10s

    Therefore, average acceleration:

    a = (v – v0)/t

    = (15.3m/s – 30.5 m/s)/10s

    = -15.2m/s2decelaration


Chapter 2 force and motion

MOTION EQUATION

  • Equation that describe the behavior of system (e.g the motion of a particle under an influence of a force) as a function of time

  • Sometimes the term refers to the differential equations that the system satisfies and sometimes to the solutions to those equations.


Chapter 2 force and motion

Motion With Constant Acceleration

When anobject moves along the straightline and velocity increase uniformly from Vo to v in time t.

constant acceleration:

a= change in velocity/time taken

= (v-u )/t

v= u+at


Chapter 2 force and motion

derivation of motion equation:

  • v = u + at

  • v= ½(u+v)t

  • s = ut + ½ at2

  • v2 = u2 + 2 as


Chapter 2 force and motion

FREE FALL

  • Objects in motion solely under the influence of gravity.

  • Expressing a=-g in the kinematics equation for constant acceleration in the y-direction yields the following;


Chapter 2 force and motion

EXERCISE

  • The speed of a car travelling along a straight road decreases uniformly from 12m/s to 8 m/s over 88 m. Calculate the

  • Decelaration of the car

  • Time taken for the speed to decrease from 12m/s to 8m/s

  • Time taken for the car to come to a halt from the speed of 12m/s

  • Total distance travelled by the car during this time.


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