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CHAPTER 2 FORCE AND MOTIONPowerPoint Presentation

CHAPTER 2 FORCE AND MOTION

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CHAPTER 2 FORCE AND MOTION

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CHAPTER 2

FORCE AND MOTION

- 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.

- Scalars & Vectors
- Speed, Velocity & Acceleration
- Motion Equation
- Newton’s Law
- Force From Newton’s Law

SCALARS & VECTORS

2 vectors are the same if :

(a) magnitude a = magnitude b

|a| = |b|

(b) a and b parallel or same direction

1)

R = A + B ?

2)

3)

the sum of vectors is independentof the order

in which the vectors are added,

as long as we maintain their length and direction

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

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

UNIT VECTOR

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

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

(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

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.

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

- 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:

EXAMPLE 2

For the vector shown in Figure above determine;

DISTANCE & DISPLACEMENT

SPEED & VELOCITY

SPEED

SI Unit: m/s

VELOCITY

SI Unit: m/s

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

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)

ACCELERATION

- rate of change of velocity.

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

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?

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/s2decelaration

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.

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

derivation of motion equation:

- v = u + at
- v= ½(u+v)t
- s = ut + ½ at2
- v2 = u2 + 2 as

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;

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.