ENGINEERING PHYSICS SEM /2012

ENGINEERING PHYSICS 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. SEM /2012

ENGINEERING PHYSICS SUBTOPICS Scalars & Vectors Speed, Velocity & Acceleration Motion Equation Newton’s Law Force From Newton’s Law SEM /2012

ENGINEERING PHYSICS SCALARS & VECTORS SEM /2012

ENGINEERING PHYSICS VECTORS 2 vectors are the same if : (a) magnitude a = magnitude b |a| = |b| (b) a and b parallel or same direction SEM /2012

ENGINEERING PHYSICS VECTOR ADDITION Example 1:R = A + B 1) SEM /2012

ENGINEERING PHYSICS VECTOR ADDITION 2) SEM /2012

ENGINEERING PHYSICS VECTOR ADDITION 3) SEM /2012

ENGINEERING PHYSICS VECTOR ADDITION ● the sum of vectors is independent of the order ● in which the vectors are added, ● as long as we maintain their length and direction SEM /2012

ENGINEERING PHYSICS VECTORS SUBSTRACTION The –ve of a vector is represented by an arrow of the same length as the original vector, but pointing in the opposite direction R = A – B = A + (–B) SEM /2012

ENGINEERING PHYSICS VECTORS 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 SEM /2012

ENGINEERING PHYSICS Vector components SEM /2012

ENGINEERING PHYSICS Magnitudes of components SEM /2012

ENGINEERING PHYSICS Phase SEM /2012

ENGINEERING PHYSICS UNIT VECTOR Unit vector has a magnitude of unity, or one, and thereby simply indicates a vector’s direction. SEM /2012

ENGINEERING PHYSICS VECTORS ADDITION BY COMPONENTS - resolve the vectors into rectangular vector components and adding the components for each axis independently y y F1F1 F2F2 Fx1 Fy1 Fx2 Fy2 x x F = F 1 + F2 Fx = Fx1 + Fx2 Fx2 Fy = Fy1 + Fy2 Fy1 Fy2 Fx1 SEM /2012

ENGINEERING PHYSICS (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 SEM /2012

ENGINEERING PHYSICS You are given two displacement vectors: 1) A with magnitude of 6.0m in the direction of 45 ° 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 C so that A + B + C equals a vector D that has magnitude of 6.0m in the + y-direction. EXAMPLE 2 SEM /2012

ENGINEERING PHYSICS ● 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 Solution: SEM /2012

ENGINEERING PHYSICS SEM /2012

ENGINEERING PHYSICS ● Calculate x – and y – components separately: x-components: Ax + Bx + Cx = Dx 4.24m + 2.5m + Cx = 0 ∴ Cx = m y-components: Ay + By + Cy = Dy m + 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: SEM /2012

ENGINEERING PHYSICS For the vector shown in Figure above determine; EXAMPLE 3 SEM /2012

ENGINEERING PHYSICS DISTANCE & DISPLACEMENT SEM /2012

ENGINEERING PHYSICS SEM /2012

ENGINEERING PHYSICS SPEED & VELOCITY SEM /2012

ENGINEERING PHYSICS SPEED SI Unit: m/s VELOCITY SI Unit: m/s SEM /2012

ENGINEERING PHYSICS EXAMPLE 4 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 300m 2.5 minutes 3.3 minutes SEM /2012

ENGINEERING PHYSICS 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) Solution: SEM /2012

ENGINEERING PHYSICS ACCELERATION - rate of change of velocity. SI Unit: meters per second squared (m/ ). SEM /2012

ENGINEERING PHYSICS 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 50km/h in 12s. What is the average acceleration of the SUV? EXAMPLE 5 SEM /2012

ENGINEERING PHYSICS Change velocities to SI unit. 1km/h = m/s v0 = 110km/h x (0.278m/s/1km/h) = 30.58m/s v = 50km/h x (0.278m/s/1km/h) =13.9m/s t = 12s Therefore, average acceleration: a = (v – v0)/t = (13.9m/s – 30.5 m/s)/12s = -1.39m/ (decelaration) Solution: SEM /2012

ENGINEERING PHYSICS 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 EQUATION SEM /2012

ENGINEERING PHYSICS When an object moves along the straight line 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 Motion With Constant Acceleration SEM /2012

ENGINEERING PHYSICS derivation of motion equation: SEM /2012

ENGINEERING PHYSICS 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; SEM /2012

ENGINEERING PHYSICS EXAMPLE 6 The speed of a car travelling along a straight road decreases uniformly from 12m/s to 8 m/s over 88 m. Calculate the: i) Decelaration of the car ii) Time taken for the speed to decrease from 12m/s to 8m/s iii) Time taken for the car to come to a halt from the speed of 12m/s iv) Total distance travelled by the car during this time. SEM /2012