0910 ENGINEERING SCIENCE 0910 ENGINEERING SCIENCE OBJECTIVES Ability
09/10 ENGINEERING SCIENCE
09/10 ENGINEERING SCIENCE OBJECTIVES • Ability to understand define scalar and vector quantity. • Ability to understand the concept of vector addition, subtraction & components and applying the analytical component method. • Ability to understand 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.
09/10 ENGINEERING SCIENCE SUBTOPICS • • • Scalars & Vectors Speed, Velocity & Acceleration Motion Equation Newton’s Law Force From Newton’s Law
09/10 ENGINEERING SCIENCE SCALARS & VECTORS SCALAR VECTOR • Quantity that only has magnitude • Does not have direction • Example: mass, time, volume, area, temperature, distance, speed 1 • Quantity with magnitude and direction • Commonly represent by boldface notation, A or arrow notation, • Example: velocity, force, acceleration, momentum
09/10 ENGINEERING SCIENCE VECTORS 2 vectors are the same if : (a) magnitude a = magnitude b |a| = |b| (b) a and b parallel or same direction
09/10 ENGINEERING SCIENCE VECTOR ADDITION 1) R=A+B? 2) 3)
ENGINEERING SCIENCE 09/10 the sum of vectors is independent of the order in which the vectors are added, as long as we maintain their length and direction
09/10 ENGINEERING SCIENCE 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
09/10 ENGINEERING SCIENCE 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 Vector components Magnitudes of components Phase
09/10 ENGINEERING SCIENCE UNIT VECTOR • Unit vector has a magnitude of unity, or one, and thereby simply indicates a vector’s direction.
ENGINEERING SCIENCE VECTORS ADDITION BY COMPONENTS • resolve the vectors into rectangular vector components and adding the components for each axis independently y y F = F 1 + F 2 F 1 F 2 Fy = Fy 1 + Fy 2 Fx 1 Fy 1 Fx 1 09/10 Fy 2 F = F 1 + F 2 x Fx = Fx 1 + Fx 2 Fy 1 x
ENGINEERING SCIENCE 09/10 (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
09/10 ENGINEERING SCIENCE EXAMPLE 1 You are given two displacement vectors: 1) A with magnitude of 6. 0 m in the direction of 45 o below the + x-axis, and 2) B, which has an x – component of +2. 5 m and a y-component of +4. 0 m. Find a vector C so that A + B + C equals a vector D that has magnitude of 6. 0 m in the + y-direction.
09/10 ENGINEERING SCIENCE SOLUTION A = 6. 0 m, 45 o below the + xaxis (4 th quadrant) • Bx = (2. 5 m)x • By = (4. 0 m)y • Find C suchxthat A+B+C=D= - components (+6. 0 m) y • y - components Ax = A cos 45 o = +4. 24 m Ay = - A sin 45 o = - 4. 24 m Bx = + 2. 5 m By = + 4. 0 m Cx = ? Cy = ? Dx = 0 Dx = +6. 0 m
09/10 ENGINEERING SCIENCE • Calculate x – and y – components separately: x-components: Ax + Bx + Cx = Dx 4. 24 m + 2. 5 m + Cx = 0 ∴Cx = - 6. 74 m y-components: Ay + By + Cy = Dy - 4. 24 m + 4. 0 m + Cy = 6. 0 m ∴ Cy = +6. 24 m • • So, C = (-6. 74 m) x + (6. 24 m) y We may also express the results in magnitude-angle form: Magnitude: Phase:
ENGINEERING SCIENCE EXAMPLE 2 09/10 For the vector shown in Figure above determine;
09/10 ENGINEERING SCIENCE DISTANCE & DISPLACEMENT DISTANCE • scalar quantity • Total path length traversed in moving from one location to another • Only +ve value DISPLACEMENT • vector quantity • straight line distance between 2 points along with the direction from the starting point to another • Can have +ve or –ve values (indicate the direction)
09/10 ENGINEERING SCIENCE
09/10 ENGINEERING SCIENCE SPEED & VELOCITY SPEED § refers to how fast an object is moving § scalar quantity § defined as the rate of motion, or rate of change in position. §fast moving = high speed, slow moving = slow speed § zero speed = no movement VELOCITY §Refers to how fast something is moving and in which direction it is moving. § vector quantity § defined as the rate of change of displacement or the rate of displacement. § velocity = 0, does not mean the object is not moving.
09/10 ENGINEERING SCIENCE SPEED SI Unit: m/s VELOCITY SI Unit: m/s
09/10 ENGINEERING SCIENCE EXAMPLE 3 A jogger jogs from one end to the other of a straight 300 m 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 300 m 2. 5 minutes
09/10 ENGINEERING SCIENCE SOLUTION Given : Δx 1= 300 m Δx 2 = -300 m a) b) c) Δt 1= 2. 50 min x 60 s = 150 s Δt 2 = 3. 30 min x 60 s = 198 s
09/10 ENGINEERING SCIENCE ACCELERATION - rate of change of velocity. SI Unit: meters per second squared (m/s 2).
09/10 ENGINEERING SCIENCE EXAMPLE 4 A couple of sport-utility vehicle (SUV) are traveling at 110 km/h on a PLUS highway. The drives sees an accident in the distance and slows down to 55 km/h in 10 s. What is the average acceleration of the SUV?
09/10 ENGINEERING SCIENCE SOLUTION • Change velocities to SI unit. 1 km/h = 0. 278 m/s • v 0 = 110 kmh-1 x (0. 278 ms-1/1 kmh-1) = 30. 5 m/s v = 55 kmh-1 x (0. 278 ms-1/1 kmh-1) = 15. 3 m/s t = 10 s Therefore, average acceleration: a = (v – v 0)/t = (15. 3 m/s – 30. 5 m/s)/10 s = -15. 2 m/s 2 decelaration
09/10 ENGINEERING SCIENCE 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.
09/10 ENGINEERING SCIENCE MOTION WITH CONSTANT ACCELERATION 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
ENGINEERING SCIENCE 09/10 DERIVATION OF MOTION EQUATION: • v = u + at • v= ½(u+v)t • s = ut + ½ at 2 • v 2 = u 2 + 2 as
09/10 ENGINEERING SCIENCE 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;
09/10 ENGINEERING SCIENCE EXAMPLE • The speed of a car travelling along a straight road decreases uniformly from 12 m/s to 8 m/s over 88 m. Calculate the a) Decelaration of the car b)Time taken for the speed to decrease from 12 m/s to 8 m/s c) Time taken for the car to come to a halt from the speed of 12 m/s d)Total distance travelled by the car during this time.
- Slides: 30