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CHAPTER 2 FORCE AND MOTION

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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  1. CHAPTER 2 FORCE AND MOTION

  2. 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.

  3. SUBTOPICS Scalars & Vectors Speed, Velocity & Acceleration Motion Equation Newton’s Law Force From Newton’s Law

  4. SCALARS & VECTORS

  5. VECTORS 2 vectors are the same if : (a) magnitude a = magnitude b |a| = |b| (b) a and b parallel or same direction

  6. VECTOR ADDITION 1) R = A + B ? 2) 3)

  7. the sum of vectors is independentof the order in which the vectors are added, as long as we maintain their length and direction

  8. 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

  9. VECTOR COMPONENTS Vector components Magnitudes of components Phase • 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

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

  11. VECTORS ADDITION BY COMPONENTS resolve the vectors into rectangular vector components and adding the components for each axis independently 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

  12. (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

  13. 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.

  14. 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

  15. 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:

  16. EXAMPLE 2 For the vector shown in Figure above determine;

  17. DISTANCE & DISPLACEMENT

  18. SPEED & VELOCITY

  19. SPEED SI Unit: m/s VELOCITY SI Unit: m/s

  20. 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

  21. SOLUTION Given : Δx1= 300m Δt1= 2.50 min x 60 s = 150 s Δx2 = -300m Δt2 = 3.30 min x 60 s = 198 s a) b) c)

  22. ACCELERATION - rate of change of velocity. SI Unit: meters per second squared (m/s2).

  23. 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?

  24. 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

  25. 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

  26. 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 Motion With Constant Acceleration

  27. derivation of motion equation: v = u + at v= ½(u+v)t s = ut + ½ at2 v2 = u2 + 2 as

  28. 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;

  29. EXAMPLE 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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