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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FORCE AND MOTION
SCALARS & VECTORS
2 vectors are the same if :
(a) magnitude a = magnitude b
|a| = |b|
(b) a and b parallel or same direction
R = A + B ?
the sum of vectors is independentof the order
in which the vectors are added,
as long as we maintain their length and direction
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
Magnitudes of components
F = F1 + F2
F = F1 + F2
Fy = Fy1 + Fy2
Fx = Fx1 + Fx2
VECTORS ADDITION BY COMPONENTS
(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
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.
Ax + Bx + Cx = Dx
4.24m + 2.5m + Cx = 0
∴Cx = - 6.74m
Ay + By + Cy = Dy
- 4.24m + 4.0m + Cy = 6.0m
∴Cy = +6.24m
For the vector shown in Figure above determine;
DISTANCE & DISPLACEMENT
SPEED & VELOCITY
SI Unit: m/s
SI Unit: m/s
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
(a) in jogging to the far end of the track
(b) coming back to the starting point, and
(c) for total jog
Δx1= 300mΔt1= 2.50 min x 60 s = 150 s
Δx2 = -300m Δt2 = 3.30 min x 60 s = 198 s
- rate of change of velocity.
SI Unit: meters per second squared (m/s2).
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?
1km/h = 0.278 m/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
Motion With Constant Acceleration
When anobject moves along the straightline and velocity increase uniformly from Vo to v in time t.
a= change in velocity/time taken
= (v-u )/t
derivation of motion equation: