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Chapter 3. Acceleration and Newton’s Second Law of Motion. Mechanics (Classical/Newtonian Mechanics): Study of motion in relation to force and energy, ie, the effects of force and energy on the motion of an object.

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Chapter 3

Acceleration

and

Newton’s Second Law of Motion


Mechanics (Classical/Newtonian Mechanics): Study of motion in relation to force and energy, ie, the effects of force and energy on the motion of an object.

Kinematics – Mechanics that describes how objects move, no reference to the force or mass.

Dynamics– Mechanics that deals with force: why objects move.


Describing Motion Involves:

  • Reference frames

  • Reference Direction

  • Displacement

  • Velocity

  • Acceleration


Reference frames and coordinates

When we say a car is moving at 45 km/hr, we usually mean “with respect to the earth” although is is not explicitly stated.

An escalator is moving at 3 m/s relative to the ground. A lady walks on the escalator at 2 m/s relative to the escalator. The lady’s speed relative to the ground is 5 m/s.


Reference Direction

  • Can use compass directions: North, East, South, West.

  • In physics, we use coordinate axes

    1-D coordinate system:

    • Specify the origin.

    • Show where x axis is pointing: to the right is positive direction of x.

    • Label the axis with the relevant units.


+

y

-

+

0

x

-

Reference Direction:2-D coordinate system:

• Specify the origin.

• Show x and y axes. To the right is positive direction of x. Vertically upward is positive y.

• Label the axis with the relevant units.


Distance

Just a length.

A Scalar quantity.

SI unit = meter (m).


y

3

A

x

3

-3

B

-3

Position

  • A vector quantity describing where you are relative to an “origin”.

  • Point A is located at x =3, y =1 or (3,1). Point B is located at (-1,-2).

10


Displacement (Dr)

  • A vector quantity.

  • Change in position relative to the starting point.

  • SI unit = meter (m).

  • Dr = rf – ri

    The displacement from A to B is

    x-direction: -1 – 3 = -4

    y-direction: -2 – 1 = -3

  • r = (-4, -3), |Dr| = (42 + 32) = 5

y

3

A

x

3

-3

B

-3

13


Example
Example

  • You drive 40 km north, then turn east and drive 30 km east. What is your net displacement?

30 km E

40 km N


2. A stray dog ran 4 km due north and then 4 km due east. What is the magnitude of itsdisplacement after this movement?

  • 8 km

  • 5.7 km

  • 16 km

  • 332 km

  • 0 km


Average Speed and Average Velocity What is the magnitude of its

Average Speed = (distance traveled)/time taken = d/t

Speed: - specified only by magnitude. It is a scalar quantity.

SI unit = m/s

- always a positive number

Average velocity = (displacement)/time taken

= (x2 – x1)/(t2 – t1) = x/t

Velocity: - specified by both magnitude and direction.

- It is a vector quantity.

- units – m/s

- Positive/ negative sign used to indicate

direction


Magnitude of average speed and average velocity: not always equal.

Charles walks 120 m due north in 40s. He then walks another 60 m still due north in another 60s.

Average speed?

Average velocity?

Average speed = Total distance/time

Average velocity = Displacement/time


Charles walks 120 m due north. He then walks another 60 m due south. He took a total of 100 s for the journey.

Average speed?

Average velocity?

Average speed = Total distance/time

Average velocity = Displacement/time


Charles walks 120 m due north. He then walks another 120 m due south. He took a total of 100 s for the journey.

Average speed?

Average velocity?


Instantaneous Velocity due south. He took a total of 100 s for the journey.

x(t)

Dx

Dt

t

x(t)

t

23

  • The average velocity is displacement divided by the change in time.

  • Instantaneous velocity is limit of average velocity as Dt gets small. It is the slope of the x(t) versus t graph.

  • v = lim t0x/ t.


Instantaneous Velocity due south. He took a total of 100 s for the journey.

  • Magnitude of instantaneous speed and instantaneous velocity are equal.

  • If velocity is uniform (constant) in a motion, then magnitude of

    Instantaneous velocity = Average speed.

uniform (constant) velocity

velocity (m/s)

velocity (m/s)

t2

t1

time (s)

time (s)


Determine the velocity of the car at times A, B, C and D. due south. He took a total of 100 s for the journey.

A Positive Zero Negative

B Positive Zero Negative

C Positive Zero Negative

D Positive Zero Negative

x(t)

A

t

B

C

D


Uniform velocity
Uniform Velocity due south. He took a total of 100 s for the journey.

v(t)

x(t)

t

t

  • An object moving with uniform velocity – means velocity stays uniform (constant) throughout the motion.

  • Its magnitude and direction stays the same.

  • Its displacement changes uniformly.

  • Moving along a straight line with constant speed.

v = slope


A car moves at a constant velocity of magnitude 20 m/s. At time t = 0, its position is 50 m from a reference point. What is its position at

(i) t = 2s? (ii) t = 4s (iii) t = 10s?


Non uniform velocity
Non-uniform Velocity time t = 0, its position is 50 m from a reference point. What is its position at

  • An object moving with non-uniform velocity – means velocity does not stays uniform (constant) during the motion.

  • Either its magnitude or direction stays the same.

  • Moving along a straight line with varying speed, or moving in a curved path.

x(t)

v(t)

t

t


Area Under velocity-time Graph time t = 0, its position is 50 m from a reference point. What is its position at

Area under the velocity-time graph = magnitude of the displacement over the time interval.

uniform (constant) velocity

velocity (m/s)

Area = v(t2-t1) = (x2 – x1) = x

time (s)

t2

t1

velocity (m/s)

time (s)


A stray dog ran 3 km due north and then 4 km due east. What is the magnitude of its average velocity if it took 45 min?


v(t) is the magnitude of its average velocity if it took 45 min?

D

C

Dv

B

E

A

Dt

t

Acceleration (a)

  • The average acceleration is the change in velocity divided by the change in time.

  • SI unit = m/s2

  • Slope of velocity-time graph.


v(t) is the magnitude of its average velocity if it took 45 min?

t

Acceleration (a)

  • Instantaneous acceleration is limit of average acceleration as Dt gets small. It is the slope of the v(t) versus t graph.

  • a = lim t0v/ t.


Uniform acceleration
Uniform Acceleration is the magnitude of its average velocity if it took 45 min?

a(t)

v(t)

v

t

t

t

  • An object moving with uniform acceleration – means acceleration stays uniform (constant) throughout the motion.

  • Its magnitude and direction stays the same.

  • Its velocity changes uniformly.

  • Moving along a straight line.

a = slope


A car moves at a constant acceleration of magnitude 5 m/s is the magnitude of its average velocity if it took 45 min?2. At time t = 0, the magnitude of its velocity is 8 m/s. What is the magnitude of its velocity at

(i) t = 2s? (ii) t = 4s? (iii) t = 10s?


Acceleration (a) is the magnitude of its average velocity if it took 45 min?

Average acceleration =v/ t

T/F? If the acceleration of an object is zero, it must be at rest.

T/F? If an object is at rest, its acceleration must be zero.


Example is the magnitude of its average velocity if it took 45 min?

Train A moves due east along a straight line with a velocity 8 m/s. Within 7 seconds, it’s velocity increases to 22 m/s. What is its average acceleration?

Train B is moves due east along a straight line at a velocity of 18 m/s. Within 10 s, its velocity drops to 3 m/s. What is its average acceleration?

When an object slows down, we say it is decelerating. In that case, the direction of the acceleration will be opposite to that of the velocity.


Example is the magnitude of its average velocity if it took 45 min?

Is it possible for an object to have a positive velocity at the same time as it has a negative acceleration?

1 - Yes

2 - No

“Yes, because an object can be going forward but at the same time slowing down which would give it a negative acceleration.”

37


Graphical Representation is the magnitude of its average velocity if it took 45 min?

Slope of the line = average velocity

Position, x (m)

Time, t (s)

If the slope is zero, the object is at rest


Graphical Representation is the magnitude of its average velocity if it took 45 min?

Slope of the line = average acceleration

velocity, v (m/s)

O

Time, t (s)

If the slope is zero, the object is moving with zero acceleration (constant velocity)


velocity, v (m/s) is the magnitude of its average velocity if it took 45 min?

C

D

E

B

A

O

Time, t (s)

Slope of the line = average acceleration = v/t

  • OA

  • AB

  • BC

  • CD

  • DE

1. When is a = 0?

2. When is a < 0?

3. When is a = maximum?


1 the area under a velocity time graph represents
1. The area under a velocity-time graph represents is the magnitude of its average velocity if it took 45 min?

  • Average velocity

  • Average acceleration

  • Total displacement

  • Instantaneous velocity


2. A stray dog ran 4 km due north and then 4 km due east. What is the magnitude of itsdisplacement after this movement?

  • 8 km

  • 5.7 km

  • 16 km

  • 332 km

  • 0 km


3 which car has a southward acceleration a car traveling
3. Which car has a southward acceleration? A car traveling What is the magnitude of its

  • Southward at constant speed

  • Northward at constant speed

  • Southward and slowing down

  • Northward and speeding up

  • Northward and slowing down


4. A train moves due north along a straight path with a uniform acceleration of 0.18 m/s2. Ifits velocity is 2.4 m/s, what will its velocity be after 1 minute?

  • 13.2 m/s north

  • 2.58 m/s north

  • 10.8 m/s north

  • 144 m/s north

  • None of these


5. A bird flew 6.00 km north, then turned around and flew 1.50 km west. If it took 25.0 min, what was the average speed of the bird?

  • 5.00 m/s

  • 0.300 m/s

  • 300 m/s

  • 4.12 m/s

  • 247 m/s


6. A bird flew 6.00 km north, then turned around and flew 1.50 km west. If it took 25.0 min, what was magnitude of the average velocity of the bird?

  • 5.00 m/s

  • 0.300 m/s

  • 300 m/s

  • 4.12 m/s

  • 247 m/s


7. A train moving along a straight path due north has a velocity of 20 m/s. Within 5.0 seconds,its velocity became 5.0 m/s. What was the train's average acceleration?

  • 15 m/s2 north

  • 15 m/s2 south

  • 3.0 m/s2 north

  • 3.0 m/s2 south

  • 5.0 m/s2 north

  • 5.0 m/s2 south


VECTORS velocity of 20 m/s. Within 5.0 seconds,

tip

Vector A

tail

+y

A

Equal Vectors:

A = B only if they have equal magnitudes and same directions.

B

+x

Displacing a vector parallel to itself does not change it


The negative of a vector: velocity of 20 m/s. Within 5.0 seconds,

Two vectors with equal magnitude but opposite directions are negatives of each other

A

-A

Vectors can be multiplied by a scalar:

A

2A

½ A


Components of a vector velocity of 20 m/s. Within 5.0 seconds,

A vector can be expressed as a sum of 2 vectors called “components” that are parallel to the x and y axes:

+y

A = Ax + Ay

Ax = x- component of A.

Ay = y-component of A.

A

Ay

+x

Ax


h = hypotenus velocity of 20 m/s. Within 5.0 seconds,

o = opposite

a = adjacent

SOH CAH TOA

sin  = opposite/hypotenuse = o/h

cos  = adjacent/hypotenuse = a/h

tan  = opposite/adjacent = o/a

Pythagorean theorem: h2 = a2 + o2


A velocity of 20 m/s. Within 5.0 seconds,

Ay

Ax

sin  = o/h = Ay/A or y-component:Ay = A . sin 

cos  = a/h = Ax/A or x-component: Ax = A . cos 

A2 = Ax2 + Ay2 or magnitude of vectorA = (Ax2 + Ay2 )

tan  = o/a = Ay/Ax or  = tan-1(Ay/Ax)


Addition of vectors velocity of 20 m/s. Within 5.0 seconds,

Three methods:

Component method (analytical).

Tip-to-tail (graphical).

Parallelogram (graphical).


Component method velocity of 20 m/s. Within 5.0 seconds,

Add vectors V1 + V2 + V3:

  • Find x and y components of each vector.

    {V1=V1x + V1y}, {V2 = V2x + V2y}, {V3 = V3x + V3y

    2. Add x-components and y-components separately:

    {Vx = V1x + V2x +V3x} and {Vy = V1y + V2y +V3y}

    3. Find the magnitude of the resultant vector using pythagorean theorem: V = {V2x + V2y}

    4. Find the angle of the resultant measured from the +x axis:  = tan-1(Vy/Vx)


Example velocity of 20 m/s. Within 5.0 seconds,

y

A

B

C

30o

x

Three displacement vectors are shown in the figure below. Their magnitudes are A = 20 cm, B = 16 cm and C = 12 cm. Find the magnitude and angle of the resultant vector.


Tip-to-tail Method velocity of 20 m/s. Within 5.0 seconds,

Draw the vectors such that the “tail” of the second vector connects to the tip of the first vector. The resultant is from the tail of the first to the tip of the second.

A

B

B

C

A

-B

A

D

C = A + B

D = A - B


Parallelogram Method velocity of 20 m/s. Within 5.0 seconds,

Draw the vectors such that their “tails” are joined to a common origin.

Construct a parallelogram with the two vectors as adjacent sides.

The resultant vector is the diagonal line of the parallelogram drawn from the common origin.

A

B

C = A + B

C

A

B


From the diagram shown in the figure below, the unknown vector is

?

B

A

  • A – B

  • A + B

  • -A – B

  • – A +B


Z vector is2

Y

Z1

X

From the diagram shown in the figure below, what are vectors Z1 and Z2 in terms of X and Y?


In the diagram below, what are the x and y-components of vector A if the magnitude of A is 40 units?

y

A

35o

x


Z vector A if the magnitude of A is 40 units? 2

A

A

Z1

B

B

From the diagram shown in the figure below, what are vectors Z1 and Z2 in terms of B and A?


Relative Velocity vector A if the magnitude of A is 40 units?

You are on a train traveling 40 mph North. If you walk 5 mph toward the front of the train, what is your speed relative to the ground?

1) 45 mph 2) 40 mph 3) 35 mph

40 mph N + 5 mph N = 45 mph N

40

5

45

25


Relative Velocity vector A if the magnitude of A is 40 units?

You are on a train traveling 40 mph North. If you walk 5 mph toward the rear of the train, what is your speed relative to the ground?

1) 45 mph 2) 40 mph 3) 35 mph

40 mph N - 5 mph N = 35 mph N

40

5

35

28


Relative Velocity vector A if the magnitude of A is 40 units?

You are on a train traveling 40 mph North. If you walk 5 mph sideways across the car, what is your speed relative to the ground?

1) < 40 mph 2) 40 mph 3) >40 mph

40 mph N + 5 mph W = 41 mph N

5

40

Relative Motion (Add vector components)

30


P2.27: Find the magnitude and direction of the vector with the following components:

  • x = -5.0 cm, y = +8.0 cm

  • Fx = +120 N, Fy = -60.0 N

  • vx = -13.7 m/s, vy = -8.8 m/s

  • ax = 2.3 m/s2, ay = 6.5 cm/s2


P2.56: A 3.0 kg block is at rest on a horizontal floor. If you push horizontally on the block with a force of 12.0 N, it just starts to move.

  • What is the coefficient of static friction?

    (b) A 7.0-kg block is stacked on top of the 3.0-kg block. What is the magnitude F of the force acting horizontally on the 3.0-kg block as before, that is required to make the two blocks start to move?


P3.47: A 2010-kg elevator moves with an upward acceleration of 1.50 m/s2. What is the tension that supports the elevator?

P3.48: A 2010-kg elevator moves with a downward acceleration of 1.50 m/s2. What is the tension that supports the elevator?


A stray dog ran 4 km due north and then 4 km due east. What is the magnitude of its displacement after this movement?


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