Units Motion Scalar / vector quantities Displacement / Velocity / Acceleration Vector components...

Translational Motion

• Units• Motion • Scalar / vector quantities • Displacement / Velocity / Acceleration• Vector components

• Forces • 1st & 2nd law• ∑ F = 0 & ∑ F = ma• Force diagrams

• Momentum • Change in Momentum• Conservation of Momentum (1 & 2D collisions)

• Impulse / Newton’s 3rd Law• Energy• Work• Kinetic & Potential Energies (gravitational & spring)• Conservation of Energy

Level 2 Translational Motion

Success criterion:- meritYOU are able to rearrange an equation and determine the units for the new subject of the formulae.

Make G the subject of the formulae and determine the units for G.

The unit of Force (F) is the Newton (N = kg m s-2)

The unit of the radius squared (r2) is m2

The units of mass squared (m2) is kg2

Show all working.

Writing the correct units.

Hence the units of G are m3 kg-1 s-2

Writing the correct units.

Make G the subject of the formulae and determine the units for G.

The unit of Force (F) is the Newton (N = kg m s-2)

The unit of the radius squared (r2) is m2

The units of mass squared (m2) is kg2

Show all working.

Success criterion:- meritYOU are able to rearrange an equation and determine the units for the new subject of the formulae.

Vectors

Vector Addition (Graphical)

AB

BAC

A

B

Vectors

Vector Subtraction (Graphical)

A

B

C BAC

BAC

A

B

A

x

y

cosAAx

sinAAy

Vector Components

yx AAA

2y

2x AAA

x

y1

A

Atan

The twin farmers are trying to drag a stubborn donkey into new pasture. The maximum force the donkey can resist the twins is 820 N.

1. Use vector calculations to determine the resistive force of the donkey; and

2. Explain why the farmers are unable to drag her into the next field of pasture.

3. Use vectors to determine how the farmers could drag the donkey into the next field of pasture without having to exert any more force than they already are.

A free body sketch of the situation.

F1 = 450 N

F2 = 375 N

150

300

Resisting Force

Success criteria:• You are able to draw neat, accurate and correctly labeled vector diagrams. • You can evaluate your answer demonstrating that you have a clear

understanding of physics.

Translational Motion Vectors

2. Explain why the farmers are unable to drag her into the next field of pasture.

F1 = 450 N

F2 = 375 N

Resisting Force

By measurement the resisting force of the donkey = 760N

The farmer boys are unable to drag the donkey into greener pastures

Vector diagram

30º

15º



F1 = 450 N

F2 = 375 N

150

300

Resisting Force

1. Use vectors to determine the resistive force of the donkey


780 N MA


Formative practice enables students to gain frequent feedback and feedforward on the progress of their education

Click q for answers


F1 = 450 N

F2 = 375 N

150

300

Resisting Force

Success criteria:• You are able to draw neat, accurate and correctly labeled vector diagrams. • You can evaluate your answer demonstrating that you have a clear

understanding of physics.


F1 + F2 = 450 + 375 = 825 N

The resistive force of the donkey has maxed out at 820 N

The farmer twins combined forces are 825 N. The twins can drag the donkey into greener pastures because the donkey's maximum resisting force is only 820 N

Vector diagram

By decreasing the 150 + 30O = 450 angle between the farmers the twins are able to increase the force that they apply to the donkey as shown below



F1 = 450 N

F2 = 375 N

150

300

Resisting Force

3. Use vectors to determine how the farmers could drag the donkey into the next field of pasture without having to exert any more force than they already are.

E

atvv if

tvvx fi 2

1

axvv if 222 2

2

1attvx i

( No t )

( No t )

( No v

)

( No a

)

( No x

)

Kin

emat

ic E

qu

atio

ns

Assume a is constant!

If no external force acts, an object maintains a constant velocity, or is at rest.

1) Newton’s First Law (Law of Inertia)

2) Newton’s Second Law

maF

yy

xx

maF

maF

Units of Force

System Mass Acceleration Force SI kg ms-2 N = kg ms-2

Applications of Newton’s Laws

F

vi= 20 m/s

mv = 0

maF advv if 222

F = -10 Nm = 5 kg

d

mF

a

dm

Fvi

20 2

F

mvd i

2

2

102205 2

Find distanceblock moves

m 100


Success criteria:You are able to draw neat, accurate and correctly labelled free body diagrams.

1. Draw the free body diagram of the space shuttle sitting on the launch pad.

2. Determine the unbalanced force (ΔF) required to lift off the space shuttle.

3. Draw the free body diagram at this instant.

When you use drawings to answer a question, you will succeed better if your drawings show that:

1. you have used a calibrated straight edge (ruler) for straight lines;

2. your lines are the correct length;

3. your lines are correctly labelled; and

4. you state the scale you have used.

1. Draw the free body diagram of the space shuttle sitting on the launch pad.

Support force = 24 MN

Weight force = 24 MN

free body diagram

Scale 1cm : 2 MN


Lift force = 4.1 MN24 MN + Δ F = \ Δ F =

Δ F = +Changing the sign and reversing the direction of the vector

−

\

Sketch your solution before you draw your answer

Unbalanced force required to lift the space shuttle off the pad:

ΔF = 4.1 + 24 = 28 MN

2. Determine the unbalanced force (Δ F) required to lift off the space shuttle.

Weight FW = 9.8 x 2.4 x 106 = 23.52 x 106 N (24 MN)

Initial acceleration = 1.7 m s-2 now F = ma F = 4.1 MN


ΔF

Scale 1 cm : 5 MN

E

A

Lift force = 28 MN

Fw= 24 MN

free body diagram

Scale 1 cm : 5 MN

3. Draw the free body diagram at this instance.


E 1


m = 20 kg

q = 60o

q

m

cos

sinmg

T2

tan

mgT2 60tan

)8.9(20

T1 sin(60)

T1 cos(60)

0Fy

0Fx

sinTmg 1

sin

mgT1 60sin

)8.9(20

cosTT 12

T1 T2

mg

N 226

N 113

Find Tensions T1 and T2


m = 20 kg

q = 60oq

m

Find Tensions T1 and T2

T1 T2

mg

T1 T2

mg

q

m

sin

mgT1

tan

mgT2

T2

mg T1

q

2T

mgtan

1T

mgsin

Another method

Hooray for Algebra....

Do the Algebra

Use Diagrams

Show Working

The Physics…

q

q

N

mg cos(q)

mg sin(q)

0Fy

y

x

amFx

mg

cos mgN

Force Components ~ Frictionless incline

sin ga

amsin mg

q

q

N

mg sin(q)

0Fy amFx

mg

fk

afk msin mg

m

fa k gsin

mg cos(q)

y

x

cos mgN

Acceleration on a rough incline

a

d

q

Frictionless

q

mg

N

mg sinqmg cosq

A block is released from restat the top of an incline.Find the final speed and thetime to slide to the bottom.

maF

masinmg

singaax2vv 2

o2

ad2v2

dsing2v2

singd2v

2at

tvx2

o

2

tsing2

atd

22

singd2

t

d

q

Frictionless

q

mg

N A block is released from restat the top of an incline.Find the final speed and thetime to slide to the bottom.

d

q

Now add Friction

q

mg

N

mg sinqmg cosq

A block is released from restat the top of an incline.Find the final speed and thetime to slide to the bottom.

Fm 3m2m

amF s

am3m2mF

m6F

a

Three mass system - find acceleration

am6

m1

m2

N

m1g

T

T

m2g

gmN0F 1y

amF 1x

amT 1

maFy

amTgm 22

amgmT 22

amgmam 221

21

2mmgm

a

Forces on m1

Forces on m2

m1

m2T

T

m1gm2g

amgmT 11

amF 1

gmamT 11

amTgm 22

amF 2

amgmT 22

Mass 1 Mass 2

amgmgmam 2211

21

12mm

gmma

m1

m2T

T

m1gm2g

amfgmT 11

amF 1

fgmamT 11

amfTgm 22

amF 2

famgmT 22

Mass 1 Mass 2

Forces Investigation ~ Add friction into the mix How much is it?

2

2

12

1122

gamagmf

gmamamgmf

f

f

Is f related to mass weight?

Units Motion Scalar / vector quantities Displacement / Velocity / Acceleration Vector components...

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Transcript of Units Motion Scalar / vector quantities Displacement / Velocity / Acceleration Vector components...