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...
• 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.
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
Translational Motion Vectors
A free body sketch of the situation.
F1 = 450 N
F2 = 375 N
150
300
Resisting Force
1. Use vectors to determine the resistive force of the donkey
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.
780 N MA
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.
Formative practice enables students to gain frequent feedback and feedforward on the progress of their education
Click q for answers
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
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
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.
A free body sketch of the situation.
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
Translational Motion Vectors
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
Translational Motion Vectors
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
Translational Motion Vectors
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.
Translational Motion Vectors
E 1
Applications of Newton’s Laws
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
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.
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